The Monty Hall problem is a counter-intuitive statistics puzzle:

- There are 3 doors, behind which are two goats and a car.
- You pick a door (call it door A). You’re hoping for the car of course.
- Monty Hall, the game show host, examines the other doors (B & C) and always opens one of them with a goat (Both doors might have goats; he’ll randomly pick one to open)

Here’s the game: Do you stick with door A (original guess) or switch to the other unopened door? Does it matter?

Surprisingly, the odds aren’t 50-50. If you switch doors you’ll win 2/3 of the time!

Today let’s get an intuition for *why* a simple game could be so baffling. The game is really about re-evaluating your decisions as new information emerges.

## Play the game

You’re probably muttering that two doors mean it’s a 50-50 chance. Ok bub, let’s play the game:

Try playing the game 50 times, using a “pick and hold” strategy. Just pick door 1 (or 2, or 3) and keep clicking. Click click click. Look at your percent win rate. You’ll see it settle around 1/3.

Now reset and play it 20 times, using a “pick and switch” approach. Pick a door, Monty reveals a goat (grey door), and you switch to the other. Look at your win rate. Is it above 50% Is it closer to 60%? To 66%?

There’s a chance the stay-and-hold strategy does decent on a small number of trials (under 20 or so). If you had a coin, how many flips would you need to convince yourself it was fair? You might get 2 heads in a row and think it was rigged. Just play the game a few dozen times to even it out and reduce the noise.

## Understanding Why Switching Works

That’s the hard (but convincing) way of realizing switching works. Here’s an easier way:

**If I pick a door and hold, I have a 1/3 chance of winning.**

My first guess is 1 in 3 — there are 3 random options, right?

If I rigidly stick with my first choice no matter what, I can’t improve my chances. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance — it doesn’t matter. The best I can do with my original choice is 1 in 3. The other door must have the rest of the chances, or 2/3.

The explanation may make sense, but doesn’t explain *why* the odds “get better” on the other side. (Several readers have left their own explanations in the comments — try them out if the 1/3 stay vs 2/3 switch doesn’t click).

## Understanding The Game Filter

Let’s see why removing doors makes switching attractive. Instead of the regular game, imagine this variant:

- There are 100 doors to pick from in the beginning
- You pick one door
- Monty looks at the 99 others, finds the goats, and opens all but 1

Do you stick with your original door (1/100), or the other door, which was filtered from 99? (Try this in the simulator game; use 10 doors instead of 100).

It’s a bit clearer: Monty is taking a set of 99 choices and *improving* them by removing 98 goats. When he’s done, he has the top door out of 99 for you to pick.

Your decision: Do you want a *random* door out of 100 (initial guess) or the *best* door out of 99? Said another way, do you want 1 random chance or the best of 99 random chances?

We’re starting to see why Monty’s actions help us. He’s letting us choose between a generic, random choice and a *curated, filtered* choice. Filtered is better.

But… but… shouldn’t two choices mean a 50-50 chance?

## Overcoming Our Misconceptions

Assuming that “two choices means 50-50 chances” is our biggest hurdle.

Yes, two choices are equally likely when you know *nothing* about either choice. If I picked two random Japanese pitchers and asked “Who is ranked higher?” you’d have no guess. You pick the name that sounds cooler, and 50-50 is the best you can do. You know nothing about the situation.

Now, let’s say Pitcher A is a rookie, never been tested, and Pitcher B won the “Most Valuable Player” award the last 10 years in a row. Would this change your guess? Sure thing: you’ll pick Pitcher B (with near-certainty). Your uninformed friend would still call it a 50-50 situation.

Information matters.

## The more you know…

Here’s the general idea: **The more you know, the better your decision.**

With the Japanese baseball players, you know more than your friend and have better chances. Yes, yes, there’s a *chance* the new rookie is the best player in the league, but we’re talking *probabilities* here. The more you test the old standard, the less likely the new choice beats it.

This is what happens with the 100 door game. Your first pick is a random door (1/100) and your other choice is the champion that beat out 99 other doors (aka the MVP of the league). The odds are the champ is better than the new door, too.

## Visualizing the probability cloud

Here’s how I visualize the filtering process. At the start, every door has an equal chance — I imagine a pale green cloud, evenly distributed among all the doors.

As Monty starts removing the bad candidates (in the 99 you didn’t pick), he “pushes” the cloud away from the bad doors to the good ones on that side. On and on it goes — and the remaining doors get a brighter green cloud.

After all the filtering, there’s your original door (still with a pale green cloud) and the “Champ Door” glowing nuclear green, containing the probabilities of the 98 doors.

Here’s the key: Monty does not try to improve your door!

He is purposefully *not* examining your door and trying to get rid of the goats there. No, he is only “pulling the weeds” out of the neighbor’s lawn, not yours.

## Generalizing the game

The general principle is to re-evaluate probabilities as new information is added. For example:

A Bayesian Filter improves as it gets more information about whether messages are spam or not. You don’t want to stay static with your initial training set of data.

Evaluating theories. Without any evidence, two theories are equally likely. As you gather additional evidence (and run more trials) you can increase your confidence interval that theory A or B is correct. One aspect of statistics is determining “how much” information is needed to have confiidence in a theory.

These are general cases, but the message is clear: more information means you re-evaluate your choices. The fatal flaw of the Monty Hall paradox is *not taking Monty’s filtering into account*, thinking the chances are the same before and after he filters the other doors.

## Summary

Here’s the key points to understanding the Monty Hall puzzle:

- Two choices are 50-50 when you know nothing about them
- Monty helps us by “filtering” the bad choices on the other side. It’s a choice of a random guess and the “Champ door” that’s the best on the other side.
- In general, more information means you re-evaluate your choices.

The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after. But the goal isn’t to understand this puzzle — it’s to realize how subsequent actions & information challenge previous decisions. Happy math.

## Appendix

Let’s think about other scenarios to cement our understanding:

**Your buddy makes a guess**

Suppose your friend walks into the game after you’ve picked a door and Monty has revealed a goat — but he *doesn’t* know the reasoning that Monty used.

He sees two doors and is told to pick one: he has a 50-50 chance! He doesn’t know why one door or the other should be better (but you do). The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering.

**Monty goes wild**

Monty reveals the goat, and then has a seizure. He closes the door and mixes all the prizes, including your door. Does switching help?

No. Monty started to filter but never completed it — you have 3 random choices, just like in the beginning.

**Multiple Monty**

Monty gives you 6 doors: you pick 1, and he divides the 5 others into a group of 2 and 3. He then removes goats until each group has 1 door remaining. What do you switch to?

The group that originally had 3. It has 3 doors “collapsed” into 1, for 3/6 = 50% chance. Your original guess has 1/6 (16%), and the group that had 2 has a 2/6 = 33% of being right.

You could explain it like this too:

If you stay with the door you picked initially you succeed if the initial door has a car, which has a chance of 1/3. If you’re strategy is to switch then you succeed if your initial pick is a goat, which has a chance of 2/3.

Welcome back!

I actually blogged about this a while back:

http://blog.amhill.net/2009/07/24/from-the-archives-monty-hall/

And one of my uncles was dead-sure that I was wrong about it, so I wrote up an application (source code included) to show the percentage differences between switching and not switching. It’s linked to from that blag post above.

I like your flash version though!

I never understood this before, but as I read your explanations, the following came to mind:

The doors can be divided into two groups; the one you picked, and the ones you didn’t pick. When you pick one out of 100 doors, chances are very high that the prize is in the group you didn’t pick, you just don’t know which of those 99 doors has the car.

Luckily for you, Monty narrows it down for you by opening all but one door. The odds of the selected door having the prize hasn’t changed at all. You just know more a lot about the doors you didn’t pick.

Great article – I especially like the visualization part with the green clouds. Sometimes I think the hardest part with mathematical concepts is being able to visualize them.

@uwe: Nice, I like seeing the win and lose probabilities next to each other like that!

@Aaron: Cool, I like the automated way to go through hundreds of trials :).

@Gary: Yes, that’s exactly it. I need to think of a good concise way to get that point across — something along the line of “getting rid of the weeds in the neighbor’s garden”, i.e. improving the choices in the items you *didn’t* pick. Great observation.

@CL: Thanks, and I totally agree. I often end up with some mental picture of what’s happening when I think about math, I wish we’d all share what we “really” think about when solving a problem!

Yes, great post. After I had time to think about this problem a bit, I thought about it similar to Aaron/uwe. But I use slightly different wording: 1st, I pick a door (1 out of 3). Then, I’m given the opportunity to either 1) keep my pick or 2) choose the other 2 doors (i.e., knowing that Monty would take one away for me later so to speak). Thanks for the articles, they are generally my favorite in the blogosphere.

Another great post. When I first encountered this problem in college it nearly drove me crazy for the first couple of minutes until my lecturer simply said.. imagine it’s 1 million doors. It immediately became clear.

I like to use the Monty Hall problem when watching ‘Who wants to be a Millionaire?’ When a question is asked and I don’t know the answer, I pick one at random, and hope the actual contestant goes 50/50.

If they do, and my randomly chosen answer is still available, the correct answer more often that not is the other one.

Funny, I came across this problem a couple of years ago and never *intuitively* understood it; I was all over the internet for a better explanation. Now, I just read the problem statement (not the entire reasoning) and I immediately understood (intuitively) that switching the door is the way to go, and I’m wondering why I got so confused back then.

The way I visualize this problem is with an adversarial scenario. First I imagine it’s 100 doors, then I think of it this way:

Imagine you’re playing AGAINST Monty Hall for a car. You each must pick a door. You go first. You are forced to pick your door at random.

Monty Hall, however, has the luxury of looking at every door other than yours, and not only can but MUST pick the car door if he finds it.

He has a wide grin on his face (99 out of 100 times) and is already celebrating his new acquisition when all of a sudden you are offered to trade places with Monty and get to pick his door! Do you switch then? I think so!

@lewikee: Awesome, I like that formulation! Monty is basically picking the best door he can out of the remaining choices. Very nice ;).

@ktr: Glad you liked it! Yes, I like thinking of it as “Pick 1 door or the best of the other 2″.

@Mike: Wow, that’s a really clever use of the paradox! I wasn’t sure if it had a ‘real-world’ use :).

Another way to think about it: If you pick a random answer (A B C D) you have a 3/4 choice of being wrong. If it’s still left after the 50-50 elimination, more often than not switching to the other choice will work out (since most of the time, your random guess is just there to be the wrong answer). Quite often your random guess will be eliminated, but if it remains it’s a good bet to go the other way. Neat!

@Srikanth: Cool, glad it clicked! Yes, sometimes it takes a second reading to have it snap into place.

Nicelly analysed and explained.

Ridiculously complicated. The probability the car is not behind ‘my’ door is 2/3. And now there’s only one door to pick. Change doors!

Interesting and crazy. I’ve read it twice and still don’t get it.

@3rojka: Thanks!

@Seldon: The 1-line explanation explains the “how” mechanics, but not the “why”. If Monty randomly revealed a door and it was a goat (i.e, he wasn’t trying to look at the other two doors and pick the best, he just randomly opened one for some reason), you’d still have an equal chance of being wrong and it wouldn’t give an _advantage_ to switch. So, the secret is in the process Monty uses to reveal, not just the fact that he leaves you with 1 other door.

@geld: It is a tricky problem. You might try playing the game several times (in real life with coins under cups, 2 pennies [goats] and a quarter [car]). You’ll eventually see that your initial guess is right only 1/3 of the time.

I think this is overly wordy. What helped me is:

if you picked the door with a goat, the other doors have a car and a goat. Monty then CAN’T OPEN the door with the car because that would ruin the game.

the probability that you picked the right door out of 3 is 1/3. so you have a 2/3 of having picked the WRONG door, at which juncture, Monty opens the door with the goat. that means the other door has a 2/3 chance, and your door has a 1/3 chance.

@thegnu: Thanks for the comment. Yep, the goal of the article isn’t to just understand why switching works. As you mention, it can be done in a paragraph.

The more interesting principle is seeing the role of the filter itself, so you can handle alternative scenarios also (like your buddy playing, Monty mixing the doors again, Monty giving you 2 sets of doors to pick from).

And from there, we can see that these filters (new information that impacts earlier choices) exist all over the place in real life, from spam filters to analyzing experimental evidence. Hope this helps!

I still don’t get it and disagree. You have no extra information? It’s not a more informed decision. Maybe I’m wrong but this seems exactly like the situation of taking past occurences into an equation that have absolute no relevance. It’s a NEW decision/outcome. If 4 red’s in a row come up on the roulette table it’s no more likely to be black than 50/50 (negating zero) than it ever was. Same situation here no?

This is how I thought about it: imagine that door A is the one you pick. There are 3 equally likely scenarios. 1: Door A is correct. In this case, you would be better off staying. 2: Door B is correct. Monty would open C and you would be better off switching. 3: Door C is correct, Monty would open B and you would be better off switching. Two out of three times, you are better off switching.

I like the way you encourage visualizing the numbers with both shape and color. Great stuff.

Or you could you know figure out the really simple system used on this and get about 70% with out all the close examining you just made me do

The combinations of your first choice are:

Car

Goat(1)

Goat(2)

That’s a 1/3 chance of winning the Car – as you would expect.

After you make your first choice the combinations behind the remaining two doors are:

Goat(1) & Goat(2)

Car & Goat(2)

Car & Goat(1)

If Monty Always picks a Goat then the remaining door contains:

Goat (1 or 2)

Car

Car

Hence if you switch there is a 2/3 chance of picking the Car.

Initially we make a random choice with a 33% probability that the car is behind the door. The other two doors contain 66% of the probability. Once the goat is revealed the 66% probability is condensed into a single door. No extra probability is handed to the first door until a new random choice is made. Thus if no new random choice is made the original choice maintains its 33% probability but the other door now has all of the 66% probability of its original set. That’s why switching to the other door wins 66% of the time. If a new random choice is made of those two remaining doors, then it’s truly a 50% chance.

Everything hinges on making a random choice. Those that argue it is a 50% chance problem after the goat is revealed fail to see where the original probability of 33% came from. It came from making a random choice. If there was no random choice initially then there’d be no 33% chance. If the contestant was told what to pick, they’d claim their chances were not even. So after the goat is revealed, the only way to have a 50% chance is to flip a coin and make another random from the two remaining doors. But since 66% chance of a win was held in the other two doors and condensed from those 2 doors into 1, it makes sense to just pick that other door that has the 66% chance and NOT make any more random choices.

Why are both doors examined before they are opened? I find that part confusing. So did he open both doors or only one?

We humans have a habit of being acquisitive. Once we choose something, we have a tendency to want to hang on to it. The act of choosing a door, makes it ‘our door’ and so there is a natural reluctance to part with it.

An easier way of intuitively understanding the Monty Hall game is to hold off selection until the ‘door’ has been opened, and to show the higher probability of switching choice.

Take a pack of cards – it contains a single Ace of Diamonds – this is the winning card – pick the Ace of diamonds and you win.

Shuffle the pack and take a single card out – lay it on the table. Put the rest of the pack on the table beside the single card.

Now, the choice is pick the single card or pick the rest of the pack – which do you thinks is most likely to give you the Ace of Diamonds. Clearly, most people would choose the pack with 51 chances that they have chosen the Ace.

Now we add in the ‘opening of the doors’

As before, pick a card and lay it on the table, but before putting the pack down, somebody sorts through the pack and if it contains the Ace of Diamonds, they put it on the top of the pack and then put the pack on the table.

Now you have the choice of taking the single card or the top card off the pack – again, automatically folks will take the top card from the pack.

Now we put it all together – ‘Choose’ a card and put it on the table (we now own it). Next the pack is sorted and if the Ace is present it is put on the top of the pack.

The top card is taken off the pack and placed beside your chosen card and the rest of the pack are turned face up to show that they are not the Ace of Diamonds.

Now you are offered the choice of changing the card you chose for the remaining card from the pack. Even though ‘ownership’ is now involved, most people would not have any doubt about changing, and understand that there is not the remotest chance that the first card has a 50:50 chance of being the Ace.

they will understand that they had a one in 52 chance of picking the Ace first time, and almost dead certainty of it being the other card – they would switch…

Ya right, i got 5 wins 5 losses both ways, 50 50. your strategy fails.

I’ve read a bunch of articles about this, but the multiple doors made it click and made me think of it in a way no one has mentioned.

The reality is that 66.7% of the time, I have picked a goat. This means that 66.7% of the time, Monty still has the car and the other goat. So, 66.7% of the time, Monty is revealing the 2nd goat I did not pick.

Another way that came to me is that there are only 2 versions of this game.

1) I have picked the car and Monty is showing me one of the two goats.

How to win: stay with my first choice because it is the car.

2) I have picked a goat and Monty is showing me the other goat that I did not pick.

How to win: change my choice because the car is behind the 3rd door.

I don’t know which version of the game I’m playing, but I DO know that “version 1″ only happens 33.3% of the time, while “version 2″ happens 66.7% of the time. Since this is true, I am going to use the winning strategy that applies to 66.7% of the games I play. I will win about 66.7% of the time.

I learned about this a while ago but never tested it until today. I tried picking a door and switching 20 times twice. Each time the score was 9 wins and 11 losses. I’ll try one more time.

One sixty seven percent win. One 50/50. It doesn’t work. The reason is because if it is predetermined that a goat will be revealed from one of the doors you didn’t pick, then you are starting out with a 50/50 chance not a 1/3 chance. No matter what you picked, it’s a 50% chance of being a goat or a car because the door you didn’t pick but will be revealed will automatically have a goat in it or be irrelevant.

Think of it like this. You have a coin which has three possible landings: heads, tails, and edge. Let’s say that landing on its edge will count it as tails and you want heads. It will never land on it’s edge, whether you know it or not, so that decision automatically starts out irrelevant. Your choice is either heads or tails because no matter what, one option will be nonexistent.

To restate it in your terms: There will be a door you will never pick, but will always be a goat. Let’s call him goat 1. There is one goat you will never choose. You will either have a goat 2 or a car.

Suppose Monty didn’t examine doors B&C first. Suppose he just opened one at random, and it just *happened* to have a goat behind it. (In other words, there was a chance he could have opened the car door, thereby spoiling the game.)

In that case, is it now 50:50 for us when choosing to switch or not?

@Frank: It would be no new information if Monty randomly revealed a door. But, Monty is purposefully filtering the other side, and providing the information that “The door that remains is the *best* of these two doors”. So, you have the choice of 1) your original door or 2) the best of the other side.

Monty provides information by picking the “winner” on the other side for you.

@Lizzy: Cool viewpoint, thanks for sharing!

@rajanKazhmin: Glad you enjoyed it.

@Domogo: I’m not sure I understand…

@Edmund: Thanks for the breakdown! I love seeing how everyone approaches this problem.

@Mike: That’s exactly it — the 66% is “collapsed” into a single door after Monty’s filter. A fresh contestant, seeing two doors, has a 50% chance. But since you saw the process, you know which door went through the filter.

@Sludgie: Good feedback, I should clarify. Monty looks behind both doors, but only fully opens one for you to see. At this point he knows where the car is since he’s seen two of the three doors and can infer the third.

There’s no reason for him to do so (in the game, he’s helping you) but it makes the question interesting :).

@DerekSmith: Great analysis into the psychology of it! I like the example too.

@Ace: As I write for @sweet…, try doing more trials, like 50… you’ll see a pattern emerge.

If you have a fair coin but only flip it twice, you could get two tails and think it was biased.

@SteaveG: Awesome, thanks for the explanation!

@sweetestsadist: You bring up a good point — how many trials do we need to be convinced of something? I need to update the article — 10 isn’t enough to really eliminate chance (in such a small sample, there is a chance that staying will be better). If you can, try doing 50 or 100 trials where you stay (just keep clicking door 1)… you should see the percentage be much closer to 1/3.

Actually, your initial guess has a 1/3 chance of being a car, not 50-50. If you have 3 choices and pick 1, that’s the best you can do, right? Monty can add and take away doors on the other side, but at the time of your initial guess, it was a 1 in 3 chance of being right.

It’s true that monty leaves you with two choices in the end, but two choices doesn’t mean a 50-50 chance between them (that’s the crux of the counter-intuitive nature of the paradox). Try giving the game a shot with 100 trials to see what happens :). Also, give SteveG’s explanation a try.

@Puzzled: Great question! Yes, if Monty randomly opened a door (and it happened to be a goat), it would indeed by 50-50. But since Monty is looking at TWO doors (and leaving you with the better one), it is an advantage to switch.

still completely do not understand this.

Ok, I think I get it. Here’s how I understand it:

There is a 1/3 chance for the door you pick to have a car, therefore there is a 2/3 chance for either of the remaining doors to have a car.

After all filtering, there is still a 1/3 chance for your door to have a car and a 2/3 chance for any other door to have a car.

Since there is now only one other door there is still a 2/3 chance for that door to have the car and still 1/3 that your door has it.

Or for 100 doors. 1/100 that you pick the car the first time and 99/100 that you didn’t. 98 doors are eliminated. It’s still 1/100 that you got it right the first time and it’s still 99/100 that you didn’t. But now, there’s only one other door, so it’s 99/100 that it’s the other door.

Awesome post.

Here’s yet another way to think about this problem.

Pretend we are dealing with a lottery that everyone in the world plays – each person is given a ticket which is equally likely to win. One person’s ticket is chosen from the world population to be the winner. Would you expect yourself to be the winner, or someone else in the world? The intuitive (and correct) answer is to expect someone else.

Now suppose you have a connection inside the agency that runs the lottery. Your “inside man” sends you a letter with the name of a person (call him James.) They inform you that if the rest of the world contains the winning ticket, then that winner is James. They then give you the opportunity to switch tickets with James. Would you?

Of course you would!

@CB: It’s a weird problem to be sure. Try playing the game above but with 10 doors instead of 3 [enter 10 for the number of doors and press reset in the game].

You’ll start to see how your initial guess is “usually” wrong (it only has a 1 in 10 chance of being right). From there, you’ll start to see that when you have 3 choices, you only have a 1 in 3 chance of being right.

@Greg: Awesome, thanks for sharing your insights! I always like seeing how other people think about the problem. I too see the “probability cloud” collapsing.

@Alex: I like the lottery example, it’s something we can all relate to!

Here’s the way I finally convinced myself that switching every time is the best strategy: if you pick a goat, Monty shows you the other goat, and when you switch to the remaining door (the car), you are ALWAYS right. If you pick the car, Monty shows you one of the goats, and when you switch to the remaining door (the other goat), you are ALWAYS wrong. So if you always switch, you will win 2/3 of the time since that is how often you will pick a goat to begin with. But if you decide to never switch, you will only win 1/3 of the time, since that is how often you will correctly pick the car to begin with. So by always switching, you will win twice as often as when you never switch!

@Bruce: Awesome, thanks for sharing!

I am taking Prob & Stats this semester, and this article does as good a job as my teacher in explaining this concept.

Very well done!

The java script game is fun and helps explain the concept. Not that it matters but when there are more than 3 doors I can almost always predict which door has the car due to how your code is written. Let’s say for 10 doors, if I always pick the first door then the other door remaining will be the car. Unless the first door is the car in which case the remaining door is the last in the series of doors. So in the scenario where I always pick the first door and the first door is the car then I know it will be a car because the last door in the series is not revealed as goat (the other scenario is the last door is the car). You should randomize which door is left unrevealed when the user picks the car on the first choice.

i still don’t get it..

Suppose we have 4 doors.

suppose the player choose door A. Considering the 4 possible cases:

A,B,C,D

i) 1,0,0,0

ii) 0,1,0,0

iii) 0,0,1,0

iv) 0,0,0,1

1 mean that car is behind the door…The solution suppose that if we don’t switch, we win for 1st case only, so 1/4. If we switch, we have 3 case to win, so 3/4…

But why case ii, iii and iv are different? I still believe that we have only 2 cases (i and ii or iii or iv) , because they will alway open the door without car..

@vichet: Great question. You’re right that situation i) has one outcome (you win), and situation ii) iii) and iv) have a different one (you lose).

However, since each situation is equally likely, it’s much more likely that you’ll lose. Imagine a dice with 4 sides — if you roll a 1 you win, if you roll a 2 3 or 4 you lose. Even though it’s only 2 cases (1 or 2/3/4), you wouldn’t say winning and losing are equally likely, right?

That’s one of the tricky things — separating the 2 possible outcomes (win or lose) from the number of ways to get those outcomes (1 way vs 3 in your case). Hope this helps!

Suppose there are three doors, A,B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind that door A, which has a chance of 1 in 3. If the prize was originally behind door B on the other hand (which has a chance of 1 in 3), then when you pick door A, door C will be removed. Hence, if you switch you will be switching to door B, and therefore you will win. Finally, if the prize was originally behind door C (which again has a chance of 1 in 3) then door B will be removed, and if you switch you will be switching to door C and therefore will win. Hence, if you stay with your original door, you win if and only if the prize was originally behind door A. If you switch though, you win if it was originally behind either door B or door C. Since the chance the prize being behind door A from the get go is 1 in 3, whereas the chance of it being behind either B or C from the get go is 2 in 3, you are better off switching!

In a way, it’s still a 50/50. “Monty” may have eliminated a door, but you’re then left with two doors. Just because he eliminated the third doesn’t mean it’d be better to switch.

By the way, from switching I won 3 times, lost 7.

Let me post 2 cases to challenge those that have assumed 2 things.

1) Initial probability of the chosen door remains unchanged no matter how much info is revealed by the host on the remaining doors.

2) Given the choice to switch, always better to switch to higher probability on the accumulated remaining choices.

My case assumptions: 4 doors, one with prize, Door1 was chosen, Host will reveal 0, 1, 2 or 3 doors ; Choice will be given to switch/stay.

(Bear with me on the slight differences to the original example)

Case A: Host decides to reveal 0 doors

: P ( Door1 = Prize ) = 25%

: P ( Door2/Door3/Door4 = Prize ) = 75%

Ambiguity: It appears that switching your choice will increase your chance by 3 times ( 25% -> 75% ). However it doesn’t make sense since no extra info is revealed to you. It be the same as choosing a different door with 25% chance of hitting jackpot.

Case B: Host decides to reveal ALL of the remaining doors.

: P ( Door1 = Prize ) = 25%?

in my opinion,

: P ( Door1 = Prize ) = 0% or 100% (base on what is revealed)

Ambiguity: Many have stated that the initial probability of the Door choice is to remain unchanged despite additional info provided by the host. In this case, 2 possibilities can happen, host reveals the prize by opening all the remaining doors or reveals all the goats. Base on what is revealed, probability of Door1 being prize or not is DIRECTLY AFFECTED.

Questions:

1) Why do many claim that the probability of the initial door choice should remain unchanged when more info is given? I have given a counter case that suggests otherwise.

2) Why do many claim switching your choice to a larger portion of aggregate probabilities always earns you a higher chance? I believe in some cases, it is not apparent that you should switch like in Case A, switching or not makes no difference if no information is revealed. Even though it would be an act of switching to a higher probability.

3) Is it right to say probability is just a measure of how confident one is of hitting the jackpot, it does not actually mean a higher number corresponds to higher occurrence in every situation?

4) Probability is indirectly proportional to the total number of possibilities unless i am mistaken. When a host reveals the location of some of the goats, the total number of possibilities of the actual location of the prize is directly reduced, how can the probability if Door 1 is a prize be unaffected?

Will the following measurement make more sense?

Case C: Host decides to reveal 2 out of 3 doors

(assuming Door2 / Door3 is revealed to be goats )

: P ( Door1 = Prize ) = 25% + Adjustment

: P ( Door4 = Prize ) = 25% + Adjustment

Adjustment = Total% from revealed Doors / # of Doors remaining.

It just makes sense that ALL the unrevealed doors should have higher probability of being the prize including the initial choice that was chosen with a 25% confidence.

Sorry for the presentation if it confuses anyone, I am rather unconvinced at what the majority is accepting which motivates me to find out if I am wrong in my understanding.

Do advise if you understand what I am trying to say in my post.

only problem with your implementation is that it always picks the leftmost door that isn’t the car so if you always choose the rightmost door and it shows the middle door as a goat then the leftmost door will always be a car

position of the door do not matter.

whichever chosen in the beginning can be assigned as door1 for reference. The remaining doors regardless of position will be considered as per stated in my previous example.

my main point is that once additional information on the remaining doors are revealed, the effect is propagated to the probability of the first chosen door. So eventually, it always end in a 50/50 chance whether your first door contains the prize. Which in my opinion is a rather true reflection of real life choices.

Much has argued that remaining doors collectively contain a higher chance, then follow through their calculations while assuming the probability calculated on the first chosen door should remain the same throughout regardless of additional information revealed on the remaining doors, which i find rather weird.

I think you’re wrong. In the game, whatever your choices are at the start, in the end you will have two doors. You shouldn’t consider the probability of 1/3 from the first stage at all because it doesn’t mean a thing; it doesn’t help; but unfortunately you do compare it to the second stage probability of 1/2, the actual game probability.

By eliminating the rest of the bad doors you are not helped at all, in the end there will always be just 2 doors with equal probabilities.

It’s like you always start the game with 2 doors: One with the car ( which might be your first choice ) and one with the goat.

If you’ve picked the car(but don’t know it) and switch, your chance of winning are 50% (because the other door, that remained after elimination can have your car or a goat – you don’t know what is behind your picked door, i.e a goat or the car).

If you stick with your first choice you have the same probability of winning the car -50%.

I think you’re letting yourself be confused by the initial probability of winning equal to 1/3 (in case of 3 doors) that is less than the actual probability (1/2) at which you play the game in the final stage.

The first stage and the last stage of the game are not interconnected, so in the final stage you’ll not be helped by the decision you made initially.

You’ll always start from scratch and have to pick between a car and a goat, hidden from you. Revealing the other doors, doesn’t mean that you picked a goat, it doesn’t give you any info about the quality of you pick, it just narrows the the possibilities from many to just 2.

That is what I i understood from reading your article, my opinion is different from yours, i believe am right, but that’s just me.

I love when this question comes up, sooooo many people never get it. The way I get most people to see it is by looking at Deal or No Deal. Say you have selected your case and there are 24(or is it 25? no matter) you have a 1/24 chance of having the million in your case, most people can agree with that. If that is true, then the horde of models have a 23/24 chance of having the million.

When it gets to the final case, and you can either keep your case or switch, you do not have a 50% shot as seems intuitive (2 options) you still have 1/24, they still have 23/24.

If you still have a problem, would you agree that your odds are 1/24 if they were all opened at the same time? Then what is the difference if you open them dramatically.

Not the same exactly as the Monty Hall, but sometimes helps people figure out what is going on.

Thanks for the great site!

@John: That’s a great example. Yes, intuitively we think about 2 options as being 50-50, even though one went through a screening process and the other didn’t! It’s hard to fight our first gut reaction sometimes ;).

My wife was not convinced of it until we did the experiment with 10 doors, haha.

@Mark: Sometimes you have to see it with your own eyes :).

Frank is right (comment #18). Sorry Kalid, but on this one you’ve misconstrued the model as a conditional probability issue. Monty is always going to show you a door with a goat, no matter which door you’ve picked initially. Therefore, your initial pick has no relevance to what he is going to do and is a completely independent event. There is a a 100% chance of that he will show you a goat and your pick has nothing to do with that. Of the remaining two doors, exactly one contains a car. The chance that your prior pick contains a car is exactly 50%. If you change your mind after he shows you what he is always going to show you, that does not influence the probability that you’ll get a car, therefore the “new information” isn’t helpful and it really isn’t “information” at all. All Monty has changed is the setting of the “great reveal” — he’s moved it from a studio without a picture of a goat to a studio with a picture of a goat. Monty is essentially asking you to flip a fair coin and then showing you a picture of a goat. The intuitive answer is correct — your probability of getting a car is 50% and the “third door” is a canard.

Reading through the comments, the key mistake is assigning the initial probability of getting a car at 1/3 just because three doors appear before you. Monty is always going to show you a goat and therefore the probability of getting the prize is 1/2 before he shows you a goat and 1/2 after he shows you a goat. Ten trials are not enough to get past the randomness — do the experiment 1000 times and almost 80% of results will be within a range of 480 and 520 “wins” out of 1000 trials.

John B (#49): In “Deal or no Deal” the probability of winning the $million does change throughout the game. That game is a demonstration of conditional probability. If the last step is to pick between 2 cases, one worth $1 million and one worth $.01, then at that point you have a .50 chance of winning 1 million.

Smith Jr. (#48) hits it spot on when he stated “The first stage and the last stage of the game are not interconnected, so in the final stage you’ll not be helped by the decision you made initially. You’ll always start from scratch and have to pick between a car and a goat, hidden from you. Revealing the other doors, doesn’t mean that you picked a goat, it doesn’t give you any info about the quality of your pick…” The rest of his statement about “narrowing the possibilities” appears to be a misstatement.

Scott (#43) stated “Suppose there are three doors, A,B and C and you originally chose door A. If you stay with your original door, then the only way that you win is if originally the prize was behind that door A, which has a chance of 1 in 3.” That is not correct, because the choice is not among three doors, it’s between two doors. The elephant (goat?) in the living room is that Monty is always going to show you a goat no matter which door you pick. You already know this. So the presence of the third door is an ILLUSION… it’s an empty shell 100% of the time that has zero bearing on your chance of picking the door with the car. The “third door” might as well be a picture of a goat, or a picture of a car or a picture of the moon. Monty is always going to leave you with the same result NO MATTER WHAT. That is the very definition of an independent event. Come on, people. Switching doors can’t possibly help you because Monty didn’t base his decision to show you a goat on the fact that you made a choice. He doesn’t care which door you chose initially. His instructions to the director say, “Show the goat.” It’s irrelevant.

I choose door #1: Monty says show the goat. I am REALLY choosing between two doors.

I choose door #2: Monty says show the goat. I am REALLY choosing between two doors.

I choose door #3: Monty says show the goat. I am REALLY choosing between two doors.

1/3 divided by 2/3 is 1/2. The choices are constrained the choices to two doors before you’ve even started, because “door number three” isn’t the door labelled “DOOR 3″…it’s whichever door Monty wants it to be…which is one of the two with a goat. The probability of winning is 50% before the event of seeing a door with a goat and after seeing a door with a goat because all you’re doing is waiting to SEE the results of your coin flip until after Monty has shown you a door with a goat and Monty is giving you a chance to change your call of “heads” to a call of “tails.”

Have a great day and thanks for an awesome site. I’m supposed to be working on my precalculus so I better get back to it..

Typo fix: 2nd to last para. above: First part should read “1/3 divided by 2/3 is 1/2. The choice is constrained to only two doors before you’ve even started, because “door number three” isn’t the door LABELLED “Door 3.” It’s whichever door Monty wants it to be..which is one of the two with the goat…”

Point being: If Monty is always going to remove from the “universe” 1 of the 2 goats, then that itself defines the condition under which the game is played. That’s your sandbox and you can’t get out of it.

Kalid has caused me to spend the day in primal scream therapy with this Monty Hall problem. Fortunately my head did not explode notwithstanding the impassioned efforts of several posters here to make that happen.

Monty is a showman, yes, with the costumes and goats and tricky switchbacks. But the key to this (and I didn’t see it until the third primal scream ) is that after our contestant has picked her door, he’s showing her that one of the remaining doors is a dud. That doesn’t mean she’s guaranteed to win if she switches, it just improves her chances. Here’s why.

The contestant is still going to lose by switching if she’s picked the car to begin with, and remember picking a car in the first place has a probability of 1/3. The other 2/3 of the time, she’s going to win by switching because Monty has, by revealing one of the goats, provided VERY useful information about the door he just showed her (It has a goat) AND the one he didn’t show her, too.

That’s because he’s NOT showing her the door with the car in those 2/3 of cases where she indeed picked the goat to begin with. If he’s not showing her a car 2/3 of the time, THAT’S WHERE THE CAR IS. The other 1/3 of the time, she’d be switching from a car to a goat.

If 2/3 of the time she would win by switching, and 1/3 of the time she would win by staying with her initial pick, SWITCH IT UP, GIRL!

BTW: I read out there in cyberspace that presented with this problem, Monty himself said that yes, if the host is always going to show a goat and give the contestant a chance to switch, then switching makes sense. And we know that Monty wouldn’t mess with our heads in cyberspace!

Let me try this approach:

G = Goat

X = Car

Before you make your choice, here are the possibilities with the probability after each.

G G X = 1/3

G X G = 1/3

X G G = 1/3

Let’s say you pick the first door. Here are the possibilities of the other 2 doors with the probability after each. Note that all I did was remove the first column to concentrate on our next choice:

G X = 1/3

X G = 1/3

G G = 1/3

Let’s go deeper into each possibility.

Let’s first look at G X = 1/3. Monty MUST open a door. That door CANNOT be a car. Under this possibility, the fact that Monty opens a door at all means the unopened one MUST HAVE A CAR. Awesome, we just figured out that the first possibility (with a likelihood of 1/3) means switch = car!

Let’s look at the second possibility, X G = 1/3. Monty MUST open a door. That door CANNOT be a car. Under this possibility, the fact that Monty opens a door at all means the unopened one MUST HAVE A CAR. Awesome, we just figured out that the second possibility (with a likelihood of 1/3) means switch = car!

So far, looks like 2/3 of the time, Monty opening a door at all (which he is compelled to do) means switch = car. This is great!

Let’s look at the third line. G G = 1/3. Here Monty picks a door randomly. So half the time (UNDER THIS POSSIBILITY which is already only 1/3 of the time) he’ll open one door, and half the time (again, UNDER THIS POSSIBILITY which is already only 1/3 of the time) he’ll pick the other. It doesn’t matter which though, since when you add each half probability, you are still left with a whole 1/3 of the time where you switching = no car.

So let’s look back. 2/3 of the time switching = car. 1/3 of the time, switching = no car.

Switch!

to Groove: I also initially refused to accept that switching improves the odds (though perhaps not as vociferously as you). Every argument (including the one I just made) felt completely wrong to me.

The only thing that convinced me was me writing a program simulating the game myself, and running it a million times. Switching gave me the car 66% of the time.

Since I was where you are now, I offer no argument to you, as I know arguments led me nowhere. I only ask (for your own curiosity) that you take some time to write the program yourself with the game exactly as it is stated, and to run it a ridiculous number of times.

And hey, it’s sort of scientific too, right? You disagree with the results of other scientists, but thanks to the reproducibility property of the scientific method, you get to check for yourself!

@Groove: If Monty Hall showed you a random goat out of the three doors (possibly your own door), then yes, it would be a 50/50 split (I think). However, because he is always showing you the “loser” in the bigger group, it is better to choose the bigger group (switch).

RE: Monty Hall Problem

Think of it this way. You have a choice of one door or two doors(which has been collapsed to one by elimination). Wouldn’t you pick two doors over one if you could?

Sorry if somebody already said this, I did not read every comment.

@bret: Awesome explanation, that’s exactly it: do you want 1 door, the best of 2 doors?

@bret: Great explanation. Thanks! still finding it a real mindcracker.

It’s only logical to suspect the game show host has more information than contestants about what is behind the doors, at least at some point, or that the game is rigged in some other way, but the question does not give enough information about these things, so there are at least a couple of different possibilities to consider.

First, either contestants are always given a second choice, or they are not. A situation where the game show host is not obligated to give contestants a second choice is not likely since for maximum entertainment purposes, and thus the game show’s popularity, so a second choice would likely always be offered. So it’s more logical to assume that you’ll be allowed to change your mind and given a second choice.

Second, if the game show is rigged because what’s behind all the doors, or even just the two remaining doors, can be switched by the game show operators after the game is started, then it makes no difference if you switch or not, since the odds are solely what the operators choose them to be, which arguably is for more winners than only one third of the contestants to keep the game more popular than if only one third, or less, of the contestants win.

Although it’s highly unlikely the game is rigged in this way since that would not be ethical, and in any case it could not be expected to remain a secret, but with the limited information given you cannot be absolutely sure, so it’s better to switch for your second choice because of the following reasoning.

If you picked the door with a goat, the other doors have a car and a goat. You can certainly reason that the host then can’t open the car door because that would ruin the game for the audience and thus the show’s popularity. The probability that you first picked the car door is one out of three, so there is a two-thirds chance that you first picked the one of the two goat doors. Thus, without knowing the history of the show, you can reasonably assume that it’s at least rigged to the extent that the host knows what is behind the two remaining doors, so he will open the goat door. That means the other door now has a two-thirds chance of having the car, and your initial door selection has only a one-third chance.

So considering everything what don’t know and the new information the host gives by selecting a goat door, you have a much better chance if you switch.

Holy crap I finally understand it and here’s now *facepalm* I hope it makes sense.

Take a pizza divide it into 3 pieces. Make 2 plain and 1 pepperoni. You want the pepperoni. Choose 2 of the 3 pieces randomly. Chances are that you picked the pepperoni (2 of 3 slices 66%). (Or 33% chance you didn’t). But you want the pepperoni, even if it’s only one slice. You trade the 2 slices 1 slice w/pep. If you did this over and over, 66% of the time = keep and 33% = trade? Simple math, no?

So that means 66% of the time you wouldn’t trade but would have 2 slices(or 2 doors if you look at it that way). Pretend those slices are instead doors. Let’s looks at Monty Hall problem from Monty’s point of view and in reverse (sort of). We know where the car(pepperoni) is, so that no longer a mystery.

Pretend the 1 pizza slice was the 1 door pick by the Monty Hall contestant. We already said before that the 1 pizza slice(door) could only be the pepperoni(car) 33% of the time and is only traded 33% of the time, right?

So if you were the contestant and knew your first choice of doors was the car (or pepperoni) only 33% of the time. Wouldn’t you trade your 1 slice(door) for the other 2 slices(2 doors)? Since we know the 2 of 3 slices(2 doors) will contain the pepperoni(car) 66% of the time? Him removing a door is a kin to you dropping a piece on the ground when exchanging the 1 piece for the 2. Even though the 1 piece(door) is not there in the very end doesn’t remove the fact that the pepperoni(car) was one of those 2 pieces 66% of the time.

Change the rules: 3 doors – 1 car 2 donkeys, you can have 2 doors or just 1 door. If you choose 2 doors (aka switch your choice after the first pick) Monty is willing to take one donkey off your hands. You pick the 2 doors even though one’s a donkey for sure.

Damn that’s a lot easier to think about than to try and put into words.

just pick the wrong door and he’ll pick the right door for you. with 3 doors,t here are two wrong ones, and with 1000 doors, there are 999 wrong ones. in any case, you have over 50% chance of having picked a wrong door.

You are amazing. That just made so much sense unlike the textbooks i’ve read with all these formulas.

@Kevin: Thanks — the textbook explanations didn’t click for me either.

Another way of explaining -

Let there be 3 doors – A,B,C. You pick A.

Let goat=0, car=1

All Possibilities:

A B C

0 0 1 – B is revealed to be a goat. C left.

0 1 0 – C is revealed to be a goat. B left.

1 0 0 – Any one is revealed. Other left.

After revealing goat,

A Other gate

0 1

0 1

1 0

A’s winning chance = 1/3

Other gate’s winning chance = 2/3

The other gate S (switched) is actually an OR function.

S = B or C (ie, best of B and C)

Since the other gate S is “BEST OF TWO”, it is a better option.

Hope you don’t mind my comparision with binary functions :D:D

Valid – very well done. I admit I get a weird pleasure from seeing the logical knots the non-switchers tie themselves into. One correction, in comment 32 you said to Frank that Monty is revealing new information with the pick. But the reason the switch probability is 2/3 is precisely because Monty has not revealed any new relevant information by showing you a goat. He can always show you a goat. So your original probability of guessing correctly of 1/3 is unchanged by the revelation of a goat behind one of the remaining doors.

I have not read all of the above, so if this is already stated then … sorry you have to read it again.

This doesn’t even have to be a question of math or probability.

Given three cards: one Ace and two 9′s

You pick one of the cards but you are not allowed to look at it.

Some bozo tells you that at least one of the other cards is a 9 (if you accept this like it means something special then you are a bozo as well … because every time at least one of the other cards is a 9).

You place the single card face down and next to it you place the other two card face down.

Now the bozo tells you that if you pick the “stack of cards” that has the ace in it you win a car.

Which of the “stacks” do you pick up … the one with a 1/3 chance of winning or the other pile that has two cards in it?

It has been twenty years since I thought of this scenario … thanks Kalid, for the opportunity to keep thinking about stuff.

@Trevor: Hah! It’s funny, I think everyone starts as a non-switcher until you slowly start to realize… =). Good point, no information was actually revealed since one of them *had* to be a goat.

@Don: Thanks for the comment, glad the site’s giving you something to chew on :). Actually I find learning most enjoyable when you come back to it, vs. having to memorize some explanation about why the Monty Hall problem works (vs. _understanding_ why it works).

I like the analogy — the key is seeing you’re given the choice between 1 card (your original guess) or the best of two other cards. Monty just does you a favor by throwing one away early. Nice insight.

Actually Monty does you a confusing (to most people) disservice by throwing one away early. If he left the unpicked two doors closed and said you could open them simultaneously, and if you see a car … you win. Almost all would then trade for the two doors.

@Don: Exactly, funny how that works right?

Groove wrote:

This is a rather peculiar argument. So my odds are 1/2 even before Monty opens a door, just because “Monty is always going to show you a goat”? Fascinating! I wonder, what happens if Monty swears on a stack of textbooks that he’ll open a door, but after I make my choice, he gets an urgent phone call and has to leave? Will my odds “go back” to 1/3?

Suppose the game worked like this: John and Sarah are playing as a team. They have to agree on which door to pick. Then, John is blindfolded and Monty opens one of the other doors, showing Sarah that it has a goat. Monty closes that door and John’s blindfold is removed, then John and Sarah’s original door is opened. What is their probability of winning? If both of them think like most people do when initially encountering this, then Sarah will think it’s 1/2 (because it’s one of two door) and John will think it’s 1/3 (because he has no idea which of the other doors was revealed, so he learned absolutely nothing while blindfolded). Yet if they play repeatedly, their win rate has to converge on one or the other of those. It will, of course, converge on 1/3 — the rigamarole about blindfolding and door-opening had no effect on their initial choice.

This may be true (I’m not a statistician), but only because you’ve widened the error bars so much, allowing for “almost 80%” to count. A more relevant statistic is the simple total of wins, and for 1000 trials, that total should be much closer to 666 than to 500. This is assuming that you made sure to always switch, of course! If you mistakenly switched 2/3 of the time, your win rate will instead be about 555, which is pretty close to 520.

And if you switched exactly half the time, you’ll win half the time, too — which is exactly the way it works if you make a half-bet on ANY binary situation! If you bet on every baseball game in a season and flipped a coin beforehand to determine which team to bet on, you would win half the time, but that doesn’t mean all the teams are equally good.

When you choose a door, there is 2/3 chance of a goat. On these occasions, the other goat is eliminated and the car is in the other door, giving a certainty of a car if you change doors with your second choice.

100%*2/3=2/3

@Alexius: Yep, exactly. If you pick and hold, you’ll lose 2/3 of the time.

The most amusing comments here are by those people who’ve twisted themselves into pretzels to convince themselves that having 3 doors to choose from and only one choice somehow gives them a 50% chance of picking the right one in the first place. Monty revealing a goat does nothing to increase your original choice from 33.333%. Whether you picked a car or goat, there will ALWAYS be a goat for Monty to reveal.

Once more: suppose that you get to choose one door, and you have a friend with you who gets BOTH of the other two doors. Who will win the car more often? Your friend, of course. By switching doors after the reveal you have done the SAME thing–you can have Door 1, or BOTH Doors 2 and 3. Which is more likely to have the car? Misters Groove and Smith Jr should take special note. Oh, and suppose that Monty still reveals one of your friends doors to have a goat–do you really believe his chances and yours are now both 50/50? You better not, because one of his doors HAD to be a goat.

I’m seriously curious about how Monty revealing a goat behind one of the doors you didn’t pick somehow changes the odds of your original pick. Please, someone, expound upon that.

@Groove (#53)

You are correct that Monty will always reveal a goat, there is no doubt about that. The question is not whether or not Monty will reveal a goat, but rather is Monty forced to reveal the particular goat that he did, is his choice of goat restricted?

Say the car is behind Door A. There are 3 scenarios that are equally likely to occur:

1. You select Door B. Monty must reveal Door C. In this situation Monty is restricted to choosing Door C. (1/3 chance)

2. You select Door C. Monty must reveal Door B. In this situation Monty is restricted to Door B. (1/3 chance)

3. You select Door A. Monty must choose randomly between Door’s B and C. In this situation Monty’s choice is not restricted. The chance of him picking Door B is 1/6 and the chance of him picking Door C is 1/6. Note that the probability of picking Door B and of picking Door C add to the one third. (1/3 chance)

If this experiment was ran 300 times with the car behind Door A everytime and the player selecting each door 100 times the results would be as follows:

Door A: Never opened because the car is behind it.

Door B: Opened 150 times, 100 times because it had to be (as in Scenario 2) and 50 times by the choice of Monty (as in Scenario 3). Note that 100 times out of 300 is 1/3 while 50 times out of 300 is 1/6.

Door C: Opened 150 times, 100 because it had to be, 50 by the choice of Monty.

We can see that for both Door B and Door C that there is a 2/3 chance that it was a restricted choice, Monty was forced to choose it. If Monty was forced to choose a door, that means that there was only one available goat for him to choose from. If there is only one available goat to choose from, the player must have chosen the goat to begin with. This means that there is a 2/3 chance that you have chosen one of the goats, therefore the car is behind the door that you have not chosen.

Summary:

If the prize is behind Door A and you select Door B, Monty’s choice of doors is restricted to Door C, the remaining goat. There is a 2/3 chance that Monty’s choice will be restricted because there is a 2/3 chance that you chose a goat to begin with. The only time you will win by staying is when Monty’s choice is unrestricted (Scenario 3). There is only a 1/3 chance of Monty’s choice is unrestricted therefore if you stay you will only win 1/3 of the times.

Yes, Monty will always pick a goat, but it is whether or not he is forced to pick a goat that matters. If he is forced to pick a goat (2/3 of the times) than switching will win.

I hope that clears it up

If I were playing the game, knowing the rules, I would decide in advance always to switch. Then the situation would be equivalent to the following. I walk on to the stage and select a door. Monty says, “Do you want to stick with that door? Or would you like to open both the other doors? If a car is behind either of them, you win the car.” I agree to open both the other doors. As I walk towards them, Monty says, “I’ll save you the trouble of opening this one. As you can see, it has a goat behind it.” I therefore open the other door. I have of course increased my probability of winning, from 1/3 to 2/3, at the time I accepted the offer of opening two doors.

Here’s an alternative way to look at it: There are 3 doors and you know only 1 has a prize. You pick a door. Then, you are given the option to pick the 1 door you already chose, or the 2 other doors. Obviously, you would choose the other 2 doors. You already know that at least one of the other two doors are empty– the fact that Monty opens one of the doors is irrelevant. That’s really a psychological trick that makes you think that your options are being narrowed. In reality, the information is exactly the same; Monty simply reinforces what you already know (that at least one of the other doors is empty). Eliminating the psychological trickery makes it intuitive.

This is an interesting probability problem. We can use tree diagram to work up the chance of getting the right choice.

Gary’s articulation from the top of the comments helped me to make the connection and understand what the problem was getting at: “The odds of the selected door having the prize hasn’t changed at all. You just know more a lot about the doors you didn’t pick.”

If you originally chose a goat (without knowing it), then if you switch you will get the car.

If you originally chose the car, and switch, you will get a goat.

If you were wrong you will become right, and if you were right you will become wrong

Therefore the probability of being right after switching is the same as the probability of being wrong with your first choice i.e. 2/3

And the probability of being wrong after switching is the same as the probability of being right with your first choice, i.e. 1/3

Therefore you double your chances of being right (getting the car!) by switching

OH. I get it now. You want to TRY and pick a goat the first time. Since there are 2 goats, and 1 car, you would have a 2/3 chance of the goat actually being your first choice. Now that Monty has removed the other goat, you switch. You had a 2/3 chance of picking a goat, so you do not stay with your choice. Monty has removed the other goat, so you should pick the last one.

Summary:

First, we need to know our basics in adding fractions. 1/3 + 2/3 is 3/3, right? Okay.

Try and pick a goat (you pick door 1). So door 1 has a 2/3 chance of being a goat.

Door 2 is removed, so it IS a goat. Door 2 is out of the question

Door 1, as previously stated, has a 2/3 chance of having a goat. So where is the other 1/3? Behind door 3. So door 3 has a 1/3 chance of being a goat, and door 1 has a 2/3 chance of being a goat. Which one will you choose now? DOOR NUMBER 3!! Enjoy your new car

I have spent countless hours online trying to explain this to people, usually using the 100 or, when really desperate, 1 billion door, analogy, often to no avail.

Another method as referred to in comments, is to look at the doors not chosen as a single block with 67% probability. Revealing which of the two is the Goat does not reduce the probability of one of them being the car. It remains 67%.

However, today, I came up with what may be the best way of having naysayers challenge their thinking. Maybe.

Ask them, if the Monty Hall show ran 100 episodes, how many cars the producers would have to allow for in the show’s budget. That’d be 33 plus a contingency of a few more, right?

This should cause the naysayer to concede that, all things being equal, in 100 shows, there will be 33 winners off a straight pick with no one switching after the first goat is revealed.

It should be a short leap of logic then for them to see that if 33 people are destined to win, then 67 people are destined to lose, and get only a goat.

The next step is to ask what would happen if each contestant was ordered, under pain of death, to change their initial selection when given the opportunity by Monty.

The inescapable conclusion should then be that the 33 people who won a car when sticking with their first choice would become losers and the 67 people who originally picked a goat would become winners, simply by switching to the only other closed door.

This means it can never be a 50/50 proposition unless the choice is made by someone who does not know which door was chosen first and/or the contestant clean forgets.

So another distinction needs to be drawn between the probability of a contestant picking correctly after being given the option to switch (which is virtually impossible to compute) and the probability of one door or another holding the prize. Evidently two different things as how people use the information they have is unpredictable. A Vulcan, of course, would always switch.

As the whole Monty Hall 3-doors game is mired in preconceived ideas, it may be simpler to ask the person to consider 3 playing cards and ask them how many times in a hundred they would expect to pick a lone Ace amongst the three cards. If they agree that they would pick “wrongly” 67 times then it should be a short step to have them see that they will double their winnings if they switch every time by swapping 67 “mistakes” for 67 wins and 33 wins for 67 losses.

I think this explanation removes the emphasis from the last two doors and puts it back on the original choice by having the person acknowledge the statistical probability of the first pick being a winner and the inability of later events to improve their chances of winning from 1/3 if they stick with their first choice.

I’ve been wrong before though . . .

Editor, could you please change that last number from 67 losses to 33 losses and delete this post, please? Thanks, J.

Perhaps an easier way to get nay-sayers to see the light is to disprove the 50/50 theory rather than attempt to prove the 33/67 theory.

A simple way might be to ask the proponent of 50/50:

“If you ran a game, such as on “Let’s Make a Deal”, where contestants are given a one-in-three chance of picking a prize but in this case they are not allowed to switch, on average, how many contestants would you expect to win the car based on their first random pick from three doors?”

Hard to imagine that anyone, even a 50/50 proponent, could answer anything other than “33″ or “About 33″.

You would agree with this and perhaps say:

“Then how can each of the two remaining doors in the Monty Hall problem offer a 50% chance of winning?

If this were so, 50 contestants would each win a car if no contestant switched and, ipso facto, 50 contestants would each win a car if all contestants switched.”

“However, as we’ve agreed, only 33 contestants on average can win a car for every 100 contestants when no one switches. So how can the number of contestants who pick the correct door in a one in three guess suddenly increase from 33 to 50?”

Presuming the dropping of the naysayer’s jaw will prevent them from attempting to explain this conundrum (and we hope they won’t persist in claiming that the order in which the unopened doors are opened somehow increases the odds of winning…) one might say:

“If 33 win when no contestant switches, 67% must win when everyone switches (because the 33 who won when no one switched will now lose and the 67 who lost when no one switched will now win).”

So not only can it not be a 50/50 proposition, it must be a 33/67 proposition as the number of contestants who win can be expected to double when all contestants switch doors.

Even the proponents of 50/50 would seem to have to agree with the proposition.

I have read many explanations of the Monty Hall problem but never one that focuses on why “50/50″ cannot be right rather than why “33/67″ is right.

It would seem that if you can get the 50/50 folks to agree that you do not have a 50% chance of winning if you do not switch, then that would seem to leave 33/67 as the only plausible theory.

If the chances of winning a 1-in-3 guessing competition cannot be 50%, they must be 33% for the first door picked (i.e. 1/n, where “n” is the number of options from which it was chosen), and 67% for the other (n-1/n), as the sum of all probability must add up to 1 or 100%.

The larger the value of “n”, the more advantageous it is to switch. Many people have shown this when they point out the analogous scenario of a game comprising 100 or one million then, usually in desperation, one billion or perhaps, a Googol doors to show that the probability of the first door being a winner very much depends upon the number of doors from which it is chosen, as with any random guess.

I’m not sure if I’m getting this exactly, but this is how I explained to myself leaving the numbers aside for a sec:

You had two choices, #1 is to pic a car (1/3) and #2 is to swap. The reason switching “work” or “gives you more probabillity” is because in your first choice, you are more likely to choose a goat (2/3), so therefore, since it’s more likely that you chose a goat the first time, then when given the opportunity to swap, you are most likely to swap out of getting a goat. Please correct me if I’m wrong though!

Hi Steve, that’s an excellent way of looking at it. What’s more likely: that you swap from goat to car, or from car to goat? Since you’re twice as likely to pick the goat first (2/3 vs 1/3), it makes sense to swap.

groove’s comment helped me understand this. it all has to do with the initial conditions of the experiment. you start off with 3 doors: 2 goats & 1 car & if monty always removes 1 goat from the 2 doors you didn’t choose, you are still twice as likely to have chosen a goat with your initial pick. so switching after 1 goat is removed makes you twice as likely to choose the car. see the helpful diagram below (and in several comments above mine ;p)

goat=g, car=c, numbers= door number

this is if you always choose door 1 with your initial pick

1|23| 1|3

g|gc| g|c

__________

1|23| 1|2

g|cg| g|c

__________

1|23| 1|2/3

c|gg| c| g

Sorry, but in real life it will be 50/50 if you check all (only 12) possible outcome to this problem. Let’s start with the car behind door A. You choose door A and Monty can open either door B or C and show a goat. In both cases it will be wrong to change door (2 misses in total). If you choose door B, Monty can only open door C to show a goat, and it will be correct to change door. If you choose door C, Monty can only open door B, and it will be correct to change door. (In total 2 good and 2 bad changes) The same will repeat with the car behind door B and door C. In total 12 possible solutions, 6 of them will give the car if you change the door, 6 will make you go away from the car. 6/12 vs. 6/12 is a 50-50 change to me.

Hi Dylan, thanks for the comment!

Hi Tom, it’s actually most convincing to try it yourself with some cards [two queens and a king], and have a friend be Monty Hall. Be careful with the counting: are there 12 possible outcomes, or 9?

Door A has the car:

* Pick A, switch, lose

* Pick B, switch, win

* Pick C, switch, win

Door B has the car:

* Pick A, switch, win

* Pick B, switch, lose

* Pick C, switch, win

Door C has the car:

* Pick A, switch, win

* Pick B, switch, win

* Pick C, switch, lose

If you play the strategy “I will always switch” there are 9 scenarios, and you win 6 of them. You only lose if you picked the car right in the beginning, which is a 1/3 chance. Try writing out all the scenarios explicitly to be sure.

Update: I see where you’re counting the 4 outcomes. When you pick the right door (A), Monty can reveal either B or C. Either way, the outcome is a “miss” — you don’t count the impact of “miss, because I switched to B” differently from the impact of “miss, because I switched to C”. A miss is a miss, and you only lose once when you switch away from the winner. That’s why you are counting 2 bad and 2 good scenarios, when it’s really just 1 bad scenario (you switched away from the winner) and 2 good scenarios (you switch to the winner). You can of course subdivide the bad decision (50% of the time I incorrectly go to B, 50% of the time I incorrectly go to C), but the subdivision doesn’t change the original chances of a bad decision.

I’ve been able to understand this thanks to the alternative ways people have shown me on here.

I found Alex’s lottery analogy really good.

The point is that Monty looks at the two doors you have not picked and he has to take away one goat. If there are two goats it doesn’t matter which one he shows to you (this happens 1/3 times) and you would have the car. But IF there was a car and a goat behind those two other doors (this would happen 2/3 times) Monty goes “Ahah! I have to open the goat door. I can’t open the car door and spoil the show”. So you KNOW that out of the remaining two doors (Yours and Monty’s), Monty’s door has a 2/3 chance of being the car and you ought to switch BASED ON THE FACT that he will ALWAYS leave the car there if it is one of the two. As I say, this happens two out of three times.

Therefore switching makes much more sense as the chance is higher.

It’s difficult to understand but if you don’t get it yet try to wrap your head around it with other examples as I did. Trust me, it works!

Thanks Georgia, everyone has different analogies which help, I love seeing what works.

I’m amazed at the number of idiots who think the odds are increased if the contestant swops. Even apparently so called maths experts conned by computor programs.

Having revealed a goat which he always does theres two options remaining a goat and a car and the contestant has to choose one of those two. 1 out of 2 =50% swopping makes no difference to the odds whatsoever.

Steve, at least these “idiots” are polite and well reasoned and know how to spell, eg “swap”.

But more to the point, the odds amongst the two remaining doors are only 50/50 if a random choice is made, e.g. by someone who does not know, or can’t recall or tell, which door was chosen first.

The first chosen door has a 33% chance of being the winning door as it was chosen from 3 doors. Those odds of winning do not change with the order in which the doors are opened. the fact is, if the contestant NEVER switches, they will win 33% of the time, on average.

So clearly, the odds are not NOT 50/50 after one goat is revealed. If it were 50/50, the contestant would win 50% of the time by not switching, and that is not possible statistically. Try it at home with a deck of cards . . . try and pick the Ace amongst three cards and you’ll win around 33% of the time. Hardly surprising is it? But by your theory you would, by some miracle, pick the Ace 50% of the time provided one of the non-Aces was revealed before your own card was flipped. Magic!

But if it is not 50/50, what is it? It is 33% for the first door chosen and 1-33% for the remaining door or doors.

Consider also a sporting contest between two contestants. Just because there are only two contestants it does not necessarily follow that each has a 50/50 chance of winning. However, if you pick either contestant randomly then you have a 50/50 chance of picking the winner.

However knowledge of form in that analogy is like knowledge of which door was chosen first in the Monty Hall Problem. If you know Federer is better than Hewitt, you back Federer and have a 90% chance of winning. If you haven no idea and make a random pick, you have a 50/50 chanced of picking the winner.

If you know which door was picked first, you have a 67% chance of winning if you pick the other door. But if you choose by tossing a coin, you have a 50% chance of the coin directing you to the the winning door.

I would never call someone an “idiot” whom I had never met, Steve, however clearly you are rude, thoughtless and mistaken based on your ill-conceived post and it seems likely, like many others of your ilk who tend to protest too much, that you may never understand the Monty Hall Problem.

But, hey; they say ignorance is bliss.

I realized the answer when I looked at it like this:

There are 10 doors and you randomly pick 1. Monty closes 8 of them. So there are 2 doors left: 1 with the car and 1 with a goat. Monty knows this. It would be in your favor to switch doors, because the odds of you picking the right door the first time was 1/10 and will stay 1/10 if you stick to your original choice. You increase your winning chances by switching.

I realized the answer to the question by looking at it like this:

There are 10 doors and you randomly pick 1. Monty closes 8. So there are 2 doors left. One with the car and one with a goat. Sticking with your original door means your chances of it being the door with the car are still 1/10. So switching doors increases your winning chances. Switch and win the car.

Ok, I think I get it. Here’s how I understand it:

There is a 1/3 chance for the door you pick to have a car, therefore there is a 2/3 chance for either of the remaining doors to have a car.

After all filtering, there is still a 1/3 chance for your door to have a car and a 2/3 chance for any other door to have a car.

Since there is now only one other door there is still a 2/3 chance for that door to have the car and still 1/3 that your door has it.

Or for 100 doors. 1/100 that you pick the car the first time and 99/100 that you didn’t. 98 doors are eliminated. It’s still 1/100 that you got it right the first time and it’s still 99/100 that you didn’t. But now, there’s only one other door, so it’s 99/100 that it’s the other door.

Regards,

Snehal Masne

http://www.iSnehal.com

Excellent article Kalid.

I’ve been reading the comments section and somebody mentioned applying the Monty Hall Problem to a Who Wants to be a Millionaire scenario as a real world application. But if I’ve understood the Monty Hall Problem correctly, it would be of no use in such a scenario.

I’m imagining the Who Wants to be a Millionaire scenario as follows – You are given a question with 4 possible answers (A,B,C &D). You have have no idea which of these answers is correct so you choose one at random (lets say answer A). You use your 50:50 life line and answers B and C are removed. You are left with answers A and D. Should you switch to answer D?

My understanding is that because answer A was not protected from the 50:50 life line filtering (unlike the original door choice in Monty Hall, answer A was just as likely to be filtered out as answers B,C &D) there is no advantage in swapping from answer A to answer D. Both have a 50% chance of being correct.

Is this interpretation correct?

Thanks,

Laurence

The winning odds when are 2/3 because those are the odds of your first guess being a goat. Switching in that case guarantees a win because the host will always leave the car – it’s that simple!

Simple problem, big explanation xD.

Think visually not with arithmetic and it makes sense instantly.

Oh and creating a program to practically prove it is very good too and simple. Luv this

@Keep: Thanks, glad you liked the simulation!

Like Steve, I am amazed at all those people who do not understand probabilities. For the first part, the probability is 1 in 3 of picking the car. After the host opens a goat door the probability of picking the car is now 1 in 2. A lot of garbage is being written here about 2/3 probability remaining with the last door. the real way of viewing the probability is that 2/3 remains with the two last doors, not transferring to one door, regardless of whether one door is open or not. Once a door is opened, you are looking at a new problem. The old situation no longer applies. With a door open, the 1/3 – 2/3 probabilities no longer apply and the choice becomes 1 out of 2, or 1/2.

The way the Monty Hall problem is phrased is intentionally confusing. I see websites that wrongly say there are only 3 scenarios for the switching, but there are actually 4. the sites that try to confuse are saying that when the car is correctly chosen at first, there is only one outcome from switching, whereas there are two outcomes. Selecting to reveal a goat can be done in 2 ways, not one this brings the odds back to 1/2. Work through it yourself.

Phil, rather than be amazed, like Steve, you should take some more time to consider what you know is counter-intuitive. That means, your first instinct is most probably wrong.

You say: “After the host opens a goat door the probability of picking the car is now 1 in 2.”

Please explain how my chances of picking the one door in three with a car behind it increase from 1 in 3 to 1 in 2 because one of the doors I did not picked is opened first?

You also say: “Once a door is opened, you are looking at a new problem.”

No, it is the one and only problem, ie the same. It is new problem if someone enters the game who does not know which door the contestant chose. He has a 50/50 chance of guessing which door has the car, even though the contestant (and her door) has a 33% chance of being right. If he knew which door the contestan picked, he would not pick the same door as the other door has odds of 1-33%.

So while the odds of someone randomly guessing the right door are 50/50, the odds of that door being the contestant’s first pick are just 33%.

You say: “The way the Monty Hall problem is phrased is intentionally confusing.”

To whom? To you and Steve? I do not find it confusing, nor do those who understand and appreciate that the solution is counter-intuitive.

You also believe that “sites .. try to confuse.”

Do you really believe people would set out to confuse people to justify an incorrect theorem?

You’re not conspiracy theorist, are you Steve.

If you spent 10 mins with a friend and a three playing cards (an Ace and Two Jacks, say) you would quickly see that if you stuck with your first guess every time you would win, as you would hopefully expect, 33% of the time, and if you switched every time, you would win 67% of the time.

QED.

You’ve got to be amazed at people like Phil and Steve who accuse others of not understanding probability when they themselves appear clueless about the subject. Not to mention that there are hundreds of sites explaining the MHP with logical and mathematical proofs , it is a standard text for students in maths textbooks worldwide, and the literally millions of computer trials demonstrating the 33.3/66.6 outcome. But these, and maths experts everywhere, are all wrong, Phil and Steve however know the correct answer. LMAO.

This is a problem I have always been stumped by (and I do understand probability very well) because of the apparent conflict between the model and the reality.

At the purported start of the game, I have no door and so no chance of having the car. If I choose 1 of the 3 doors I have a 1/3 chance of choosing correctly (~33%). Due to the nature of the game, the actual game doesn’t start until Monty reveals the door with the goat behind it and asks if I want to switch.

At this point – the true start of the game – I have a 1/2 chance of having the car (50%).

If I switch, I have a 1/2 chance of choosing the correct door (50%) – but I already have a 50% chance of having the correct door.

Since probabilities are multiplicative:

Stick with original choice: 0.50

Switch: 0.50 x 0.50 = 0.25

So mathematically – if I switch it reduces my likelihood to end up with the car from 0.5 to 0.25. However, it never works out that way in the simulations.

I guess I need to spend the time to program it the way I conceptualize it and see what the results are.

Ableger, I do not mean to be rude, but just about every significant premise in your post is incorrect.

Firstly, the game actually starts when you make your initial 1 in 3 pick. This sets your odds of picking the car at 33% as you correctly state, or 1/n, where n = the number of doors when you made your first guess.

No matter what happens thereafter, you will win 33 times for every time you play this game, no matter what Monty does, if you stick with your first choice.

For instance:

- If the set burns down every time, you will still win 33 times;

- If a of wind blows both of the other doors open, you will still win 33 times;

- If a cyclone knocks the set down, you will still win 33 times in 100 attempts,

if you do not change your first pick.

The corollary is that if you switch every time, you will win 67 times in 100 attempts.

The game only starts, and so moves to 50/50, when there are just two doors remaining for:

(a) a contestant that forgets which door they picked first (which has a 33% chance of winning) and so have a 50/50 chance of picking the car when they make a random guess the second time; or

(b) someone who enters the game after the first door has been opened, and so have a 50/50 chance of randomly picking the winning door from the remaining two doors.

While the chances of

(a) the person who forgets; and

(b) the person who comes in late,

are 50/50, however, the chances of the door the contestant first picked being the winner are 33% and the changes of the only other door remaining being the winner are 67%.

But when you make a RANDOM pick between two outcomes with unequal probabilities of eventuating, the chances of picking the “winner” are 50/50.

For instance, randomly pick the winner of a tennis match between Roger Federer and Anna Kournikova and you have a 50/50 chance of picking the winner. You will win 50 times in 100.

Make an educated guess and you have a 99%+ chance of picking the winner, because you know which selection is more likely to win. You will win virtually every time.

Similarly, if you know which door was chosen randomly from the three available options, you know which door has a 33% chance of winning, and that the other door not picked must, by a process of elimination, have a 67% chance of winning.

So if you are the contestant in the game, and not suffering memory loss or entering the game at the two-door stage with no knowledge of which door was picked from the first three, you can use this information, this knowledge of which door was picked first, to improve your chances of winning from 50% to 67%.

If there are even more doors to choose from originally, like 10, the improvement in your odds by switching increases even more, as your odds go from from 1/n, where n = the number of doors, to n-1/n.

Just as you can improve your odds of picking the winner of the Federer vs Kournikova tennis match by using your knowledge of these two tennis players’ relative abilities, you can improve your odds of picking the winning door by knowing the chances each door being the winner.

Of course, if you make a random choice every time, without knowing or applying this information, your chances of winning are 50/50 and you will win, on average, 50 times for every 100 times you play.

QED.

Jonathan,

No offense at all – but if you consider the premise of what I am suggesting, you will see that the “choice” of 3 doors is actually a false choice. Since after you choose Monty is going to show you which of the doors you didn’t choose has a goat behind it, the actual odds of having the correct door is 50%.

(This is the difference between Choosing the car, and Having the car). Since the ultimate goal of the game is to Have the car, you start the game with a 50% chance of having the car.

As I said – this is a conflict between the statistical calculation and the reality of the situation. You actually start the game with a 50% chance of having the car – your initial choice is nullified by Monty’s interference.

Abelger

It is not a “false” choice. It is an actual choice. That your first choice is irrelevant is a false premise as your first choice tells you two things:

- Which door has a 33% chance of concealing the prize; and

- Which door has a 67% chance of concealing the prize.

Valuable information if you know how to use it. Just like knowing which tennis player is more likely to win is valuable information when picking between two contestants in a tennis match and sparing you the 50/50 odds of a random choice.

Anyway, it is axiomatic that the chances of the first guess concealing a prize choice are 1/n, where n = the number options when the first guess was made.

If there were 1 million doors, the chances of the first choice concealing the prize would be 1/n = 1/1,000,000.

Ergo, the chances of the “other” remaining door concealing the prize are n-1/n, = 1,000,000 – 1/1,000,000 = 999,999/1,000,000,

certainly not, and much better odds than 50/50.

Abeger

It’s difficult to know where to start with your posts as practically everything you’ve written is incorrect. You clearly don’t understand probability half as well as you think you do.

If your model doesn’t agree with reality then your model is wrong. There is no conflict between statistical (I take it you mean probability) calculation and reality, they are both in agreement: the probability the prize is behind the door you picked initially is 1/3, the probability it’s behind the door Monty didn’t open is 2/3. If you could be bothered to test it yourself (using 3 playing cards) you’d realise that’s it true.

Your premise that you “start the game with a 50% chance of having the car” is plainly absurd – there are 3 doors to choose from at the start of the game. Whichever door you choose at least 1 of the 2 doors you didn’t choose doesn’t have the prize. Using your “logic” if you pick a card from a full deck you have a 50% chance of picking the Ace of Spades, because Monty can always reveal 50 of the remaining 51 cards that are not the Ace of Spades.

Do yourself a favour and do some research before you make another posting.

I never said you had a 50% chance of PICKING on the first pick. You have a 1/3 chance.

Once you see the goat behind the other door and there are 2 doors remaining you then have a 50% chance of HAVING the car. If it were a deck of 52 cards as you suggest, if you pick one you have a 1/52 chance of having the Ace of Spades. Turn over all other cards but yours and one other – what are your odds of HAVING the Ace of Spades? 50%.

You might want to brush up on your reading comprehension before you criticize someone else.

What you said was “you start the game with a 50% chance of having the car”. You start the game when you pick a door. After you’ve picked a door and after Monty has opened a door is NOT the start of the game – you’re halfway way through the game by that stage.

If you think the pack of cards example is a 50/50 proposition then try it out yourself. Get a pack of cards and pick a card at random – don’t look at it and put it face down in front of you. Now, pick up the remaining 51 cards and remove 50 cards that are NOT the Ace of Spades. You now have 1 card left in your hand, and the card you picked to begin with face down on the table. Which one is the Ace? Repeat as many times as necessary for you to realise that the card left in your hand is 51 times more likely to be the Ace than the one face down on the table.

You might want to brush up on your math skills before criticizing someone else’s

Kalid, this is really great work and I think your explanation is very clear. The outcome of the puzzle is highly counter-intuitive at first, but once explained clearly as you have done I think it can be readily grasped by most people.

” Repeat as many times as necessary for you to realise that the card left in your hand is 51 times more likely to be the Ace than the one face down on the table.”

Palmer – so you are now saying keeping the card is more likely to get you the win? I thought switching was the winning strategy?

I won’t clog up Kalid’s message board with more back and forth, so good luck to you.

BTW – this is a great blog Kalid!

Ableger

What I’m saying is that the card you picked to begin with (the one face down on the table) has a 1/52 chance of being the Ace, the ‘other’ card, the one not turned over from the 51 cards you didn’t pick, the card left in your hand, the one you’d switch to given the opportunity has a 51/52 chance of being the Ace. They’re not equi-probable as you assert.

Albeger

You wrote:

“I never said you had a 50% chance of PICKING on the first pick. You have a 1/3 chance.”

This tells us you know that you have a 33% chance of wining if you make a 1 in 3 guess.

Imagine you were the Producer of the Monty Hall show which was planning 100 episodes and you wanted to budget for the cost of cars you might have to give away IF CONTESTANTS WERE NOT ALLOWED TO SWITCH.

It would be 33 cars, wouldn’t it?

BUTG HANG ON, you’ve been telling us that after the FIRST DOOR IS OPENED by Monty, the chances of the contestant winning are 50/50.

THis means , JUST BECAUSE MONTY OPENS ONE OF THE DOORS AFTER THE CONTESTANT HAS MADE A ONE IN THREE GUESS, I am going to have to budget for 50 cars, not just 33 as logically one would think. Damn Monty for opening that door and turning Goats into Cars!

Good thing you’re the produce and not some idiot who thinks they only need 33 cars! The show would go broke!

BUT HANG ON A SEC, If statistically only 33 contestants will have picked the car on their first guess in a one in three guess, HOW DO 17 GOATS TURN INTO CARS JUST BY MONTY OPENING A DOOR WITH A GOAT BEHIND IT?

A theory is supposed to explain reality, but 50/50 only explains fantasy, because as we both know, ONLY 33 CARS WILL BE WON IF NO ONE SWITCHES.

Your assertion that CONTESTANTS HAVE A 50/50 CHANCE OF WINNING AFTER THE FIRST DOOR IS OPEN would mean that 50 contestants would win and as a producer you would have to budget for 50 cars not just 33.

Not only does it not happen, IT WILL NOT HAPPEN.

WHY?

Because, as you said in your post above, “on the first pick. You have a 1/3 chance” and if you have a one in three chance it stays one in three NO MATTER WHAT MONTY OR ANYONE ELSE DOES.

And while this may not be an intelligence test, it is certainly a test of your logic and reasoning ability and ability to see that a logical conclusion will survive all manner of irrelevant events, like doors blowing open, or Monty opening a door with a goat. The conclusion will survive intact to because it is an inevitable conclusion.

Finally, Albeger, consider this:

This problem is COUNTER-INTUITIVE. That means the correct answer goes against instincts, not maths. So if your intuition is telling you 50/50 you know you gotta think again.

Secondly, there are millions of people who thought like you and realised they were wrong. However, I will give a thousand dollars to anyone who can find ONE PERSON who used to think it was 33/67 and then became convinced it was 50/50.

NO SUCH PERSON EXISTS!

Finally, in any such forum, you generally only find the 50/50 proponents getting abusive and calling the 33/67 idiots. There is simply no need to get emotional when the answer is logical and clear underpinned by maths rather than emotion or intuition and yelling at people acting on instinct simply does not work. We simply use logic.

If you cannot understand the logic of Monty Hall, the logic of those three facts should give you A LOT OF PAUSE before you send your next post.

I understand if you do not share with us the bittersweet realisation that you got it wrong but at least you can enjoy explaining Monty Hall to people you know who don’t get it.

In that regard, Monty Hall is not just a test of reasoning, it is a test of character.

Good luck with your next post.

@Jonathan -

“Finally, in any such forum, you generally only find the 50/50 proponents getting abusive and calling the 33/67 idiots. ” – well you are certainly the exception to that rule.

Albegler

Please provide direct quotes from my post(s) showing evidence of any “abuse” as opposed to forthright facts and opinion.

God luck with that too, Ablegler.

Another way of looking at it..

You have only three choices…

Goat Goat Car If you Switch If do not Switch

Choice1 X Win Lose

Choice2 X Win Lose

Choice3 X Lose Win

Hence, if you switch you have a 2/3 chance of winning

Apologies if someone already pointed this out.

My boss gave me an explanation that makes it clearer to me than the 100 or 1 million boxes even.

Same scenario, but instead of Monty taking one box that always has a goat away, he says, you can either stick with your choice, or you can have whatever is under BOTH the other two boxes.

Would you stick or change.

As before if you stick you have a 1/3 chance of getting the car.

If you change you have a 2/3 chance of one of the other 2 boxes having the car.

Monty just made it simpler by removing the goat. Same probability though.

Great explanations! Funny how hardened academics can’t see it. It’s not like flipping a coin… the second choice is NOT independent! You are not a machine and you have information from the first selection. You are just more likely to have a goat. Period. More doors = more confidence in the switch. Forget what is removed.

You choose a door – 66.6% chance you have a goat – ALWAYS.

The fact that a door is removed is immaterial – it does not reset your odds.

All you can say is you are 66.6% sure you have a goat – Period.

Forget that he removed the door or that there are now two doors. Nothing will change the initial odds of you having a goat which are 66.6% on your INITIAL CHOICE. He can paint the doors blue or dance the Macarena.. you still have a 66.6% goat door, or 99% if there were 100 doors. On that you must agree. So of course you switch..!

There is never a 50:50 choice involved anywhere, that would only happen if he re-scrambled the doors or someone else chose randomly.

I don’t think there are any “hardened academics” who still don’t get it. When the problem was originally posed it was poorly worded and open to different interpretations. Without all the conditions that need to be present clearly defined (as Kalid does here) the odds for each door can indeed be 50/50 after a goat door is opened

Albegler,

I invite you to a friendly game! We will do the monty hall problem with a full deck of cards. Here’s how it works: The ace of spades always wins the hand. You draw one card randomly from the deck, our friend (Call him M.H.) will take the remaining 51 cards and sort them, giving me the ace of spades if it’s still in the deck or a random card if you’ve already selected it. MH will then lay out the remaining 50 cards between us on the table to show that the ace of spades is missing. Now we each have a 50% chance of holding the ace of spades, so we can play all night and not too much money will ever change hands. I want at least 10 rounds at $100 each, but I’ll play as many more as you like. Be careful though, if we use the same calculation that gives a 66/33 split for monty hall, I’ll win this game 98.07% of the time.

Looking forward to it!

Ok guys, Groove almost got me with his argument, but only things that actually conviced my is this short clip by BBC ,

http://www.bbc.co.uk/news/magazine-24045598

It’s explained for dummies….Cheers!

I think “Albegler” probably gets it, but as for Ablegler, he has been silent amidst an insurmountable onslaught of reasoned opinion and logic.

Is this silence because:

(A) Ablegler now understands the MH Problem but he is:

(i) not prepared to admit his error; or

(ii) too busy showing his friends how clever he is?

(B) Ablegler still does not understand MHP because he has:

(i) not been reading the latest posts; or

(ii) read the posts but his thinking remains rigid and so still does not “get it”?

This is the first time I have read these 127 posts, but I am surprised at the number of ways folks have of complicating what is at heart a simple issue. You originally make a random choice (meaning by definition that the probability of choosing any of the three doors is 1/3), and so in the long run you will have chosen the car about 1/3 of the time. This is clearly true at the moment you make the choice, and absolutely nothing that happens later can affect it in any way. Thus if you keep the choice, you will have the car with probability 1/3. So it must be elsewhere with probability 2/3 (the only two possibilities must have probabilities that add to 1). If Monty removes one choice, it is still elsewhere with probability 2/3, and so must be behind the only remaining door with probability 2/3. Incidentally, of course, Monty does not make a random choice – he cheats in the interest of the game, but nothing he could possibly do affects the a priori probabilities in any way.

Thanks Bob, I like that way of looking at it.

If Monty *randomly* opened a door, without looking, then switching makes no difference. But the rules of the game are that Monty opens the “worst” door (always a goat), and therefore is filtering the other options to your benefit.

Here’s my explanation.

You choose one door, and Monty gets the other two doors. One of the three doors has a car, so there’s a 1/3 chance you have a car, and a 2/3 chance Monty has a car, right? He always gets a goat, since he gets two doors, so every game, he will remove one door, which will always be a goat. Since it’s not random, that means he still has a 2/3 chance of having the car, it’s just that the 2/3 chance is all vested in his one door.

Think of it this way! No doors! Whatever choice you make, you have a 1/3 chance of having the car, and Monty has a 2/3 chance. When he shows the goat, he still has a 2/3 chance, right? He didn’t give anything to anyone else. He’s just shown you a cute goat while you decide. When you do decide, you’re really choosing between your 1/3 or his 2/3.

It’s tricky because he opens a door and your brain wants to think “Hey, I’m choosing between two doors, so it’s 50/50 like a coin flip!”, but in reality, since he ALWAYS removes his inevitable goat, he’s not at all lowering the odds of him having the car behind that other door, it means that door is twice as likely to have a car, since he had twice the doors, and the game rules dictate he has to show a goat. In essence, you’re just choosing between your one door, and his two; one opened and one closed. Now you don’t even have to try to guess which of his doors would have the car, he shows you which one DOESN’T for sure.

Another way to look at it. The door you choose is a random person, and when you choose your random person, Monty is given a world-renowned chef. They will do battle in the kitchen, making spaghetti! It’s possible that your person is actually amazing at making food and can out-cook the world class chef, but it’s much more likely the chef is better, right? Comparing to doors, the “goat” door he reveals would be like revealing that the chef has a culinary degree and showing the business license and registration to his high-tier restaurant. (3 Michelin stars for you restaurant savvy people). It’s just proving he’s more likely to win, he has TWICE the probability of the random person.

At the end, he’s basically asking if you want to hedge your bet on the Chef or the random person. I love cheering for the underdog, but if a car is on the line, I’ll take the Chef, thank you!

@kalid: Actually, if Monty *randomly* opens a door (instead of using his knowledge of where the cars and goats are), then if Monty still shows a goat, it’s still best to switch in this case: the reason is, that what “Monty does next” doesn’t affect the original probability of choosing correctly (your 1/3 chance of the initial choice being unaffected hasn’t changed).

Monty choosing at random affects the basic scenario only in the times that Monty’s random choice brings up the car. In that case, you’ve lost your only chance of winning – and sticking or switching become irrelevant.

@Jason: Whoops, great call — you’re correct. In a way, Monty choosing randomly makes your odds “double or nothing”:

* If Monty reveals the car, you lose automatically (your original 1/3 odds become 0)

* If Monty reveals a goat, you switch (your original 1/3 odds become a 2/3 chance of a car).

@Jason. That’s incorrect. If Monty randomly opens one of the two unpicked doors and it’s a goat, then the probability of your door containing the car increases from 1/3 to 1/2

@Kalid, you were right 1st time, if Monty doesn’t know where the car is it makes no difference if you stick or switch when he opens a goat door.

@PalmerEldritch: Actually I think you’re correct, there. Though, the chance of having picked correctly at the start is still 1/3, it doesn’t rise to 1/2. What happens is there’s an overall reduction in your chance of winning of 1/3 due to the chance of Monty revealing the car rather than a goat.

Going over the percentages, remember that all outcomes add to 1:

1/3 of the time, you picked the car at the start. Then, Monty reveals a goat at random. So, 1/3 of the time it’s best to “stick”.

2/3 of the time, you picked a goat at the start. Now, Monty opens a door at random. Since Monty is choosing between 2 doors and there’s 1 goat, 1 car, he has a 1/2 chance of revealing either a goat or the car. 2/3 * 1/2 = 1/3, so there are two outcomes here, each with 1/3 probability.

So, 1/3 chance you should stick, 1/3 chance you should switch, and 1/3 chance that Monty accidentally revealed the car, thus blowing the prize. In this scenario, then, there’s no benefit to switching, though you’ll still only win 1/3 rounds of the game by sticking.

When Monty reveals a goat door at random then it makes no difference whether you stick or switch doors. in other words it’s 50/50 or equal probabilities of 1/2 and 1/2. Overall you win only 1/3 of the time, but in the specific cases where he you get to choose, the probability of your door having the car is 1/2 and not 1/3.

That was my point: 1/3 you started with the car, 1/3 the remaining unopened door had the car, and 1/3 Monty opened the car door himself. Since the 1/3 times where Monty opened the “car door” himself are removed (instant loss), the other two are equally likely, e.g. at that point it’s the flip of a coin.

Of course, the above isn’t relevant to the standard rules of the MHP, where switching is always preferable by a margin of 2/3 to 1/3.

In other words, if you didn’t know whether Monty was random or always opened a goat door, you would still be better off switching – since sticking never has better odds that switching. It’s equal, at best.

Ableger, you wrote:

“You start the game with a 50% chance of having the car.”

Patently wrong, whether when you are referring to when you make your first pick or after Monty has opened the door.

You also wrote:

“As I said – this is a conflict between the statistical calculation and the reality of the situation. You actually start the game with a 50% chance of having the car – your initial choice is nullified by Monty’s interference.”

There is only a “conflict” between the statistics / maths that prove the 33/67 theory and YOUR view of the reality of the game. If you do not switch, you will only win 33% of the time, that is the reality. If you always switch, you will win 67% of the time, that is the other reality of the game.

You went further:

“If it were a deck of 52 cards as you suggest, if you pick one you have a 1/52 chance of having the Ace of Spades. Turn over all other cards but yours and one other – what are your odds of HAVING the Ace of Spades? 50%.”

You honestly believe that turning over 50 cards changes your chances of having picked an Ace from a deck of 52 cards from 1/52 to 1/2 or 50%?

And that if you play that game 100 times you are likely to have picked the Ace of Spades 50 times when statistically you would be lucky to pick it twice?

I see what you mean by a “conflict between the statistical calculation and the reality of the situation.” The maths says one thing but Ableger’s distorted reality says another. My advice: go with the maths and try and adjust your view of reality.

Apropos my comments re the 33/67 deniers aka 50/50 proponents getting personal, you wrote:

“You might want to brush up on your reading comprehension before you criticize someone else.”

even though the person had not misapprehended or misrepresented your position (albeit you claimed a distinction without difference between “picking” and “having” the car).

One of your co-deniers, Steve, also wrote:

“I’m amazed at the number of idiots who think the odds are increased if the contestant swops. Even apparently so called maths experts conned by computor programs.”

Poor comprehension, poor spelling and rude. The Holy Trinity!

Like you and another denier Phil, who were equally affronted and “amazed” at everyone’s else’s stupidity, Steve has fallen strangely silent, as if the penny has dropped.

@Jonathan: you’re correct. Another way to write the 52-cards version:

The dealer shuffles the deck, and asks you to pick a card at random, without looking at it. You place this card face down on the table. Next, the dealer looks through all 51 remaining cards, and selects one. Then, he places that one card face down next to yours. He throws the remaining 50 unused cards away. The dealer then says “I guarantee you that one of these two cards is Ace Of Spades, pick one”.

Now, should you bet that the card you picked originally is the Ace Of Spades or the card the dealer picked is the Ace Of Spades, which is more likely? Ableger would say that it’s “50/50″.

It’s actually quite simple. 1 in 52 times the player correctly picks the Ace Of Spades at the start. In that case the dealer selects any “junk” card as his offer. The other 51 out of 52 times, you didn’t originally pick the Ace Of Spades, so 51 out of 52 times the dealer offers you the Ace Of Spades as your “switch”.

In the above case, “switching” only loses if you originally picked the Ace Of Spades out of an unmarked deck (1/52 chance).

In the same way with the Monty Hall Problem, switching only loses if you originally picked the car out of 3 unmarked doors at random – (1/3 chance).

@ Jason: Random does not mean “equally distributed amongst all choices”. If you played 52 times, randomness won’t guarantee you pick the Ace of Spades even once; nor does it guarantee you will fail to pick it more than 5 or 8 or 10 times..

@Jonathan: Monty deliberately removes a losing outcome (door C). Your original choice has no bearing on what Monty chooses, except that he can’t choose your door.

Same thing with the cards. Your odds of picking the Ace of spades was 1/52 initially. The possibility that you have the Ace of spades is only 1/52.

The dealer turns over a 2 of diamonds.

He then puts down one card face down.

You put your card face down. next to it.

Now you must “pick again”. Due to randomness, there is no advantage to taking your card vs the dealer’s card — unless you think you are lucky/unlucky.

@ZenasDad: your analysis of the 52 card example is incorrect. You pick a card at random and, as you say, you have a 1/52 (1.923%) of having selected the Ace of Spades.

he remaining 51 cards have a 51/52 chance of containing the Ace of Spades (obviously). The dealer turns over the 2 of Diamonds (I assume he deliberately avoids the Ace of Spades if he has it). The remaining 50 cards still have a 51/52 chance of containing the Ace of Spades. He picks a card from the 50 cards and puts it face down – that card has a 51/52 * 1/50 (or a 1.9615%) of being the Ace of Spades

Therefore it is advantageous to take the dealer’s card.

@ZenasDad:

> Now you must “pick again”. Due to randomness, there is no advantage to taking your > > card vs the dealer’s card — unless you think you are lucky/unlucky.

That would only be true if you don’t know which card was yours originally and were picking again completely at random.

If, however, you know which was your card originally and which one the dealer placed there, then it’s advantageous to choose the dealer’s card. That’s because you picked *at random* from 52 cards, so your card is a 1/52 chance of being the Ace of Spades. The dealer however did not choose his card at random – he chose the Ace of Spades if it was available.

Now when the two cards are placed side by side, in one out of 52 games, the card you chose will be Ace of Spaces, and in the remaining games, 51 times out of 52, you will have failed to choose the Ace of Spades initially, thus the dealer MUST have it.

Think of it this way: you pick 1 card from the deck first, and place it in front of you, unseen. That card has a 1/52 chance of being any particular card, e.g. Ace of Spades.

The dealer then manually looks through the remaining cards and picks a card. You know that he always picks Ace of Spades if possible (i.e. if you don’t have it).

The dealer then places his card face down in front of him. You can then swap cards if you want, or just reveal both cards as they are.

Two observations:

1) who is more likely to have the Ace of Spades at this point? Don’t tell me it’s 50/50.

2) if you “do nothing” how can your odds of having Ace of Spades magically jump from 1/52 to 1/2 just because the dealer threw away unused cards AFTER your choice?

@ZenasDad. Further to my previous comment: I’m assuming that after the dealer has turned over the 2 of diamonds (avoiding the Ace of Spades), he picks a card at random from the remaining 50 cards.

Of course if the card he puts down is deliberately chosen, then it’s a 51/52 chance of being the Ace of Spades (whether he turns over another card beforehand or not). If you didn’t realise this and chose between your card and the dealer’s card at random, then you’d have a 50% chance of ending up with the Ace of Spades.

* @ZenasDad: when we say e.g. “win 1 game in 3″ it’s short-hard for “in the long run, for every 3 games you play, the average number of wins will be 1 per 3″. But we shorted than to “1 in 3″ because repeating disclaimers every time we are talking about probabilities just obfuscates the points people are discussing under piles of jargon. Nobody is claiming that if you play 3 games you will always win 1 of those 3. That’s a red herring / strawman.

@ZenasDad: Let’s go back to the classic Monty Hall problem.

Your initial chances of picking the car are 1/3rd. Agreed? This includes scenarios where you are not offered a “swap” or you refuse the swap. In short, if you don’t take any actions after your initial choice, your odds do not change:

Scenario 1: No swap allowed. If you’re never offered the swap, your first pick has a 1/3 chance of winning.

Scenario 2: Swap offered, but rejected or ignored. If you don’t swap when offered, then your chance of winning is exactly the same as Scenario 1. No matter what order Monty opens doors, that doesn’t change the chance that you picked correctly at first.

Scenario 3: Monty reveals a goat after you choose your door but says “I’m not going to offer you a swap AT ALL. I will reveal MY remaining door, and if it’s the car, then *I* am driving it home!”

Now, in Scenario 3, since no swap is offered, you still have the 1/3 chance of winning the car that you always do, because you were stuck with your first choice. No matter what Monty says or does after you choose the car doesn’t change the chance that you picked correctly the first time.

But what is also clear is that if you lose, Monty wins the car for himself. Since you lose 2/3rds of the time, Monty wins the car for himself 2/3rds of the time. Hence if you were allowed to “swap” your prizes, then you would get the car as often as Monty does, 2/3rds of the time.

@Jason: Absolutely correct. Nobody is claiming that if you play 3 games you will always win 1 of those 3. Because the winning location is “random”.

PalmerEldritch states: If you didn’t know anything about the selection process then you’d have a 50% chance of ending up with the Ace of Spades.

@PalmerEldritch: How can that be true? If the chance were 50/50 for a new player with your same card choices… The chance *must* be 50/50 for you as well.

Whether or not you “know” how the two choices were derived, there are in fact two choices.

“Stay or swap” is the same thing as “Choose between these 2 options”.

Consider a random set of numbers:

A A C B A B A A — imperfect distribution but still random. In this random sequence, you will win five times out of eight by “staying” with A. Random equal distribution.

At best you can only make an approximation. Accounting for randomness R, the initial probability of getting the winning card/door during any given game should be:

R * 1 / # choices

I admit that R is imaginary/incalculable value; I think it must lie somewhere between 0 and the # choices. Consider the “lucky” person who picks correctly ten times in a row. Unlikely but possible.

@Jason: Correct me if I’m wrong. When you say ““in the long run, for every 3 games you play, the average number of wins will be 1 per 3″ I think you are making a flawed assumption about randomness vs ‘equal distribution’.

BTW: I’m not trying to be obstinate — I find this discussion fascinating.

@Jason: I agree that your initial choice is r * 1 / 3 precicely because there are 3 choices.

Do you agree with PalmerEldritch that a new player, given the 2 remaining choices and no knowledge of the selection process, would have a 50/50 probability? Why?

@ZenasDad: If the chance were 50/50 for a new player with your same card choices… The chance *must* be 50/50 for you as well.

faulty logic. here’s how it works for the 3 doors:

player 1 picks door 1. 1/3rd chance of being correct. there is a 2/3 chance that the car is actually behind EITHER door 2 or door 3. monty then eliminates the least-valuable door from 2 and 3. call that remaining door “monty’s door”. “Monty’s door” has the car in every game where the player’s door does not – 2/3 chances. That’s because player 1 just has “door 1″ whereas monty has “whichever door from 2 or 3 holds the car” – the statement “door 2 or door 3 hold the car” is true in 2/3 games.

Ok, now consider a “player 2″ who just came in. he chooses at random as he doesn’t know which door player 1 picked, and which door Monty left. let’s assume the 1/3 vs 2/3 chance is correct:

1/3 chance door “A” is the winner

2/3 chance door “B” is the winner.

if player 2 picks door “A” he wins 1/3 * 1/2 = 1/6 games

if player 2 picks door “B” he wins 2/3 * 1/2 = 2/6 games

1/6 + 2/6 = 1/2 chance of winning.

Let’s do it for a 1/10 to 9/10 chance for the two doors:

1/10 chance door “A” is the winner

9/10 chance door “B” is the winner.

if player 2 picks door “A” he wins 1/10 * 1/2 = 1/20 games

if player 2 picks door “B” he wins 9/10 * 1/2 = 9/20 games

1/20 + 9/20 = 1/2 chance of winning.

So, you see the relative chance of the car being behind either door doesn’t benefit someone who doesn’t know which door is more likely. But if you DO know the odds per door, you can win that often.

in the classic Monty Hall problem the player knows which door he chose first, thus he knows that that specific door had a 1/3 chance since he picked it at random from a group of 3. Since Monty always leaves a winning door unopened, the door he didn’t open must have a 1 – 1/3 = 2/3 chance of being the correct one, since probabilities always add up to 1.

But, someone who doesn’t know which is monty’s door vs the player’s door has no better than blind luck in choosing.

here’s how it works with cards:

player 1 draws a card, and lays it face down on the table. he’s not allowed to swap in this version. the dealer then looks through the deck and CHOOSES a card and lays it face down on the table. the “winner” is the one with Ace of Spades.

player 1 correctly chose the Ace of Spades 1/52 times. The dealer thus gets the Ace of Spades in any game where the player failed. Simple logic.

Player 1 wins 1/52 times.

Dealer wins 51/52 times.

Consider a Player 2 that saw the draw and know which card is which. He can win as often as the dealer does by betting on the dealer’s card.

Also consider a Player 3 that didn’t see the draw – he has no information about which card was completely random, and which card was selected. he chooses each card 50% of the time:

Chooses Card 1: wins 1/52 * 1/2 = 1/104 times

Chooses Card 2: wins 51/52 * 1/2 = 51/104 times

1/104 + 51/104 = 52/104 = 1/2 of all times.

See, the skewed chances of either card being right do not chance the nature of random choice, but if you know the relative odds on each card, you can win proportional to those odds.

Gee, Jason, you take the prize, if not the cake, for the most complicated attempts at explaining the odds of the random player scenario. The fact is, it is typically the simple explanations that are the correct ones and those positing incorrect hypotheses rely on complex faulty formulae to support reasoning that itself is wrong.

Via your own formula, for instance, you claim the random person who comes into the game without knowing which door the contestant first chose has a 1/6 chance of winning if they pick the contestant’s door and a 2/6 chance of winning if they pick “Monty’s” door.

That means if the latecomer is allowed to pick both doors his chances of winning will be just 3/6 or 50%.

Now, call me an optimist, but if I picked both doors I would be thinking my chances were a lot better than 50%, I’d be thinking 100%, in round numbers.

The formula Jason proposed was:

“if player 2 picks door “A” he wins 1/3 * 1/2 = 1/6 games

if player 2 picks door “B” he wins 2/3 * 1/2 = 2/6 games”,

clearly is flawed.

The reason the two doors are a 50/50 proposition is that it is a random pick made from two options by someone with no information about which door is more likely to hold the car. Put simply, Jason, if the player with no knowledge plays 100 times he will, on average, win 50 times. Just like you would win, on average, 50 times, if you picked heads or tails in a 100 game coin toss.

It is the same as if sporting contests were set up between the top players and complete novices. A sports fan will pick the winner nearly every time whereas a guy from Mars who has no knowledge of sport will pick the winner, on average, 50 times.

So, Jason, you have two problems:

1. Your hypothesis about “Ignorant Player 2′s” chances, which clearly is incorrect; and

2. The maths you use which is faulty and does not even support your incorrect hypothesis as it provides that the random picker has half the chances (1/6 and 2/6) of picking the doors as the contestant (1/3 and 2/3) and only has a 50% chance of picking the car even if he picks both doors.

I think not.

@Jonothan. Do you understand probability trees? That’s where the 1/6 etc come from:

Start by assuming option A is correct 1/3 of the time and option B is correct 2/3 of the time.

Then, the person picks randomly, i.e. they pick option A 1/2 times and option B 1/2 times.

The chance of winning if you picked A = 1/3 * 1/2 = 1/6.

The chance of losing if you picked A = 2/3 * 1/2 = 2/6.

The chance of winning if you picked B = 2/3 * 1/2 = 2/6.

The chance of losing if you picked B = 1/3 * 1/2 = 2/6.

Now, these 4 possible situations add up to 1 (unity), with half the outcomes being wins and half the outcomes being losses. This is like year 8 mathematics.

^ sorry, typo: it should read “The chance of losing if you picked B = 1/3 * 1/2 = 1/6″

>2. The maths you use which is faulty and does not even support your incorrect >hypothesis as it provides that the random picker has half the chances (1/6 and >2/6) of picking the doors as the contestant (1/3 and 2/3) and only has a 50% >chance of picking the car even if he picks both doors.

Nope, you fail at understanding probability trees.

1/6 = door A picked, Winner (1/3 * 1/2)

2/6 = door A picked, Loser (2/3 * 1/2)

2/6 = door B picked, Winner (2/3 * 1/2)

1/6 = door B picked, Loser (1/3 * 1/2)

Total probabilities add up to 1. branches with “door A picked” or “door A picked” are equal at 1/2 each. Branches where Door A had the car add up to 1/3, Branches where Door A had the car add up to 2/3.

Therefore all the maths adds up and your criticism is unfounded.

@Jonathan, reading your previous posts it seems you’re not a Monty Hall denier … neither am I, but you need to look at the context of my analysis. It was in response to Zenasdad, who wrote:

“@PalmerEldritch: How can that be true? If the chance were 50/50 for a new player with your same card choices… The chance *must* be 50/50 for you as well.”

My posts were in response to that conversation. I was pointing out that just because a “random” player has no better than 50/50 chance doesn’t mean that ALL players have no better than 50/50 chance. Probability trees show how that works.

Jason, I do understand probability trees and, as someone else will chime in shortly and explain to you, they are for calculating the probability of outcomes dependent upon multiple sequential decisions (ie more than 1) not single decisions in once-off scenarios like picking from one of two doors.

A decision tree has no place in explaining a situation where there is only a single, random decision to pick either door A or door B.

There is an equal probability of an outsider picking the door with the car from the two remaining doors each time they guess. Without even doing the math, you know, instinctively, that the random guesser should “win” 50% of time.

You should also know that your decision tree, which provides that the outsider has chances of 1/6 and 2/6 of winning for each of the two doors instead of the correct 1/2 for each (totaling 1 not 1/2), has no place in understanding said probability.

So yes, I do understand decision trees, what they are, what that are for and how to use them. I should use them more often, eg chances you will read this post:

100% (no change in odds of ensuing possibilities there);

Chance of you realising you are wrong: 50%

Chance you will realise you were wrong: 10%

Chance you will respond if you have not figure it out: 100%

Chances your next post will be to refute everything i have said: 50%

Chances your next post will be to tell me you were mistaken and you now get it: 5%.

So that’s a good example of how you use a decision tree.

The question is, if you understand decision trees, why are you using one for a problem involving a single decision when they are designed solely for use in situations involving multiple sequential decisions?

You might have realised the tree was not a useful tool for this problem after finding, using the tree, that the probabilities of all possible outcomes added up to just a 1/2 when you know they must add up to 1?

As for your second set of even more flawed calculations that now show that it does add up to one:

“The chance of winning if you picked A = 1/3 * 1/2 = 1/6.

The chance of losing if you picked A = 2/3 * 1/2 = 2/6.

The chance of winning if you picked B = 2/3 * 1/2 = 2/6.

The chance of losing if you picked B = 1/3 * 1/2 = 2/6.

Now, these 4 possible situations add up to 1 (unity), with half the outcomes being wins and half the outcomes being losses. This is like year 8 mathematics.”

No, this is like Year 1 MBA done by someone in Year 8.

It only now adds up to one because you created 4 possible outcomes when there are only two! The facts is, in your own words and by your own calculation, above, the chance of winning if you pick A and B is 1/6 + 2/6 = 50% when the chance of winning if you pick A + B is 100%.

Why do you keep multiplying the 1/3 and 2/3 odds from Monty Hall by 50% when MH odds have nothing to do with a random selection from, to the outsider, two doors equally likely to provide the car?

Does the fact that your combined odds of winning for BOTH doors add up to 50% not tell you, instinctively, that you’re approaching this incorrectly?

Definitely the most imaginative yet completely misguided post on this forum so far. But enough with the decision tree for a single decision event lest anyone think you’re more than just misguided.

Here’s hoping against the odds for that 5% outside chance mea culpa instead of the 50% chance of another flawed rebuttal.

PS – sorry, that should have read: Chance you will respond if you realise you were wrong: 10%, not chance you will realise you were wrong.

Just guessing of course.

I don’t think your really getting what I was talking about, which was in response to Zenasdad.

He was saying that because a random chooser only has 50/50 chance of successfully picking the right door after the initial selection then that disproves the 33/66% chance of winning for the person who actually knows which door has which odds.

You need to follow the context of the discussion to know why i multiplied everything by the 1/2 factor.

>It only now adds up to one because you created 4 possible outcomes when there are >only two! The facts is, in your own words and by your own calculation, above, the >chance of winning if you pick A and B is 1/6 + 2/6 = 50% when the chance of >winning if you pick A + B is 100%.

What does “pick A and B” mean and what does “pick A + B” mean here? That is as clear as mud. Remember the context of the discussion that lead here was Zenasdad claiming a “dumb” player only getting 50/50 odds disproves that the doors are really skewed (i.e. a “smart” player can’t get more than 50/50).

This gives two independent variables, where the car is, and what choice the “dumb” player makes. A probability tree proves that:

/ 1/2 (“dumb” player picks door A) – 1/6 (Win)

1/3 (door A=car)

/ \ 1/2 (“dumb” player picks door B) – 1/6 (Lose)

O

\ / 1/2 (“dumb” player picks door A) – 2/6 (Lose)

2/3 (door B=car)

\ 1/2 (“dumb” player picks door B) – 2/6 (Win)

Note, as said this only refers to a “dumb” player that doesn’t know the odds, not a normal Monty Hall player. what mathematical error have I made here exactly?

Oops it doesn’t like spaces for formatting:

___________________ / 1/2 (“dumb” player picks door A) – 1/6 (Win)

___1/3 (door A=car)

__/________________ \ 1/2 (“dumb” player picks door B) – 1/6 (Lose)

O

__\ _______________ / 1/2 (“dumb” player picks door A) – 2/6 (Lose)

___2/3 (door B=car)

___________________\ 1/2 (“dumb” player picks door B) – 2/6 (Win)

Obviously, a player who knows which is door A and which is Door B gets better odds, but Zenasdad disputed that. Mapping it out as two independent decisions shows why that is flawed. Someone who can’t tell the doors apart gets 50/50 odds of winning no matter what the actual odds-per-door are.

>Why do you keep multiplying the 1/3 and 2/3 odds from Monty Hall by 50% when >MH odds have nothing to do with a random selection from, to the outsider, two >doors equally likely to provide the car?

because the doors aren’t equally likely to have the car, outsider or not. The doors were selected via the MH system, thus they carry the MH odds (for a smart player). Zenasdad’s contention was that the 50/50 chance for a random guesser proves that the doors themselves were 50/50 for the smart player. Doing the maths proves that skewed doors still give 50/50 odds if one is chosen at random.

@Jason: No, it was PalmerEldrich’s idea that a “dumb” player would have 50/50. I *asked* if you agree with that supposition.

You are playing two separate games; The first game only “sets up” the second, by eliminating a losing possibility. Therefore it is correct to think:

‘I probably chose a loser, so I should choose a different card’

and equally correct to think, “Now I feel more sure about choosing the same card again”

@Jason: You are stating as a fact, things that are suppositions:

-”Obviously, a player who knows which is door A and which is Door B gets better odds, but Zenasdad disputed that. “

Untrue, unless you subscribe to the idea that randomness == equal distribution of outcomes

-”Zenasdad’s contention was that the 50/50 chance for a random guesser proves that the doors themselves were 50/50 for the smart player.”

My contention is not of a “random guesser,” but a “dumb” player — one who doesn’t know how the two choices were derived.

Your second choice (you are making two) is in no way dependent upon your first choice.

>You are playing two separate games; The first game only “sets up” the second, by

>eliminating a losing possibility. Therefore it is correct to think:

>‘I probably chose a loser, so I should choose a different card’

>and equally correct to think, “Now I feel more sure about choosing the same card >again”

Nope, you may “feel” more certain, but it’s delusional.

Say there’s no swap allowed, you pick a card, and you win if it’s Ace of Spades. 1/52 chance, right?

Then, add to that the dealer offering you the Monty-Hall type swap. What that’s really saying, is that if you picked the Ace, he offers you junk, and if you picked junk, he offers you the Ace.

But … the dealer just “offering” another card doesn’t change what you picked in the first place. Think about this: you have a 1/52 chance of your card being ANY card. It breaks down like this:

Your Card ______________ Dealer’s Card

2 of Diamonds _________ Ace of Spades

3 of Diamonds _________ Ace of Spades

4 of Diamonds _________ Ace of Spades

5 of Diamonds _________ Ace of Spades

6 of Diamonds _________ Ace of Spades

[...]

Queen of Spades ______ Ace of Spades

King of Spades ________ Ace of Spades

Ace of Spades _________ (some other card)

Each of these 52 outcomes is equally likely.

Now, since you’re just as likely to have chosen any specific card, when you flip your cards over, you will only have the Ace of Spades 1/52 games, and the dealer will have the Ace of Spades 51/52 times.

@Jason I meant to write:

>> Therefore it is correct to think:

>> ‘I probably chose a loser, so I should choose a different DOOR

>> and equally correct to think, “I feel more sure about choosing the same DOOR again”

And no, it isn’t delusional to feel more confident after a goat is revealed.. here’s why.

GAME 1. Before you pick a door, math, knowledge, truth is:

A=1/3 –Monty knows IF the car is behind A; we don’t ; Neither changes where the car IS.

B=1/3 –Monty knows IF the car is behind B; we don’t ; Neither changes where the car IS.

C=1/3 –Monty knows the car isn’t behind C; we don’t ; Neither changes where the car IS.

A + B + C = 1 (Because ONE of them does have the winning door).

B + C = 2/3 –> Sure, I agree; also

A + B = 2/3

A + C = 2/3

We choose ‘A’. That doesn’t change where the car IS, it ONLY means Monty cannot reveal door A.

Monty reveals/eliminates C (a goat). That doesn’t change where the car IS.

That is where differ. According to you, my chance stays the same:

A = 1/3

The article discussed “collapsing” such that Monty’s probability B {magically} goes to 2/3. That is completely arbitrary –see the math/knowledge/truth table. The game state (or ‘world’) HAS changed.

A=1/n –Monty knows IF the car is behind A; we don’t ; Neither changes where the car IS.

B=1/n –Monty knows IF the car is behind B; we don’t ; Neither changes where the car IS.

GAME 2 (C is revealed to be a goat). You are choosing between two. We all agree:

A + B MUST be 1

3 is an inappropriate denominator. There are only 2 choices!

If A remains 1/3, then B must remain 1/3 — why? The fact that C contained a goat, DOES NOT affect where the car is.

That he told/showed us a goat DOES NOT affect where the car is.

Neither Monty’s smile, good nature, *knowledge* or beliefs affect where the car is….

The car IS behind either A or B. My chance of choosing a goat in GAME 2 is 1/2.

Nope, each door starts with the same odds of 1/3.

say, we always call the door that the player picks as “door A” for convenience. There are three possible outcomes, all equally likely:

1) A Car, B Goat, C Goat

2) A Goat, B Car, C Goat

3) A Goat, B Goat, C Car

Now, the “Switch” door wins 2/3 of the time, because it’s EITHER doors B or C, it has the same probability as the statement “EITHER door B or C holds the car”, which is true 2/3 of the time, even though each door has a 1/3 chance.

In situation #1, 1/3rd of the time, switching loses.

In situation #2, Monty eliminates the “C” door goat, switching wins

In situation #3, Monty eliminates the “B” door goat, switching wins

It’s a 2/3 to win on switching, because 1/3 you’re switching to winning door B, 1/3 your switching to winning door C, and 1/3 door A actually had the car all along.

@Zenas.

Your chance of choosing the correct door from the 2 remaining is only 50/50 if you pick one at random, but if you think the problem through as follows you’ll realise why it’s better to switch doors:

Say you picked door1 and Monty opened door3 knowing it contained a goat. Now the car is obviously behind either door1 or door2, and both were equally likely at the outset.

If the car is behind door2 then the probability Monty opened door3 = 1 (he had no choice)

If the car is behind door1 then the probability Monty opened door3 = 1/2 (as he could equally have chosen to open door2)

Therefore it is twice as likely he opened door3 because he had to than because he chose to, consequently it is twice as likely the car is behind door2 as door1. So you switch to double your chances of winning. (That’s a non-mathematical explanation. A mathematical solution using Bayes Theory produces the same answer)

I was also thinking of another thought-experiment that shows that the “odds per door” can in fact change as different doors are opened:

Say there are 100 doors, and one prize is initially placed at random. Each door has an equal 1% chance of having the prize.

Player #1 is given Doors 1-10

Player #1 is given Doors 11-100

In this game, there is no second choice, no switching.

Imagine that all the doors are opened at once, revealing where the prize is. Clearly, when all doors are opened, we can say it’s a 10% chance than Player #1 has the prize, and 90% chance that player #2 has the prize.

Next, lets ask what happens when the doors are opened one at a time ? This should give the same result as opening them all at once, right? Imagine, the host opens 98 of the doors but he never opens the prize door and he leaves one closed door per player. So, 9 out of 10 of player #1′s doors are opened, and 89 out of 90 of player #2′s doors are opened.

There are now two doors left, one of player 1 and one of player 2. Nobody has switched or made another choice. What are the odds of it being behind either remaining door? 50/50? But wouldn’t that contradict the fact that when all doors are opened, Player#2 should win 90% of the time?

Remember, nobody has done any switching here, so when “all doors are opened” Player#2 should still win 90% of the time. That means the odds of it being behind Player#1′s last door are 10%, and the odds of it being behind Player#2′s last door are 90%

^ correction: Should read: “Player #2 is given Doors 11-100″

“Your chance of choosing the correct door from the 2 remaining is only 50/50 if you pick one at random,”

According to Jason and the rest, this is not true. Could you please explain why picking one of two choices at random gives you 50/50.

If you choose completely at random the odds are always 50/50 of picking the right one. It’s independent of the actual odds. This can be shown mathematically very easily, and in fact I already laid it out. Here we go again. If you can’t process this maths, you don’t belong in this discussion:

Imagine two horses are racing, and Horse A wins 90% of the time, and Horse B wins 10% of the time. If you know this, betting on Horse A wins 90% of the time.

But if you don’t know which horse is better and you flip a coin to lay your bets, you will pick the winning horse 50% of the time. Here is the breakdown of all possible combinations of win/lose and bets:

Horse A wins, and you picked A = 0.9 * 0.5 = 0.45 probability (win)

Horse A wins, but you picked B = 0.9 * 0.5 = 0.45 probability (lose)

Horse B wins, but you picked A = 0.1 * 0.5 = 0.05 probability (win)

Horse B wins, and you picked B = 0.1 * 0.5 = 0.05 probability (lose)

Adding all the probabilities together gives 1.0, all possible outcomes. “Win” outcomes add to 0.5 and “Lose” outcomes add to 0.5. The maths works out the same no matter what the different odds are.

A “Coin flip” always gives the right answer 1/2 the time. The actual odds of the two answers being correct are not related to it. e.g. Door A could be right 100% of the time, and door B 0% of the time, flipping a coin still gives you a 50/50 chance, the same as if the doors were actually 50/50, 33/66, 66/33 or any other combination adding up to 100%.

Sorry for a couple of typos, but there’s no edit button -_-. this is a poor place for such discussions.

The problem is equivalent to choose 1/3(one door has car) and 2/3 (two doors). if you view the switch is equivalent to as long as the car in two of the other doors (totally elimates the Monty) you will win, this should be more apparent to you.

@Jason:

> If you choose completely at random the odds are always 50/50 of picking the right

> one. It’s independent of the actual odds. This can be shown mathematically very

> easily, and in fact I already laid it out. Here we go again. If you can’t process this

> maths, you don’t belong in this discussion:

We don’t need to go again, you are talking about cards, horses, stars, whatever.

I ask you to stick with the 3 door problem.

>> “If you choose completely at random the odds are always 50/50 of picking

>> the right one. It’s independent of the actual odds. “

Let’s just agree that door C is revealed/eliminated.My position, succinctly is:

–”I don’t see the MHP as one game with choose, then stay and switch — I see it as two INDEPENDENT choices.”

–”It doesn’t matter HOW you derive two choices; there are TWO”

–”If a ‘dumb’ person has 50/50 odds then so does a knowledgeable person.” Whether or not you KNOW where the car is does not affect the outcome.

If you pick A it doesn’t magically move the car.

Monty reveals B OR C, it doesn’t magically move the car.

@ZenasDad:

Your statement “Whether or not you KNOW where the car is does not affect the outcome.” is clearly incorrect. If you KNOW the car is behind Door1 (say) then you have a 100% chance of picking the car, if you don’t know where the car is and pick a door at random you only have a 50% chance of picking the car.

In the MHP your choice is not so clear cut, but from your perspective the door you didn’t pick has a 66.7% chance of containing the car and the door you picked to begin with only has a 33.3% chance, so if you switch doors you’re twice as likely to win the car. If you pick a door at random then you have a:

(1/2*33.3%) + (1/2*66.7%) = 50% chance of picking the car.

@PalmerEldritch: I need to slow down (doing surveillance at the time).

Please refer to my comment # 166 – “Math/Knowledge/Truth Table” and refute it. My position is,

–”Whatever Monty knows is irrelevant, because it (does|did) not affect where the car actually is.”

–”Whatever WE know is irrelevant, because it (does|did) not affect where the car

actually is”

–”Choosing and revealing are irrelevant.”

Let us not degenerate into ” The world is FLAT!!! So many of us agree so we must be right!!” That’s Hogwash.

>> In the MHP your choice is not so clear cut, but from your perspective the door

>> you didn’t pick has a 66.7% chance of containing the car and the door you picked

>> to begin with only has a 33.3% chance, so if you switch doors you’re twice as

>> likely to win the car. If you pick a door at random then you have a:

>> (1/2*33.3%) + (1/2*66.7%) = 50% chance of picking the car.

And I don’t agree. Where did the “3″ come from? The number of doors. There is no “switch”. There is no “stay”. There are two non-dependant (individual) choices being made.

>> If you KNOW the car is behind Door1 (say) then you have a 100% chance of picking the car,

Yes, I withdraw that comment. see above.

>> if you don’t know where the car is and pick a door at random you only have a

>> 50% chance of picking the car.

The picking need not be random to have a 50% chance!

I agree that the initial likelyhood is in Monty’s favor (he has 2 to 1 odds!).

No one has yet explained the “collapse” phenomenon satisfactorily. You’re not choosing between A and BC. I am not a denier, I am unconvinced.

Here’s a thought experiment. What facts do we KNOW based monty revealing a goat (C).?

(A is not *that goat*) and

(B is not *that goat*)

If door ‘B’ becomes more likely to have the car (independent of the revealed goat C) why wouldn’t door ‘A’ also become more likely to have the car **by the same amount**. Again, refer to comment #166.

@ZenasDad

Re-read my comment #168 and tell me where you disagree with the analysis of the example given (You pick Door1, Monty opens Door3), because that’s the reasoning that gives the door you didn’t pick a probability of 2/3.

BTW: Thank you all for this discussion. Especially Jonothan Jason and Palmer Eldridch.

How about this:

GAME CAR CHOOSE SHOW SWAP RESULT

1 A A B – WIN*

1 A A B C LOSE

1 A A C – WIN*

1 A A C B LOSE

1 A B C – LOSE

1 A B C A WIN*

1 A C B – LOSE

1 A C B A WIN*

GAME 2, I CHOOSE “A”; WINNER IS “B”

GAME CAR CHOOSE SHOW SWAP RESULT

2 B A C – LOSE

2 B A C B WIN*

2 B B A – WIN*

2 B B A C LOSE

2 B B C A LOSE

2 B B C – WIN*

2 B C A – LOSE

2 B C A A WIN*

GAME CAR CHOOSE SHOW SWAP RESULT

3 C A B – LOSE

3 C A B C WIN*

3 C B A – LOSE

3 C B A C WIN*

3 C C A – WIN*

3 C C A B LOSE

3 C C B A LOSE

3 C C B – WIN*

6 Stay and Win

6 Stay and Lose

6 swap and Win

6 swap and LOSE

Can anyone double-check / verify this data?

It appears to me that there is a 50/50 chance when you iterate through all possibilities.

I will be in the field for 2 days but would love to hear back from you. If requested I’ll post source code.

Your mistake is to give each of the 4 outcomes the same weight.

Say, the player’s choice is called Door A:

Outcome 1: Car is Door A, Monty reveals goat in door B, swapping loses

Outcome 2: Car is Door A, Monty reveals goat in door C, swapping loses

Outcome 3: Car is Door B, Monty reveals goat in door C, swapping wins

Outcome 4: Car is Door C, Monty reveals goat in door B, swapping wins

Now, you can say 2 of 4 the outcomes lead to a loss and 2 of 4 the outcomes lead to a win, but there’s a very obvious error. in 2/4 scenarios the car was behind door A in the first place. So it only works if you assume the car was twice as likely to be behind the exact door that the player chose than the other doors.

In truth, Door A was no more likely to have the car than the other doors, so Outcomes #1 and #2 add up to 1/3, i.e. both have a chance of 1/6. Outcomes #3 and #4 both have a 1/3 chance of occurring.

Thanks.. I am looking for (hoping someone can find) flaws in the data. I am trying to explore all possible outcomes; If we can agree on the data-set that will be a good start. I remain unconvinced by maths tainted with intuition.

If we can agree on a set of all possible outcomes in each scenario/game, every individual outcome must be equally likely.

BTW: I realize you guys know far more math(s) than I. I’m like Alan Cox; Linus Torvalds said “he doesn’t use git because… it’s some kind of defect.”

Formatting of the table was messed up… I used \t tabs in python. The ‘-’ indicates no swap; it should look more like this:

GAME CAR CHOOSE SHOW SWAP RESULT

1 A A B – WIN*

1 A A B C LOSE

(hard to edit on zte-concord phone)

If we plot ALL possible MHP outcomes, each result should be equally likely.

If the conclusion you guys draw from maths (“over the long run”) is true, wouldn’t the possible outcomes will be weighted towards a player “switching”?

In other words, we should NOT see equal totals for staying and switching.

@ZenasDad

You said

“In other words, we should NOT see equal totals for staying and switching.”

And that’s true, we don’t see equal totals.

(Jason has already explained this, but anyway…..) Take the example where the car is behind Door1. The only possible outcomes and the probability of each outcome occurring are:

1) You pick Door1, Monty opens Door2. Switch LOSE. Probability = 1/3*1/2 = 1/6

2) You pick Door1, Monty opens Door3. Switch LOSE. Probability = 1/3*1/2 = 1/6

3) You pick Door2, Monty opens Door3. Switch WIN. Probability = 1/3*1 = 1/3

4) You pick Door3, Monty opens Door2. Switch WIN. Probability = 1/3*1 = 1/3

Probability of switch and WIN = 1/3+1/3 =2/3

Probability of switch and LOSE = 1/6 + 1/6 = 1/3

And since the only alternative to switch is stay, then

Probability of stay and WIN = 1-2/3 = 1/3

Probability of stay and LOSE = 1-1/3 = 2/3

(And the same scenarios hold true if the car is behind Door2 or Door3)

So you can see we do not see equal probabilities for staying and switching.

“If we can agree on a set of all possible outcomes in each scenario/game, every individual outcome must be equally likely.”

There is no mathematical rule that guarantees that. e.g. if you flip two coins, you can say there are 3 outcomes:

#1) 2 heads

#2) 1 head, 1 tail

#3) 2 tails

Outcome #2 happens 50% of the time, the other two happen 25% of the time.

@ZenasDad if you want to look at all possible outcomes in an exhaustive way, we have to look at every step of the game. I’ll use letters for doors

Step#1 – placing the Car -1/3 chance of A, B, C

Step#2 – player chooses a door – 1/3 chance of A, B, C

At this stage, that 3 x 3 = 9 possible outcomes. We can use 2 letter abbreviations for each outcome here, where the first letter is the “true door” and the 2nd letter is the door the player picked. this gives us a set of 9 outcomes. each outcome at this point is 1/9 likely.

AA, AB, AC, BA, BB, BC, CA, CB, CC

Next, we add a third letter, which is the door Monty opens revealing one of the goats. If the door with the car is not the door the player picked, this is guaranteed to be the same door every time on that branch, but if the player has the car, it’s 50/50 which other door Monty opens. I’ll put the pairs in brackets, that shows they’re really part of the same branch:

(AAB/AAC), ABC, ACB, BAC, (BBA/BBC), BCA, CAB, CBA, (CCA/CCB)

Now, we still have the 9 basic outcomes but the ones in brackets have two possible sub-outcomes each, both of those have a total probability of 1/18. The ones in brackets are also the ones where switching loses.

You might be tempted to say “but now there are equal 12 outcomes” but you can apply a reality check on that showing it’s not possible:

Every branch starting with “A” is where the car is behind Door A. The second letter is the player’s initial choice. If you count AAB, AAC, ABC, ACB as equally likely outcomes you notice that the player selected door A 50% of the time when the car was behind door A, and similarly, chose Door B 50% of the time when the car was there, and the same with door C. This is clearly impossible, because it would mean the players choice was dependent on where the car was, which contradicts how the game was set up.

@Jason:

I stated the same thing. Note that flipping 3 coins would produce:

considering first coin:

HHH THH

HHT THT

HTH TTH

HTT TTT

consider middle coin:

HHH HTH –both dupes

HHT HTT –both dupes

THH TTH — both dupes

THT TTT –both dupes

and last coin:

HHH HHT –both dupes

HTH HTT — both dupes

THH THT –both dupes

TTH TTT — both dupes

Those are ALL possible combinations when you flip three coins one time. Dupes must be eliminated because there’s only one coin toss. This yields 8 equally probable outcomes.

..back to MHP

I intended my comment to be read as:

If you plot all possible outcomes:

—If the conclusion you guys draw from maths (“over the long run”) is true:

——a. Wouldn’t the possible outcomes will be weighted towards a player “switching”?

——b. We should NOT see equal totals for staying and switching.

… and that is exactly what modelling the problem shows. What’s the question exactly?

Fire away, if you have source code that says otherwise, I’ll check it out and prove that it contains an error.

@Jason: I didn’t intend it for public consumption but it’s on pastebin:

http://pastebin.com/TKVPKqqH

Thanks,

@Jason I introduced a bug to cleanup formatting. Corrected version here:

http://pastebin.com/upN7duFD

The error causes display to read ‘win=false’ when you did win by staying… No other side effects that I can see.

If you really want to test it, I just wrote this bit of .vbs that does 10000 games and tallies the output. Chuck it in a file called monty.vbs on your desktop then double click it (Windows only sorry), it will spit out a file giving details of every game played, and totals.

Dim fso, f

Set fso = CreateObject(“Scripting.FileSystemObject”)

Set f = fso.OpenTextFile(“monty_wins.txt”, 2, 1)

rounds = 10000

stick_wins = 0

switch_wins = 0

for i = 1 to rounds

car = int(rnd * 3 + 1) ‘ random choice 1-3

player = int(rnd * 3 + 1) ‘ random choice 1-3

elim = get_other_door(car, player)

switch = get_other_door(elim, player)

f.Write “Car = ” & car & “. Player = ” & player & “. Eliminated = ” & elim & ” Switch = ” & switch

if (car = player) then

stick_wins = stick_wins + 1

f.Writeline ” switching loses”

End If

if (car = switch) then

switch_wins = switch_wins + 1

f.Writeline ” switching wins”

End If

next

f.writeline (“stick wins = ” & stick_wins & ” out of ” & rounds & ” games”)

f.writeline (“stick ratio = ” & 100*(stick_wins/rounds ) & “%”)

f.writeline (“switch wins = ” & switch_wins & ” out of ” & rounds & ” games”)

f.writeline (“switch ratio = ” & 100*(switch_wins/rounds ) & “%”)

function get_other_door(door1, door2)

get_other_door = int(rnd * 3 + 1)

do while get_other_door = door1 or get_other_door = door2

get_other_door= int(rnd * 3 + 1)

Loop

end function

@Jason:

Please can you post that to pastebin? This thread removes extra whitespace and tabs.

It looks like you have a function called get_other_door ; I don’t see where its defined.

Later you use a variable named get_other_door

@ZenasDad let me cite your output:

Car: A Choice: A Reveal: B Change: – win= True .

Car: A Choice: A Reveal: B Change: C win= False .

Car: A Choice: A Reveal: C Change: – win= True .

Car: A Choice: A Reveal: C Change: B win= False .

Car: A Choice: B Reveal: C Change: – win= False .

Car: A Choice: B Reveal: C Change: A win= True .

Car: A Choice: C Reveal: B Change: – win= False .

Car: A Choice: C Reveal: B Change: A win= True .

Car: B Choice: A Reveal: C Change: – win= False .

Car: B Choice: A Reveal: C Change: B win= True .

Car: B Choice: B Reveal: A Change: – win= True .

Car: B Choice: B Reveal: A Change: C win= False .

Car: B Choice: B Reveal: C Change: – win= True .

Car: B Choice: B Reveal: C Change: A win= False .

Car: B Choice: C Reveal: A Change: – win= False .

Car: B Choice: C Reveal: A Change: B win= True .

Car: C Choice: A Reveal: B Change: – win= False .

Car: C Choice: A Reveal: B Change: C win= True .

Car: C Choice: B Reveal: A Change: – win= False .

Car: C Choice: B Reveal: A Change: C win= True .

Car: C Choice: C Reveal: A Change: – win= True .

Car: C Choice: C Reveal: A Change: B win= False .

Car: C Choice: C Reveal: B Change: – win= True .

Car: C Choice: C Reveal: B Change: A win= False .

6 stay and win

6 stay and lose

6 switch and win

6 switch and lose

=====================

There is a clear error here. According to your code, The player always chooses the winning door 50% of the time. i.e. if Door A has the car, they originally pick Door A twice as often as either Door B or Door C. But if Door B has the car, they are more likely to have chosen Door B at the start than the other doors.

A correct model will always have the player’s first choice as independent of where the car is.

VBS doesn’t worry about indentation like python does, you can just paste it into a text file name “monty.vbs” and run it.

The last block is the function “get_other_door”. all that does is that you give two door numbers and it spits out a door guaranteed to be different. if you give it two of the same (e.g. 1 and 1) it will spit out either 2 or 3 at random.

VBS uses a variable with the same name of the function to store the return value from the function

@Jason

>> A correct model will always have the player’s first choice as independent

>> of where the car is.

Yes. The output (attempts to show) every possible combinations of stay/switch, monty’s reveal and car locations.

I think it’s useful to examine without inferring anything from the data.

Once (if?) we agree that the table is correct, we can disprove/prove my idea.

The table should be unbiased: If choice=car, monty has two goats he could reveal. You can still stay or switch. That’s why there are 4 outcomes when you first correctly pick the car.

For every possible car location in our 3-door problem,

– there are eight (8) possible outcomes

Since there are 3 doors, there are 24 total possible outcomes.

The table is clearly biased if you’re using it to try and prove something about probabilities. you’ve doubled the probability weights of the choices where choice=car.

Consider this version with coins:

- Flip a coin

- If it’s heads, flip it again

Outcomes:

1) HH

2) HT

3) T

Ok, this is basically using the same logic as your python program, I can show that you only flip “Tails” on the first flip 1/3 of the time. It’s clearly wrong though, because I’ve given too much weight to the outcomes where you flipped heads first. Having to make an additional choice on one of the branches doesn’t make the other branch less likely.

In the same sense, you’ve artificially made the branches where the player picks the same as the car more likely because Monty made an additional choice on those branches. The player actually picks each door an equal amount of the time, but your table claims that the player picks the car door originally 50% of the time, and picks the other two doors 25% of the time each.

Show me a version where the player is equally likely to pick any door regardless of where the car is, and we can talk. It’s bullshit otherwise.

Here is the crux of your error, ZenasDad. You seem to simultaneously hold two contradictory views, yet have avoided confronting the contradiction.

Let me make an uncontroversial statement. The following three possibilities are all equally likely:

Contestant Door Choice, Winning Door

AA

AB

AC

Do you disagree? (If so, please tell me why.)

If we introduce Monty’s reveal, we get this outcome set:

Contestant Door Choice, Winning Door, Monty’s Reveal

AAB

AAC

ABC

ACB

You also claim that these four outcomes are equally likely. This is in direct contradiction to the first outcome set. AAB and AAC are equally likely, but both are only half as likely as ABC or ACB. This error has been repeatedly pointed out by others. By making this claim you build into your assumptions a 50% chance of choosing the correct door to begin with, which obviously biases your final conclusion.

@Janus:

Welcome. I am actually trying to prove myself wrong. (No, really… But the ‘evidence’ -aka- all possible outcomes — does not support the math that the majority seems to try to impose. [I'm trying to be diplomatic]

I can cede my worldview which is “two, separate decisions,” but only for the purpose of brevity.

If you objectively look at *ALL* possible outcomes, which I did enumerate, the outcome is 50/50 to win whether you stay or switch. So the conventional (majority?) application of math(s) fails.

Moreso: Show me math(s) that explain 24 possible [Begin][Choice][Switch] endstates. Therein lies a revelation. So far, none do.

The all_possible_outcomes table I generated (unbiased), shows 50/50. Run it yourself. Remember we’re not calculating the incalculable (RANDOM).. I am calcualting *all* begin/choice/reveal/switch_or_not states.

The complete set does not support the math everyone espouses. Just look at it!

What is your position when you are presented with data that refutes your preconceived/learned ideas?”

@ZenasDad:

You have not answered any of my questions. Please tell me how you reconcile your competing ideas about likelihood.

The data does not refute my (or the majority’s) ideas. You are assuming that all outcomes are equally likely, which is prima facie absurd, something my simple example above demonstrates. It is akin to saying — well, if I buy a lottery ticket, there are two outcomes: win or lose, so I have 50:50 odds. Just because two outcomes exist does not mean they are equally likely, a fact you have yet to come to terms with.

@Janus:

What exactly is wrong with the truth table I generated? There is no unrepresented path through the game. If you accept that, which I think you should… All paths are equally likely.

If your idea (maths) were correct you would see, in the the all_possible_outcome distribution — a bias toward losing without switching.

That is not the case.

You insist on repeating the bold and patently false claim that every possible outcome is equally likely, let’s simplify things and go one question at a time. Are the following outcomes equally likely?

Contestant Door Choice, Winning Door

AA

AB

AC

Not every path has the same probability, that’s the problem ZenasDad. Your output from your program is correct in that it shows 24 combinations of outcomes, but if you look at it logically AT ALL you see it’s flawed to say all 24 outcomes are equally likely. You have not provided any maths that backs that up.

As proof, your “player” is making different choices depending on which door has the car:

- In the 8 “Door A” branches, the player picks door A, 4 times, door B, 2 times and door C, 2 times

- In the 8 “Door B” branches, the player picks door A, 2 times, door B,4 times and door C, 2 times

- In the 8 “Door C” branches, the player picks door A, 2 times, door C, 2 times and door C, 4 times

That’s enough to show that there’s an error in your logic of assuming “all outcomes” implies “equally likely outcomes”

for any position of the car there are exactly 8 outcomes. not 4. if you show all possible outcomes you end up with a 50 50 chance.

no one here can refute the truth table because it is the truth.

we’re having a disconnect because of peoples preconceived notions about what is.

you’re calculating possibilities. I have shown you every shown every possibility.

the result regardless of what you want to believe seems to be 50/50. if you can state your questions have a question I will answer.

I have asked a simple question, but will post it again:

Are the following outcomes equally likely?

Contestant Door Choice, Winning Door

AA

AB

AC

Question: Why is the player more likely to pick the car door at the start than the other doors?

e.g. you have 8 outcomes where the car was behind Door A, and 4/8 of those outcomes have the player choosing Door A at random, but only 2/8 where they pick door B and 2/8 where they pick door C.

This would seem to contradict the idea of the player choosing from 3 doors at random.

@Janus

@Jason

in my earlier post I said that randomness does not mean ” with 99 iterations of the game, 33 will fall into each door.

this was summarily dismissed by most here.

when we say probabilities, are we not discussing possible outcomes?

@Janus

if every possibility is not equally as likely, how do you explain your idea that my initial likelihood is one of three.

Plus, another contradiction is the chance of the car being behind each door changes based on what the player chooses, e.g. of the 8 branches where the player chose “A” he made the right call 50% of the time, but it’s actually behind door B 50% of the time when you picked “B” and door C 50% of the time when you picked “C”.

you have 3 equally likely things, and each of them is 50% likely, plus the car seems to move based on what the player guesses.

A clear failure of probabilities. The probabilities of all possible outcomes need to add up to 100%

@ZenasDad

I agree, all three (AA, AB, AC) are equally likely.

Now, introduce Monty’s reveal (indicated in parentheses):

AA maps to:

AA(B)

AA(C)

AB maps to:

AB(C)

AC maps to:

AC(B)

Now, if you claim that all four of these outcomes are also equally likely (1/4), you contradict your initial claim. Why? You said P(AA) = 1/3. And yet, if all four detailed now are equally likely, then P(AA(B) or AA(C)) = 1/2 = P(AA). Yet you just agreed P(AA) = 1/3. This is why you are wrong.

AA maps to:

AA(B) — P=1/6

AA(C) — P=1/6

AB maps to:

AB(C) — P=1/3

AC maps to:

AC(B) — P=1/3

These are the correct probabilities, not an equal likelihood of 1/4 each.

> if every possibility is not equally as likely, how do you explain your idea that my

> initial likelihood is one of three.

The car is equally likely to be behind any door:

A (1/3)

B (1/3)

C (1/3)

The player chooses a door at random:

AA (1/3 * 1/3 = 1/9)

AB (1/3 * 1/3 = 1/9)

AC (1/3 * 1/3 = 1/9)

BA (1/3 * 1/3 = 1/9)

BB (1/3 * 1/3 = 1/9)

BC (1/3 * 1/3 = 1/9)

CA (1/3 * 1/3 = 1/9)

CB (1/3 * 1/3 = 1/9)

CC (1/3 * 1/3 = 1/9)

Ok, at this point we have 9 outcomes, all outcomes are equally likely. if Monty makes a 50/50 choice on one of those outcomes, but not others, that outcome splits into two equally-likely outcomes each with 1/9 * 1/2 = 1/18 probability:

AAB (1/3 * 1/3 * 1/2 = 1/18)

AAC (1/3 * 1/3 * 1/2 = 1/18)

ABC (1/3 * 1/3 = 1/9)

ACB (1/3 * 1/3 = 1/9)

BAC (1/3 * 1/3 = 1/9)

BBA (1/3 * 1/3 * 1/2 = 1/18)

BBC (1/3 * 1/3 * 1/2 = 1/18)

BCA (1/3 * 1/3 = 1/9)

CAB (1/3 * 1/3 = 1/9)

CBA (1/3 * 1/3 = 1/9)

CCA (1/3 * 1/3 * 1/2 = 1/18)

CCB (1/3 * 1/3 * 1/2 = 1/18)

I think this is the first time I’ve been happy that we couldn’t locate a fugitive and had to remain in the office… I need some time to come up with a succinct answer. hopefully you guys

are n’t getting in trouble at work.

I think my table could be flawed – hence request for critique. if you read the source code you will see the door that the player chooses is never depends upon where the car is.

@Janus: if it is absurd to think that all possibilities are equally likely, it’s also absurd to think that the first choice has 1/3 probability.

> I think my table could be flawed – hence request for critique.

Your table isn’t flawed, it lists all the things that could happen. What is flawed is interpreting it as giving information about probabilities, when it doesn’t actually attempt to do so.

> if you read the source code you will see the door that the player chooses is never

> depends upon where the car is.

That’s just wrong, and ignoring the problems with your programs output: The outcomes where “Car= Door A” has the player picking A 50%, B 25%, C 25%, but if “Car = Door B” the player chooses A 25%, B 50%, and C 25%, and if “Car = Door C” the player chooses A 25%, B 25%, and C 50%. The chances of the player choosing the different things in your table are not equal, they are dependent on the location of the car, this biases the data.

Also, if you add up all the times the player chose “Door A” for example (8 times), it turns out to be right 4/8 of those times. The same for Door B and Door C. So, Door A is right 50% of the time when you chose it, but so is Door B and so is Door C, which clearly contradicts common sense when picking between 3 doors, they can’t all possibly have a 50% chance of being right. The chances need to add up to 100%.

>@Janus: if it is absurd to think that all possibilities are equally likely, it’s also

>absurd to think that the first choice has 1/3 probability.

that logic is the absurd thing. One thing being unequal doesn’t disprove another thing being equal and vice-versa. You’re just saying it’s absurd without any rhyme or reason.

This is an equivalent setup:

1) Randomly choose a number from 1-3

2) if you chose number 1 – flip a coin.

Just like the Doors example, there are 4 basic outcomes:

#1) 1 / Heads

#2) 1 / Tails

#3) 2

#4) 3

Mathematically, this is identical to the doors example. If you say each is 25% likely, then that contradicts the part where you chose randomly from 3 numbers. Multiplying the probabilities together in the branches solves the contradiction, the first two branches have a chance of 1/3 * 1/2 = 1/6 each, and the other two have a 1/3 chance.

@ZenasDad

Did you even read my post #184? If so where do you disagree with the analysis that showed that staying and switching are not equally likely outcomes?

@PalmerEldrich:

Your comment #186 is missing possibilities. I have added them below.

Take the example where the car is behind Door1. The only possible outcomes and the probability of each outcome occurring are:

1) You pick Door1, Monty opens Door2. Switch LOSE. Probability = 1/3*1/2 = 1/6

+) You pick Door1, Monty opens Door2. STAY WIN.

2) You pick Door1, Monty opens Door3. Switch LOSE. Probability = 1/3*1/2 = 1/6

+) You pick Door1, Monty opens Door3. STAY WIN.

3) You pick Door2, Monty opens Door3. Switch WIN. Probability = 1/3*1 = 1/3

+) You pick Door2, Monty opens Door3. Stay LOSE.

4) You pick Door3, Monty opens Door2. Switch WIN. Probability = 1/3*1 = 1/3

+) You pick Door3, Monty opens Door2. Stay LOSE.

When you consider 1 & 2 together it appears I am favoring the player when his first choice IS the car. I am not. This is a consequence of two game rules:

– Monty never reveals the car

– Monty cannot reveal your choice.

choice1 -> monty 2 -> stay

choice1 -> monty 2 -> switch

choice1 -> monty 3 -> stay

choice1 -> monty 3 -> switch

Really simple. IF you choose the car, monty has goats # 2 and #3. IF you don’t choose the car, monty has only one goat.

@ZenasDad

My comment was missing no possibilities if you’d read it all the way through as I’d included the stay option as simply the opposite of the switch option and calculated the probabilities accordingly.

Anyway, add up the probabilities in YOUR previous comment:

Switch WIN – probability = 1/3+1/3 = 2/3

Switch LOSE – probability = 1/6+1/6 = 1/3

Stay WIN – probability = 1/6 + 1/6 = 1/3

Stay LOSE – probability = 1/3 + 1/3 = 2/3

Is that clearer?

this is a longthread but I’m glad i stumbled on it. Since 2009??? wow.

it took me 2 days for me to read and think. One thing for shure, jason and palmer have a s*load of patience. They demonstrate intelligence. ZenasDad might is wrong, but he did try and he put out every possibility. Total of all possibility he showed are 50/50, 50/50, 50/50. I think firstchoice is 1/3 vs 2/3. Palmer say zenasdad matrix show correctly all possibility.

Sofar no maths have “8″ or “24″ in them. So i see why zenasdad thnks your math is wrong. Someone can explain why he is wrong to show all possibility. There no game he did not show.

—excuse my english

Great to see such violent agreement about whatever you’re each doing.

Interesting to note that in the simulation atop this page, computerised Monty always chooses the left most door when he has the option of doing so (ie if you have chosen the door with the car car or if the car is behind the door to the right).

This means that if computerised Monty opens the door to the right of the other door you did not choose, it is because the other door conceals the car, and so in this situation switching wins 100% of the time.

Of course, if you switch every time, you should still win 67% of the time.

Perhaps the simulator needs some extra code to randomise Monty’s selection of the door to open when the player has chosen the door concealing the car.

It is mistake to say zensdad is wrong because of maths we are taught. He say there are two choices, after first choice, every tim, we can stay or switch. zenadad right when he say monty can not open first choice door. zenadad also right monty do not open car door.

No mistake in his program http://pastebin.com/upN7duFD

only mistake is what we think about possible results.

Interesting to note that in the simulation atop this page, computerised Monty always chooses the left most door when he has the option of doing so (ie if you have chosen the door with the car car or if the car is behind the door to the right).

@Jonathan:

I think you’re erroneously inferring from the sort-order of all possible outcomes. If you modify line 1 of my code to:

DOORS=['C','B','A']

The totals are the same…

Computerized Monty (in my code) follows two rules:

-I cannot reveal player’s door

-I cannot reveal winning (car) door

@Jason:

>> Show me a version where the player is equally likely to pick any door

>> regardless of where the car is, and we can talk. It’s bullshit otherwise.

I thought I did..

1. My code should put 2 blank line between car locations a,b,c

2. My code should put 1 blank line between stay, switch.

Change source code line 18 from:

print “Car: %s Choice: %s Reveal: %s Change: %s win=%s” \

to:

print “Car: %s Choice: %s Reveal: %s Switch: %s win=%s” \

Should clarify my point. Program output :

Car: A Choice: A Reveal: B Switch: – win=True

Car: A Choice: A Reveal: B Switch: C win=False

Car: A Choice: A Reveal: C Switch: – win=True

Car: A Choice: A Reveal: C Switch: B win=False

Car: A Choice: B Reveal: C Switch: – win=False

Car: A Choice: B Reveal: C Switch: A win=True

Car: A Choice: C Reveal: B Switch: – win=False

Car: A Choice: C Reveal: B Switch: A win=True

Car: B Choice: A Reveal: C Switch: – win=False

Car: B Choice: A Reveal: C Switch: B win=True

Car: B Choice: B Reveal: A Switch: – win=True

Car: B Choice: B Reveal: A Switch: C win=False

Car: B Choice: B Reveal: C Switch: – win=True

Car: B Choice: B Reveal: C Switch: A win=False

Car: B Choice: C Reveal: A Switch: – win=False

Car: B Choice: C Reveal: A Switch: B win=True

Car: C Choice: A Reveal: B Switch: – win=False

Car: C Choice: A Reveal: B Switch: C win=True

Car: C Choice: B Reveal: A Switch: – win=False

Car: C Choice: B Reveal: A Switch: C win=True

Car: C Choice: C Reveal: A Switch: – win=True

Car: C Choice: C Reveal: A Switch: B win=False

Car: C Choice: C Reveal: B Switch: – win=True

Car: C Choice: C Reveal: B Switch: A win=False

6 stay and win

6 stay and lose

6 Switch to and win

6 Switch to and lose

12 total wins

12 total losses

Because of rule 2 (cannot reveal car), computerised Monty only opens the right-most door of his two options when the car is definitely behind the other non-contestant door.

This is because computerised Monty always opens the ‘first’ or left-most of the two options when each door conceals a goat.

The certainty arising when computerised Monty opens the right-most door would be eliminated if computerised Monty alternated randomly between the left- and right-most options when each of the two non-contestant’s doors concealed a goat, ie whenever the contestant had initially chosen the winning door.

Here we go again. (I know, humor me)

Assertion: I am biased toward the player initially choosing the car.

Answer: No, Not true.

We need only examine game #1 in truth table,

where the car IS behind door A.

IFF (If and only If) the player initially chooses the car,

there are two outcomes where he/she can STAY and win.

–Car: A Choice: A Reveal: B Switch: – win=True

–Car: A Choice: A Reveal: C Switch: – win=True

IFF the player initially chooses the car,

there are ALSO two outcomes where he/she can SWITCH and lose:

–Car: A Choice: A Reveal: B Switch: C win=False

–Car: A Choice: A Reveal: C Switch: B win=False

**No situation exists where a player can initially choose the car,

and switch, and win.

I DO NOT favor the player initially choosing the car!

That is a mistaken inference. Above, I account for facts/rules:

..Monty has two goats (he can only reveal a goat).. and

..Monty cannot reveal player’s choice.. and

..player can switch/not switch.

We’re not done. Other possibilities with Car behind Door A:

If you choose goat(B), monty cannot reveal it; he MUST show goat(C).

–Car: A Choice: B Reveal: C Switch: A win=True

–Car: A Choice: B Reveal: C Switch: – win=False

If you choose goat(C), monty cannot reveal it; he MUST show goat(B).

–Car: A Choice: C Reveal: B Switch: A win=True

–Car: A Choice: C Reveal: B Switch: – win=False

It should be obvious why, when Choice != car:

.. (player did not choose the car initially) … another way to say it is:

.. (player chose a goat initially)

… Again, thank you guys for your time.

What!?

I give up.

Guys, my boss istampering with the thread (I left my machine logged-in while I went to the bathroom). Somehow, bounty-hunters don’t think this is serious. Editors, pls erase comment # 227 snd # 228

Hell. 226 is my post too — I didn’t see they were messing with it.

@ZenasDad

Have you actually run your program (say 10,000 iterations of the game) and recorded the results?

I don’t know why you’re bothering with computer programs though – my post #219 (in conjunction with post #218) explains that switching wins twice as often as losing, and staying loses twice as often as winning (when considering all 8 options if the car is behind Door1, and their relative probabilities)

@Janus:

I don’t claim that

>> all four of these outcomes are also equally likely (1/4), you contradict your initial claim.

>> Why? You said P(AA) = 1/3. And yet, if all four detailed now are equally likely,

>> then P(AA(B) or AA(C)) = 1/2 = P(AA). Yet you just agreed P(AA) = 1/3. This is

>> why you are wrong.

There are not 4 outcomes. There are 8, for any

location of the car & choice & swap.

Maybe it’d be easier for you to grok the all_outcome_table if “CAR” was the last field. I’m really not sure why you don’t see it.

@Jason:

Did I run 10000 iterations -> no

(Cant run VBS on linux I’m guessing it comes with MS Office?).

Post 218 & 219

Quote: (referring to all_outcome_table):

Also, if you add up all the times the player chose “Door A” for example (8 times), it turns out to be right 4/8 of those times. The same for Door B and Door C. So, Door A is right 50% of the time when you chose it, but so is Door B and so is Door C, which clearly contradicts common sense when picking between 3 doors, they can’t all possibly have a 50% chance of being right. The chances need to add up to 100%.

Rethink your math. EVERY post that claims 1/3, 2/3 depends upon a fanciful equal distribution of outcomes. I said:

–”you can’t calculate random”.

–”random does not mean equivalent distribution of outcomes.

–if you run 99 times the car will NOT be 33*A, 33*B, 33*C.

The outcomes my program generates are untainted by my ideas/beliefs, yet I’m accused of “favoring the player when he first chooses correctly”

I’d be happy if anyone could explain simple questions.

–Aren’t we evaluating the likelihood of outcomes?

–Do we agree that with 3 unknowns, our FirstChoice probability is 1/3?

–Why (besides intuition or education) is your formula correct when it does not reflect the “reality” of possible outcomes?

Purpose of model is to represent reality. So last question:

–You acknowledge there are 8 game paths for any car_location A|B|C. Three doors * 8 yields 24 possible game-paths (irrespective of first_choice, car_location, monty_reveal and switch/stay).. So how do you justify maths without values/expressions [equal to or evaluating to] “8″ or “24″?

Are these two different outcomes? Switch:- means DidNOTSwitch

–Car: A Choice: A Reveal: B Switch: – win=True

–Car: A Choice: A Reveal: C Switch: – win=True

@ZenasDad

“The outcomes my program generates are untainted by my ideas/beliefs,”

We haven’t seen any output generated by your program – you haven’t actually run it.

The outcomes you talk about are documented in (your) post #218, and the probabilities for each outcome actually occurring in my post #219, and these 2 posts clearly show switching wins with a probability of 2/3, staying wins with a probability of 1/3.

I’m not sure what you’re still disagreeing with.

I did, in fact run it. Here’s the output

mackbolan@siductionm90:/opt$ python monty.py

Car: A Choice: A Reveal: B Change: – win=True

Car: A Choice: A Reveal: B Change: C win=False

Car: A Choice: A Reveal: C Change: – win=True

Car: A Choice: A Reveal: C Change: B win=False

Car: A Choice: B Reveal: C Change: – win=False

Car: A Choice: B Reveal: C Change: A win=True

Car: A Choice: C Reveal: B Change: – win=False

Car: A Choice: C Reveal: B Change: A win=True

Car: B Choice: A Reveal: C Change: – win=False

Car: B Choice: A Reveal: C Change: B win=True

Car: B Choice: B Reveal: A Change: – win=True

Car: B Choice: B Reveal: A Change: C win=False

Car: B Choice: B Reveal: C Change: – win=True

Car: B Choice: B Reveal: C Change: A win=False

Car: B Choice: C Reveal: A Change: – win=False

Car: B Choice: C Reveal: A Change: B win=True

Car: C Choice: A Reveal: B Change: – win=False

Car: C Choice: A Reveal: B Change: C win=True

Car: C Choice: B Reveal: A Change: – win=False

Car: C Choice: B Reveal: A Change: C win=True

Car: C Choice: C Reveal: A Change: – win=True

Car: C Choice: C Reveal: A Change: B win=False

Car: C Choice: C Reveal: B Change: – win=True

Car: C Choice: C Reveal: B Change: A win=False

6 stay and win

6 stay and lose

6 switch and win

6 switch and lose

12 total wins

12 total losses

mackbolan@siductionm90:/opt$

Code is at pastebin. http://pastebin.com/upN7duFD

Non-obvious lines:

#35

switch_list = [(g) for g in DOORS if g my_door and g revealed]

line 35 returns a single element list ['A'], which is later converted to a string . switch_list[0]

#53

monty_choices = [(g) for g in DOORS if g winning_door and g choice]

This says monty can’t reveal the car(winning_door) or MY_DOOR(choice)

#53

man. No preview. Please refer to code on pastebin; the NOTEQUAL, GREATERTHAN and LESSTHAN operators were stripped from post #236.

I should have used the word “Switch” instead of “Change” in line # 18.

@ZenasDad

That’ s quite remarkable.

So you ran the program 24 times: exactly 8 times the car was randomly placed behind Door A, 8 times behind Door B and 8 times behind Door C.

Not only that you randomly picked Door A exactly 8 times (and precisely 4 times when the car was behind Door A), Door B 8 times (4 times when the car was behind Door B ), and Door C 8 times (4 times when the car was behind Door C)

What are the odds of that happening?

What’s more remarkable is that in ZenasDad’s model, the car is placed differently based on what the player chooses. If you look at only the branches where the player choose Door A (i.e. an “always pick door A” strategy), there are 8 such branches. and 50% of them have the car behind Door A, 25% behind Door B and 25% behind Door C. But if the player chooses a “Door B only” strategy, the car shifts to be behind Door B 50% of the time, and similar with a “Door C only” strategy.

@PalmerEldritch:

See comment #232: I cant run the VBS. (I don’t have Windows, Visual Basic or MS-Office)

@PalmerEldritch:

“My program” (written python) simply shows all possible outcomes. It does not “play the game”.

What the data (output) implies is:

–”No matter where the car is, In the long run, if you ALWAYS switch, you will win 50% and lose 50%”

Here is output from ZenasDad’s program modified so the player always chooses Door A, by creating a separate array Doors2 which only has “A” in it for the choices.

Car: A Choice: A Reveal: B Change: – win= True .

Car: A Choice: A Reveal: B Change: C win= False .

Car: A Choice: A Reveal: C Change: – win= True .

Car: A Choice: A Reveal: C Change: B win= False .

Car: B Choice: A Reveal: C Change: – win= False .

Car: B Choice: A Reveal: C Change: B win= True .

Car: C Choice: A Reveal: B Change: – win= False .

Car: C Choice: A Reveal: B Change: C win= True .

2 stay and win

2 stay and lose

2 switch and win

2 switch and lose

Note, that the car was behind Door A 50% of the time, twice as likely as the other doors. If you switch the program to only allowing the player to choose one of the other doors, then the car obediently (and magically) moves behind whichever door is needed to make the numbers come up 50/50.

You data is bullshit ZenasDad because it doesn’t take outcome probability into account, thus it gives ludicrous and impossible results, like the location of the car changes dependent on the player’s guess.

I mean it’s patently ludicrous, that if you pick “A” you’re right 50% of the time, but also if you pick “B” you’re right 50% of the time, and also a 50% chance of “C” being right if you pick that.

Any sane person would have warning bells going off that they made a mistake at this point. EACH of 3 possible results can’t be correct 50% of the time.

@Jason: You are correct. The overall statistics have me puzzled. It can’t be disputed that IF you choose the car, monty has two goats g=2; and you can switch or not c=2 ;; # possible_outcomes = c * g = 4

When I have time I will modify the code. Remember though, it simply shows possibilities — it does not “play the game”.

What factors do you want me to control for?

The fix is very simple to explain. the times Monty has to choose between two doors to open, each choice gets 50% of the probability of that branch.

That one change fixes all the issues with your table.

“””I mean it’s patently ludicrous, that if you pick “A” you’re right 50% of the time, but also if you pick “B” you’re right 50% of the time, and also a 50% chance of “C” being right if you pick that.

Any sane person would have warning bells going off that they made a mistake at this point. EACH of 3 possible results can’t be correct 50% of the time.”””

Since I am insane [a blasphemer?] …

The data does NOT say there are 3 results each of which are correct 50% of the time.

The data shows: “If you pick A and always switch, you will win 50% of the time”

The data shows: “If you pick A and always -stay-, you will win 50% of the time”

This is in no way influenced by monty revealing ‘B’ or ‘C’.

Every pair of Car Door / Player Choice has a probability of 1/9. Simple to under stand right? 3×3 = 9.

On the branches where Monty only has one door he can open, he opens that door 100% of the time so it’s 1/9 * 1/1 = 1/9 still

On the branches where Monty has to choose between two doors to open, he opens each door 50% of the time so it’s 1/9 * 1/2 = 1/18, per door. 1/18 times two is equal to 1/9, the original chance of being in this branch. Making a 50/50 choice on one branch splits the probability of that branch into two halves, it doesn’t affect the probability of unrelated branches.

@ZenasDad: seriously? This is what your program claims :-

-If the player picks A, 50% chance of the car being behind door A

-If the player picks B, 50% chance of the car being behind door B

-If the player picks C, 50% chance of the car being behind door C

So, whatever I pick I’m guaranteed to get it right first go 50% of the time?

Ok I modified ZenasDad’s program to propagate the actual probabilities from the choices. What I did what multiply the size of each choice array together as the calculation goes on. Here is the output:

Car: A Choice: A Reveal: B Change: – win= True . chance = 1/ 18

Car: A Choice: A Reveal: B Change: C win= False . chance = 1/ 18

Car: A Choice: A Reveal: C Change: – win= True . chance = 1/ 18

Car: A Choice: A Reveal: C Change: B win= False . chance = 1/ 18

Car: A Choice: B Reveal: C Change: – win= False . chance = 1/ 9

Car: A Choice: B Reveal: C Change: A win= True . chance = 1/ 9

Car: A Choice: C Reveal: B Change: – win= False . chance = 1/ 9

Car: A Choice: C Reveal: B Change: A win= True . chance = 1/ 9

Car: B Choice: A Reveal: C Change: – win= False . chance = 1/ 9

Car: B Choice: A Reveal: C Change: B win= True . chance = 1/ 9

Car: B Choice: B Reveal: A Change: – win= True . chance = 1/ 18

Car: B Choice: B Reveal: A Change: C win= False . chance = 1/ 18

Car: B Choice: B Reveal: C Change: – win= True . chance = 1/ 18

Car: B Choice: B Reveal: C Change: A win= False . chance = 1/ 18

Car: B Choice: C Reveal: A Change: – win= False . chance = 1/ 9

Car: B Choice: C Reveal: A Change: B win= True . chance = 1/ 9

Car: C Choice: A Reveal: B Change: – win= False . chance = 1/ 9

Car: C Choice: A Reveal: B Change: C win= True . chance = 1/ 9

Car: C Choice: B Reveal: A Change: – win= False . chance = 1/ 9

Car: C Choice: B Reveal: A Change: C win= True . chance = 1/ 9

Car: C Choice: C Reveal: A Change: – win= True . chance = 1/ 18

Car: C Choice: C Reveal: A Change: B win= False . chance = 1/ 18

Car: C Choice: C Reveal: B Change: – win= True . chance = 1/ 18

Car: C Choice: C Reveal: B Change: A win= False . chance = 1/ 18

0.33333333333333337 stay and win

0.6666666666666667 stay and lose

0.6666666666666667 switch and win

0.33333333333333337 switch and lose

Use view page source to see the formatting:

http://jplaylist.net/py/test.html

The changes are to keep track of how many things are being chosen from at each step:

for winning_door in DOORS:

chance1 = len(DOORS)

for choice in CHOOSE:

chance2 = chance1 * len(CHOOSE)

monty_choices = [(g) for g in DOORS if g != winning_door and g != choice]

for reveal in monty_choices:

chance3 = chance2 * len(monty_choices)

stay(choice, reveal, winning_door, chance3)

switch(choice, reveal, winning_door, chance3)

print

print

@ZenasDad

What is the point in writing a program to list all the possible outcomes of the MHP?

Listing the outcomes is trivial (see post #184) and can be done manually in less than 10 minutes, the hard part is getting you to understand that not all outcomes are equally likely.

Apart from being pointless, your program fails to take the probability of each outcome into consideration, and is therefore WRONG. The data it produces is INVALID; it doesn’t match reality (as evidenced by millions of trials run by computer programs and countless manual trials), and it doesn’t agree with formal mathematical proofs.

You should disregard the data, throwaway the program – it’s a complete waste of time.

@Jason:

Tell me if you think this tree is correct:

http://www.curiouser.co.uk/monty/diagram.gif

@ZenasDad

Yes that tree is correct

Okay… how about this one:

http://en.wikipedia.org/wiki/File:ModelOfEvents-Probabilities.jpg

There are no errors in that.

It’s the most simplified version possible though, because it doesn’t worry about what each door is named, just whether you correctly selected the car door or not. The maths works out the same, either way.

The second one does not contradict the first one.

that’s because you multiply the number of splits together to work out how probable a choice is. the left and middle branches have a 3-way choice, followed by a 2-way choice. Each leaf of the tree is therefore 1/6th

But the right-hand path has a 3-way split, followed by a 2-way split, and another 2-way split, so each leaf of the tree there happens 1/12th of the time.

That is how probability trees work, the number of splits determines the likeliness of an outcome, and you multiply the number of splits at each level. A branch with more splits has more outcomes, but they’re individually less likely than the outcomes on branches with less splits.

===

If you try and claim that each leaf happens 1/8th of the time, then you have a logical problem … 50% of the outcomes require you to have correctly selected the car from the 3 possible choices, whilst you only picked Goat-1 25% of the time and Goat-2 25% of the time each, rather than equal chances of choosing any of them.

Applying my understanding of the way computerised Monty behaves:

Chooses the leftmost door when he has two goat doors from which to choose or the car is behind the rightmost door; and

Chooses the rightmost door when the car is behind the leftmost door,

I achieved seven correct guesses in a row and a 94.12% success rate over 17 games – only game 8 wrong – way more success than the expected outcome based on odds that do not take into account the anomaly that gives rise to a higher degree of predictability.

I stopped at game 20 with a 90% success rate and only 2 games wrong.

*Individual results may vary.

Tried again:

Stats:

Wins: 17

Losses: 3

Win %: 85.00

@Jason: ref #257. I am wrong. (Worse, I have to call my daughter and eat crow).

Please don’t misconstrue this as “I’m wrong and so are you”

I asked a professor. (I know, ‘appeal to authority’ fallacy) . Disclaimer: He’s a friend. He’s going to enumerate & annotate falsehoods throughout this discussion. He said, “off the cuff, odds of correctly choosing the car, staying and winning should be closer to 1/3 * 1/3 *1/2″

This means 17/18 or 94.555% you switch and win. Seems to match Jonathan’s results.

@Zenasdad

I’m afraid your professor friend is totally off the mark. How he comes up with 1/3*1/3*1/2 is beyond me. The answer is, and has always been, 66.7% for switch and win, 33.3% for stay and win.

@Jonathon

100 trials always switching (always picked Door1)

57 wins

43 losses

57% win rate

Program always opens Door2 if it contains a goat, only opens Door3 when Door2 hides the car.

@Jason #250: Thank you.

Can you find/correct my error below (or did I get it this time):

Two independent and one dependent probabilities:

1/3 chance of car behind A,B,C

1/3 chance of player choosing A,B,C

1/2 because monty reveals a not-choice door

so 1/3 * 1/3 * 1/2 would be my stay odds?

@Jason: I mean to ask if

1/3 * 1/3 *1/2 = “I picked the car first time”, ‘goat 1′ revealed, I stay & win

1/3 * 1/3 *1/2 = “I picked the car first time”, ‘goat 2′ revealed, I stay & win?

@ZenasDad

“so 1/3 * 1/3 * 1/2 would be my stay odds?”

No they wouldn’t. 1/3 would be your stay chances of winning. This has already been explained a gazillion times. Look at the tree structure in your post #253 for a visual representation.

This thread used to be a forum for a debate between people who understood the Monty Hall Problem and those who didn’t and a competition for the simplest explanation.

Anyway, @Zenasdad (do we have to use the “@”?), there are much simpler ways of proving the MH Problem and disproving your professor’s theorem. My results do not reflect the true probabilities.

@PalmerEldritch, my above average wins were achieved by always choosing the door that last won the car (after initial guess). I think I also elected to not switch on some occasions when the leftmost of the two remaining doors was opened by Monty.

I think both results were aberrations (flukes).

“so 1/3 * 1/3 * 1/2 would be my stay odds?”

You can drop the 1/2 factor, that’s saying you switch or stay at random. If you’re analyzing what happens when you choose on or the other it never appears.

The first 1/3 is the game placing the car behind one of the doors, the second 1/3 is the player randomly choosing a door. That gives 9 combinations each with a 1/9 chance:

AA, AB, AC, BA, BB, BC, CA, CB, CC

1/9 is the chance the the car is behind Door A, and the player also chooses Door A at random, but since you win by matching any of the 3 doors, the chance of winning like this is 1/9 * 3 = 1/3

Jason: Back from dinner & drinks with P. Rosser;

His explanation was eerily similar to your posts #250, 267, except he waited until I “got it”, then he said the same thing about “dropping the 1/2″.

By “eerily” I mean word choice, not content.

Doors 1,2 and 3

Suppose car behind door 1

Choose 1 M opens 2 don’t switch – success

Choose 1 M opens 3 don’t switch – success

Choose 2 M opens 3 switch – success

Choose 3 M opens 2 switch – success

Choose 1 M opens 2 switch – failure

Choose 1 M opens 3 switch – failure

Choose 2 M opens 3 don’t switch – failure

Choose 3 M opens 2 don’t switch – failure

Fifty fifty chance of success or failure in switching.

Is ignoring the fact that Monty has TWO choices of doors to open when the contestant chooses the right door giving misleading logic?

Hi Pamela, great question. Yep, the hard part is remembering that while Monty may have 1 or 2 choices when opening a door, it doesn’t change the initial chances of what door you pick.

In your table, you’ll see there’s 4 starting points where you pick Door 1, 2 where you pick Door 2, and 2 where you pick Door 3.

Hrm… something is fishy here! Without looking at the full sentence (just “Choose 1…” not “Choose 1 M opens 2 don’t switch”) we know something isn’t balanced, because we’re already biased towards picking the right door (#1) from the getgo. Covering up the 2nd part of the sentence (just leaving “Choose 1″ “Choose 2″ “Choose 3″) should look the same, no matter what M does, or whether we switch later.

Here’s an updated diagram: http://imgur.com/Xd422PF

You’ll see that the 3 initial branches need to be equal. If Monty is “forced” to pick a door, then all of that 33% probability is concentrated in that option. Otherwise, it’s split between his decisions.

@pamela: kalid is perfectly correct. that logic only works if the player is biased to selecting the correct door.

…in 50% of your cases, the player chose the correct door, and only 25% chance of either of the other doors. Try the table again with car behind door 2 or 3, and the players initial choice “magically” skews towards whichever door is correct. A real warning sign that there’s a mistake.

The logical error is because what someone does later doesn’t change the probability of something that has already happened. There’s a 1/3 chance of you picking the correct door at the start. Monty them makes a 50/50 decision on which other door to open. Each of those outcomes then has a 1/3 * 1/2 chance, i.e. they happen in 1/6 games.

Originally, you pick one door out of 3. You have exactly a 1/3 chance if you choose at random of picking the Car door. And that is the odds you are stuck with if you don’t switch.

Hi Kalid

Yes you are right. I thought it through and realised that because Monty is constrained not to choose either the door with the car behind it nor the door already chosen by the contestant then it makes it more likely that the remaining door has the car behind it. That this probability is 2/3 can easily be shown. I now feel intuitively that this is true. A bit annoyed with myself for not seeing it earlier but thanks for helping me to think further. Thanks also to Jason

probability doesnt work that way

when a door is opened the chances of winning are 1/3 and when monty opens a door with goat it becomes obvious that the car is inside one of the two doors & thus probability changes to 1/2…

@bajivarma

Erm….. probability does work that way.

Just because there’s only 2 doors remaining doesn’t mean they are equally likely to contain the prize, as has already been clearly explained in many previous comments.

Yep, it only goes to 50/50 if Monty is unconstrained in what door he can choose to eliminate. Here’s a 10-door variant to demonstrate:

- a prize is placed behind one of 10 doors, with a 10% chance of being behind any door. A player then chooses 1 door out of 10. For the sake of argument they always choose Door 1. Monty then eliminates 8 other doors, but he’s not allowed to eliminate either the player’s Door 1, or the prize door.

Now, at this stage of the game there will always be two doors to choose from – “Door 1″ or one other door. Does door 1 *REALLY* have a 50/50 chance of holding the prize now? What if the player chose Door 2, would that then have a 50% chance of being correct?

@PalmerEldritch

yeah..your correct.i just realized that..monty hall problem..respect

Yet another strange way to understand this:

1st Choice:

There is a p of 2/3 that you do not pick car.

2nd Choice:

There is then a p of 1/2 that you do not pick the car after switching.

Multiply the two probabilities to get p of 1/3 that you will not pick the car at the end.

@Donnie

That is indeed a very strange way to look at the problem – it’s also the wrong way to look at it.

Using the same argument:

1st Choice:

There is a p of 1/3 that you do pick car.

2nd Choice:

There is then a p of 1/2 that you do pick the car after not switching.

Multiply the two probabilities to get p of 1/6 that you will pick the car at the end.

Your error is in the 2nd choice, there is no “p of 1/2″ if you decide to switch, there’s only a “p of 1/2″ if you make a random choice between switching and not switching which results in a 50% chance of winning the car.

“2nd Choice:

There is then a p of 1/2 that you do pick the car after not switching.”

I disagree with your above statement. The probability of picking the car by not switching is different than the probability of picking the car by switching. By not switching you are stuck with the prior probability of 1/3 since the act of deciding not to switch is a nonevent.

I flipped the discussion on its head to try and reconcile the math behind multiplying the probabilities to end with 1/3 and 2/3.

If you do not agree with me, how do you reconcile the math at each event?

Thanks,

Donnie

2/3 * 1/2 does in fact equal 1/3. But the chain of logic you present doesn’t connect that to the chance of winning the car.

The biggest proof that the logic you used is invalid, is that “win the car” which you put a 1/3 and “don’t win the car”, which would be 1/6 as Palmer showed, do not add up to 1

The set of all probabilities must add up to 1.

Consider switching after your first choice. If you already had the car, switching has a 100% chance of losing it. There is no “p of 1/2″ then. If you didn’t have the car, switching has a 100% chance of WINNING it, here, also there is no element of chance, so no “p of 1/2″.

After you pick your initial door, the player does not know whether their initial door held the car or not, but Monty already knows whether you picked correctly, because he knows where the car was placed for that specific round of the game.

Hence, after your initial choice, no further probabilities are involved, and the proof is that Monty already knows for 100% certain whether switching will get you the car, the very second you chose your initial door.

This also shows how different observers can know different probabilities for the same choice, dependent on their level of knowledge:

-> Monty already knows that the two remaining doors have a 100% chance and a 0% chance (he knows where the car is). Monty can guess correctly 100% of the time.

Anyone who knows which door The Player selected knows that the “switch” door has a 66% chance and the initial door has a 33% chance -> they can guess correctly 66% of the time.

-> someone who just sees two doors and doesn’t know which was the players first choice from the three doors -> they can guess correctly 50% of the time.

Jason, “win the car” are your words not mine. I said pick the car meaning pick the door with the car behind it. Picking the car at the first step does not mean win the car. We all agree that picking the car at the first step is 1/3p. This means that picking a goat is 2/3p.

I still think my math works. How do you reconcile the math of the probabilities for the example I gave in my first post?

You have a major problem with the logic of your “maths” for want of a better word.

You said originally:

“1st Choice:

There is a p of 2/3 that you do not pick car.

2nd Choice:

There is then a p of 1/2 that you do not pick the car after switching.

Multiply the two probabilities to get p of 1/3 that you will not pick the car at the end.”

This is mathematically incoherent. The fact that it “adds up” to the right answer is 100% coincidental.

In your version the “2/3″ represents the times you started with a goat, and the “1/2″ represents the chance that the switch door does not give you the car. But that’s logically incorrect, because if your door has a goat, then the other door is guaranteed to give you the car – i.e. when you have a goat, switching is a guaranteed win.

But, in your version if you get a goat first (p=2/3), you say that after that (p=1/2) of the time switching doesn’t get you a car.

But that means that in 50% of games you’d be switching from “goat” to “other goat”, which we know is impossible in the Monty Hall problem, since there will always be one goat and one car left to choose from.

You need to enumerate ALL branches on your tree, not just one, and add up the probabilities of the outcomes. That will show you the problems in your analysis. You have stated that there are two independent choices, 2×2 = 4 outcomes. You’ve only calculated one branch, and incorrectly at that.

ok, @donnie: summary, you claim that there is a 1/3 chance of not getting the car by switching, if you pick a goat first,right? 2/3 (no car) * 1/2 (no car) = 1/3

Aside from the problem that this calculation allows the possibility of some having picked a goat first, then switching to another goat, after Monty revealed yet another goat (That’s 3 goats, right) you implicitly assume that the other 2/3 means “win the car”. That means that in every other situation, switching MUST get you the car, even if you chose the car in your first choice.

Now, I contend that the chance of switching to the car whenever you picked the car as your first door is a bloody 0% chance no matter how you look at it. Since you pick the car first try (p=1/3) of the time, this 1/3 needs to be added to your 1/3 from the above calculation, which now states that the player loses 2/3

===

so, to summarize, your version requires the player to always win by switching to a car after initially picking a car (two cars in that case) but also, if they picked a goat first, to have a 50% chance that the switch door also holds a goat (that would be 3 goats, counting Monty’s revealed goat).

The upshot is that improperly calculated probabilities lead to paradoxes and contradictory statements.

@Don

To add to what Jason has already posted, take your statement

“There is then a p of 1/2 that you do not pick the car after switching.”

IF this statement were true (it isn’t but anyway) then since you must either switch or stay and probabilities have to add to 1, this statement would also have to be true

“There is then a p of 1/2 that you do not pick the car after staying.”

Also since you can only ‘not pick’ a goat or a car then these 2 statements must also be true

“There is then a p of 1/2 that you do not pick the goat after switching.”

“There is then a p of 1/2 that you do not pick the goat after staying.”

In other words all 2nd choices in your example are 50/50.

In response to your question “How do you reconcile the math of the probabilities for the example I gave in my first post?”. They don’t reconcile. You arrive at a figure of 1/3 by incorrectly applying the probabilities involved.

If you want to see how the probabilities of 1/3 and 2/3 are derived mathematically you need to look up Bayes Theory and how it is applied to the MHP.

According to http://formalisedthinking.wordpress.com/2010/10/05/bayes-theorem-and-the-monty-hall-problem/#comment-109 where Bayes theorem is used to solve the problem, the p of Monty opening door B given that you originally chose A is 1/2.

Palmer, I noticed that you asked the author why this was so but he never responded. This p of 1/2 is used to solve the problem mathematically using Bayes Theorem.

Are you both saying that a p of 1/2 never enters the calculation and that the formentioned author is wrong?

Thanks for the continuing and interesting debate,

Donnie

I finally found a probability tree from MIT that shows where the 1/2 p fits in. I was thinking of what this shows but admittedly it came out jumbled.

https://www.google.com/url?sa=t&source=web&rct=j&ei=iqo9U6iQKIXuyQGOuYHgCg&url=http://courses.csail.mit.edu/6.042/fall05/ln12.pdf&cd=8&ved=0CEAQFjAH&usg=AFQjCNGfilWzSVYlb-M0DIoS-dTu_JbeoA&sig2=-wow37ntE7tR96JTZ5Y-yw

@Don

In that blog (http://formalisedthinking.wordpress.com/2010/10/05/bayes-theorem-and-the-monty-hall-problem/#comment-109) the author states that P(Open(B)|Choice(A)) = 1/2. This is correct but the author doesn’t explain how this value is derived – hence my question.

In order to arrive at this value it is necessary to assume that given a choice of 2 goat doors to open, the host picks one at random (this often isn’t stated in the problem definition). If this isn’t the case then P(Open(B)|Choice(A)) doesn’t = 1/2.

It is possible for P(Open(B)|Choice(A)) to take any value between 1/3 and 2/3 depending on what basis the host chooses to open Door B.

The question is really: Would you rather have one door or two doors? Monty is letting you have two doors with your second guess. Who would not take this deal?

Yes, one of the two doors will be a goat, but Monty will tell you which of your two doors it is. He is giving you a 100% chance of a door with 2/3 probability of winning. That’s better than the original 100% chance of a door with 1/3 probability winning.

Conversely, why would you stick with a door that has a 2/3 chance of being a goat?

You will choose the goat 66% of the time

Therefore, 66% of the time, the host has the choice to open only one door.

Therefore, 66% of the time, the door that the host does not open contains the goat.

I haven’t followed the explanation or read through the comments because there is a fundamental flaw in the opening explanation. It is not clear!

I heard the problem as being that the host KNOWS where the car & goats are & is therefore able to open the one out of the two possibilities to reveal a goat.

There never is a 1 in 3 chance of winning the car because the opening ‘game’ is interrupted by the host so that the first choice is not carried out.

Andrew, the player picks first. Out of 3 doors. Given an equal chance of the game’s producers having placed the car behind any of the 3 doors, there is a 1 in 3 chance of that first choice being correct.

If you “stick” to your first choice in every game played, then you will win 1 in 3 games, because those are the odds of that choice being correct.

What the host does NEXT has no effect on the odds of that first selected door being the car. He could open no doors, open both doors, or pick a door and waggle it open and closed repeatedly. You already had either a car or goat as soon as you said “I choose door X”.

… and since the “stick” strategy always wins 1 game in 3, and the host is constrained so that he never can reveal the car himself, then any time the “stick” strategy has a goat, the “switch” strategy must have the car: i.e. 2 out of 3 games.

short version: “stick” wins 1 game in 3. “switch” wins any time “stick” loses, i.e. 2 games in 3.

This game is rigged. Why is that the first door that is opened always has the goat and not the car? As a true game of probability, the chances will be 50/50.

Completely wrong, just read through the posts if you don’t get it. no point repeating what’s been said 1 million times for the mathematically illiterate.

Didn’t get it at all until SteaveG (comment number 28) expressed it in terms of what you haven’t picked rather than what you have. For the longest time, I was confused by the fact that Monty’s filtering acts the same way whether you have the car or not (he has to pick a goat) but the fact that your 33.3% chance of picking the right thing on the first round means there’s a 66.6% chance the car is behind one of the two doors you haven’t picked. Monty opening one door transfers that 66.6% likelihood to the other one.

Yeah, there’s also a big hang up that a lot of people have about this too: since originally each door had a 33% chance of having the car behind it, then each door “must” stay equal chances with every other door for the entire game. But that thinking just isn’t correct. As soon as the car is placed (which happens before the game starts), that actual door has a real chance of 100% and the other doors are actually 0% chance each. There was no point in time when all doors were equal, that’s just an illusion. The percentages we place on the unrevealed doors are just probability estimates that can change as information is revealed.

For an example, say there are 100 doors with one randomly-placed car. Player 1 wins if the car is behind doors 1-10 and player 2 wins on doors 11-100. Now, if we reveal the car *right away* player 2 will win 90% of the time because he has 90% of the doors. Then, say the host opens all but two doors – one for each player, without revealing the car. Each player has two doors. Just opening doors in a different order can’t change the likeliness of the car being in the range 1-10 or 11-100. Hence Player 1′s last door wins 10% of games and Player 2′s last door wins 90% of games.

The odds are never 1/3 at the outset if Monty always picks a door with a goat. Otherwise he would pick a car 33.33% of the time … but he never does. I would say there is faulty reasoning here.

Another question is, at what point does the randomization take place, before Monty opens the door or after.

So if I don’t commit to a choice before he opens a door and delay my choice until after he opens a door, my chances are 50/50?

After 10 plays always choosing Door 1 and then switching, I scored more than 80% correct.

It’s better than that, if you don’t know which door the player chose you only have 50% chance of picking correctly, regardless of how the final 2 doors were selected. You’d also have a 50/50 chance if you just flipped a coin to decide between stay or switch.

It’s all about knowledge – to the random guy off the street selecting right at the end each door is 50/50. Anyone who knows the order doors were selected thinks the doors are 66/33, and anyone who knew what Monty knows thinks they’re 100/0.

Jason (#302):

Please Explain: “You’d also have a 50/50 chance if you just flipped a coin to decide between stay or switch.”

From the general consensus, that statement should read: “You’d have a 33/66 chance if you just flipped a coin to decide between stay or switch.” Where the 33 would be ‘stay’.

I mean, for example, heads=stay; tails=switch

You “do” have a 50/50 of getting heads or tails… However landing on tails would give you twice as many ‘WINS’ as landing on heads. (if that makes sense).

think about it this way, if you flip a coin you’re only half as likely to “switch” than the person who deliberately switches. hence, you don’t get the full 66%. It breaks down like this:

Case #1 – stick door had the car (1/3 probablity). If this happens and you flip a coin you’re 50% likely to choose the “stick” door and 50% likely to choose the “switch” door. So you win 50% of the time in “Case #1″ scenarios.

Case #1 – switch door had the car (2/3 probablity). If this happens and you flip a coin you’re again 50% likely to choose the “stick” door and 50% likely to choose the “switch” door. So you win 50% of the time in “Case #1″ scenarios.

So if you choose blindly, you win 50% of the time in either possible scenario, even though scenario #2 is more likely.

Remember no door *actually* ever has any probability except for 100% or 0%. We’re not talking about a quantum car that materializes magically behind a door as soon as it’s opened. The car was already behind a specific door before you started playing.

The variable probabilities are estimates based on current levels of knowledge. If you choose completely at random on the final choice, you’re not taking advantage of the small amount of knowledge you did gain.

If someone comes in right at the end without knowing which door is which (either one could be the switch door or the stay door), how are they magically going to pick correctly 66% of the time by flipping a coin? Think about that.

Another way to look at is is that if you choose switch 50% of the time and stay 50% of the time, you’re longterm average winning rate will be halfway between the two extremes:

Stayers win 33% of the time, switchers win 66 of the time. If you go with either choice half-half your win rate will be half-half:

(33%+66%)/2 = 50%

Jason, thanks, but you just confused me. Please correct #305, or reconcile (#305-#307)

#305 seems to be saying the same thing as

ZenasDad #147: If the chance were 50/50 for a new player with your same card choices… The chance *must* be 50/50 for you as well.

Given 2 options, any choice made by flipping a (fair) coin is by definition 50/50. The probability of picking the correct option is determined by the probability of the coin landing heads (or tails), and that’s the case even if you know with 100% certainty that Option1 is the right one (why you would flip a coin in that instance is another matter).

ZenasDad#147 is just wrong logic.

Monty knows for sure which door it is: To him they’re 100% and 0%. He can “guess” the car correctly 100% of the time. Anyone who knows which door the player locked in in round 1 can guess the car correctly 66% of the time. Neither are constrained by what happens to a coin-flipper.

Think about a horse race. Horse 1 is better and wins 90% of the time. Horse 2 wins 10% of the time. If you knew that you pack horse 1 and also win your bet 90% of the time. Someone picking completely at random will not win 90% of the time no matter what. They’re equally likely to back the winner as the loser.

A coin flip picks either option half the time. So you get the average win rate of between options. And when they add up to 100% (100/0, 66/33, 90/10 etc) the average is always 50%

Btw i dont need to reconcile #305 and #307, they’re saying the same thing, but in different words.

To try and clarify #305 for you we have two “scenarios” :

- Stick wins (33%)

- Switch wins (66%)

And then we flip a coin. Flipping a coin is independent of the chances of the two scenarios happening. Look at it this way:

Scenario #1 (stick wins, 33% of games). you flip a coin, 50% chance of stick-or-switch therefore 50% chance of winning.

Scenario #2 (switch wins, 66% of games). you flip a coin, 50% chance of sticking therefore 50% chance of winning.

So, you see no matter which of the two scenario you end up in, you flip a coin and win 50% of the time. This in no way constrains the likelihood of the two scenarios occuring, because they’re independent events to the coin flip.

If you study probability trees you can map out the different steps and see where they’re independent.

I came to understand the problem like this:

There are 100 doors and you select one of them but do not open it.

You have just divided all possible doors into two groups.

One group has 1 door in it, (the one you chose) and the other has 99 doors in it.

The group with 1 door in it has a 1% chance of having the car in its group and the group with 99 doors in it has a 99% chance of having the car in its group.

If Monty knows where the car is and removes every door in the large group except for 1 by proving they don’t have the car behind, it does not change the fact that there was a 99% chance of the car being in that large group.

We are still left with two groups, each having 1 door but we know that the first group had a 1% chance of having a car and the second group had a 99% chance of having a car.

If we choose the second group, we have a 99% chance of getting the car.

I understand this in following way:

Suppose this game is played between two players A and B,

A is give chance to select a door and he wins if car is behind that door.

B will have both remaining doors and wins if car is behind any of the two doors not selected by A.

Lets examine the probability of winning for A and B.

A: 1/3

B:2/3

Now lets see it in context to Monty Hall problem. If you switch you are B (one is already shown to you and only other door has randomness in case of Monty Hall), if you stick to your choice you are A. Pretty simple :p

I understand this in following way:

Suppose this game is played between two players A and B,

A is give chance to select a door and he wins if car is behind that door.

B will have both remaining doors and wins if car is behind any of the two doors not selected by A.

Lets examine the probability of winning for A and B.

A: 1/3

B:2/3

Now lets see it in context to Monty Hall problem. If you switch you are B (one is already shown to you and only other door has randomness in case of Monty Hall), if you stick to your choice you are A. Pretty simple :p

#311 – “Scenario #1 (stick wins, 33% of games). you flip a coin, 50% chance of stick-or-switch therefore 50% chance of winning.”

No, surely it’s 50% chance of sticking, not 50% chance of winning. Put another way, it’s 50% chance that ‘Box 1 will be opened’ on the coin toss, but not 50% chance that ‘Box 1 will be opened and the car will be inside’.

There are two scenarios, and flipping a coin has a 50% chance of winning the car in EITHER scenario. The fact that one of the scenarios is less likely than the other doesn’t change your chances.

Let me elaborate with a 10% / 90% split:

- Door A has the car 10% of the time

- Door B has the car 90% of the time

If someone “flips a coin” are they REALLY going to get it right 90% of the time. That would imply that flipping a coin gave them Door B 90% of the time.

What actually happens is that if the car was behind Door A, then flipping a coin has a 50% chance of giving you either A or B (Heads = pick A, Tails = pick B).

On the other hand, if the car was behind Door B, then flipping a coin has a 50% chance of giving you either A or B (Heads = pick A, Tails = pick B).

So, even if it’s 10/90, 33/66 or 1/100, flipping a coin matches the right door half the time.

Do a probability tree, and it’s clearer.

1/3 chance “Car = Door A”

In these 1/3 times, you then flip a coin getting1/2 Heads (Pick A), 1/2 Tails (pick B)

1/3 * 1/2 = 1/6 chance of winning, 1/6 chance of losing

2/3 chance “Car = Door B”

In these 2/3 times, you then flip a coin getting1/2 Heads (Pick A), 1/2 Tails (pick B)

2/3 * 1/2 = 2/6 chance of winning, 2/6 chance of losing

To calculate your total odds of winning add the “win” probabilities for all possible outcomes: 1/6 + 2/6 = 3/6 = 50%

The 10/90 example works the same. 5% chance of wins with Door A, 45% chance of wins with Door B, adding to 50%. “Yeah but door B wins more” is balanced out by “but Door A therefore loses more, and you were just as likely to choose that losing door with your magical coin flip”.

In other words, you can never “beat the odds” by flipping a coin.

#317 – “”… door B wins more” is balanced out by “but Door A therefore loses more …””

You’re right – I mapped this out assuming an even distribution and it resolved to 50/50 really quickly.

So in response to the original problem, ‘stick or switch?’:

Stick = 1 in 3 chance of winning the car;

Flip a coin to pick a door = 1 in 2 chance of winning the car;

Flip a coin to stick or switch = 1 in 2 chance of winning the car;

Switch = 2 in 3 chance of winning the car.

Another way to understand the original problem:

The whole game is a two-door problem, never a three-door one. Everything before the final choice is simply about altering the chances of what is behind those final two doors (the human intervention in the chances stops them remaining completely random). If I were Monty’s stage hand, this is how it might go:

I put three empty doors out and the contestant chooses one. Behind the scenes, I reach into a giant bag of cars and goats (containing twice as many goats as cars), randomly pull one out and put it behind the chosen door. Then I empty out the bag, pick up a goat and put it behind one of the remaining doors. Monty opens this door, thus eliminating it. Remembering which I put behind the chosen door, I now pick the opposite (a goat or a car) and put it behind the last door. The chance of me pulling out a car for the chosen door was 1/3, therefore the chance of me needing a goat for the last door is also 1/3 – the other 2/3 times I need a car. You know the rest.

Holy moly.

Two thirds of the time, the car is behind the door Monty doesn’t open.

Done.

When Monty reveals a goat it is no help whatsoever.

He shows something that was known to be there – in any set of two doors there will always be at least one goat – one third of the time two goats – but always one goat.

Given that the chosen door is door 1 – that leaves two other doors – numbers 2&3.

Monty shows a goat behind door 2 say – well we knew he could show a goat – what does it matter if we see a goat behind door 2?

Perhaps instead of opening door 2 he opens door 3 and reveals a goat – again we knew he could do this.

We only learn that a goat sits behind either door 2 or door 3 – this information has no significance at all – 2 or 3 / 3 or 2? So what?

If for instance Monty does not bother to reveal a goat – nothing changes – we know absolutely that one third of the time there are two goats and that two-thirds of the time one goat and a car – seeing a goat that we knew to be there tells us nothing.

The goat-door is not eliminated from the mix just because we can see a goat- it still carries its original probability of a 1/3 chance of the car – similarly the door with the car behind it also has a 1/3 chance of the car.

The probability of each door hiding the car is fixed by the number of doors and the number of prizes – no matter what happens this 1/3 chance per door – open or closed – car or goat – does not change.

The real choice that Monty offers is the original 1/3 chance door or the other two 1/3 chance doors together – he craftily shows a goat before making his offer – it makes no difference.

As Alchemy says – It’s a Group Thing

Group of one versus group of two

Not a difficult choice – not rocket science.

This is so true I truer it like 100 times with every door because I set the door count to one hundred. Every time I did i switched and every time I got the car

The simples way for me to explain this is:

Car is behind 1/3 doors.

That is

Door A: 33.3%

Door B: 33.3%

Door C: 33.3%

You choose Door A, therefore you have a 33.3% of being correct.

Your 33.3% chance of being right also means that there is a 66.7% chance that you are wrong and the car is in the field (remaining doors). This is true because each door independently has a 33.3% chance (as stated above), meaning that the field is (Door B + Door C) = 66.7%.

Monty hooks you up and eliminates Door C.

Seemingly making it 50/50? Not so.

There is still a 33.3% chance that door A is correct, and a 66.7% that the field is correct (which has now been reduced to Door B thanks to Monty).

Thus explaining the improvement in odds. All of the 66.7% attributed to the field is now resting on door B.

Door A = 33.3%

Door B = 66.7%

Would you stick with door A still?

Monty does not “eliminate” a door when he opens it to show a goat – the two doors not selected will contain:

Goat & Goat

Goat & Car

Car & Goat

There is always a goat to be seen – a goat may surprise a child playing peek-a-boo but it should not be of interest to an adult – because – there is always a goat that can be shown – we do not need to see the goat to know there’s one there.

The rules of the Monty Hall Problem oblige Monty to show a goat – all well and good – but he need not bother doing so because the goat is a known quantity – 100% certain to be there weather Monty shows it or not.

Monty’s offer – he does not say so – is to swap from one 33% door to two 33% doors together – before he makes his offer he deviously reveals a goat – the goat he need not bother revealing – and this confuses most observers – they think it’s a one for one swap – they think the goat-door has mysteriously gone away and even more mysteriously its 1/3 chance of the car magically transferred to the other door – why not to the chosen door? Why not in proportion to both closed doors?

There is no explanation for this magic transfer of probability from one place to another – each door has a 1/3 chance of the car no mater what lies behind it – that’s what the 1/3 car & 2/3 goat probability means – it allows for either outcome in proportion – the probability is not changed based on the eventual outcome – what MIGHT happen is entirely different to what DOES happen.

The goat might just as well not show up at all and if it does we should just ignore it – Monty’s offer is to exchange one door for the other two together.

@richard buxton Why argue semantics if the point is correct? The door obviously did not vanish, and that was not implied. Instead the door was “eliminated” as a selection choice.

I know if I was on a game show and Monty shows me a goat behind a door I think it’s safe to say that the door has been “eliminated” as a possible choice 100/100 times.

The point is not correct – the door is a selection choice because the offer is to exchange one door for the other two together – this is how the swap achieves a 2/3 chance of success 1/3 + 1/3 = 2/3. One component from each door.

If switching from one door to the other two there will always be at least one goat between them – sometimes two goats – but always one – seeing the goat is no assistance – of no consequence – Monty might just as well not show it – a contestant faced with the goat does best to ignore it.

But when Monty first shows this completely useless goat the casual observer imagines something significant has happened – it hasn’t – a goat appears from where a goat ought to be.

The swap is from one door to two doors – Monty doesn’t say it directly because if he did there would be no Monty Hall problem – his timing – the revealed goat and the style of his offer all combine to complete his crafty ruse.

You swap from one door with a 1/3 chance to two doors each with their own 1/3 chance – and there’ll always be a goat even if Monty doesn’t bother to show it – and leave the goat door closed.

When you swap you exchange a closed door with a 1/3 chance for two doors together each with their own 1/3 chance – the fact that one of those doors has a goat behind it is a predetermined fact – we do not need to see the goat to be aware of it – we know it’s there without seeing it – the goat makes no difference.

The door is a selection choice even though it’s a goat-door – it gives a 1/3 chance. Your choice is effectively that single door or these two together – but Monty does not say this…

@Richard,

Simply put the 2 doors (B and C) you didn’t pick have a combined 2/3 chance of containing the car. The 2 doors retain that 2/3 chance after Monty shows there’s a goat behind Door C because Monty has to show a goat, and seeing one tells you nothing new about your door. Since Door C now has a zero chance, Door B must therefore now have a 2/3 chance.

As another poster quite succinctly said “Monty is effectively offering you 1 door or the BEST of the other 2 doors”. 2/3rds of the time “the BEST of the other 2 doors” is the car.

“… seeing the goat is no assistance – of no consequence – Monty might just as well not show it …”

If the goat is not revealed the outcome of switching remains 1/3 chance-of-car, the same as sticking. Revealing the goat increases the outcome of switching to 2/3 chance-of-car.

If you have a devious and suspicious mind as I do this problem is not a problem at all.

If you’re a generally trusting person it requires some hard thought because Monty leads you astray.

The real deal – crafty Monty fails to spell it out – in fact he tries hard to hide what’s happening – is to swap the original chosen door for the other two together.

Now – between any two doors there will always be at least one goat to show – 100% certainty – the swap offer is give up one door and accept two in exchange – the game rules oblige Monty to show a goat and then make his offer – difficult to work it through for many people but naughty Monty shows the goat – the goat 100% certain to be there – and when we see it it’s so easy to fall into his trap of thinking the swap is a simple one-for-one and that the goat-door tells something important and/or because it’s a goat it’s removed from the mix – nothing of the sort.

Each door has a 1/3 chance of the car – two doors together have a 2/3 chance and yet it’s an absolute guaranteed 100% certainty that between any two doors there will always be at least one goat – we do not need to see the goat to know the truth of it being there.

Monty might just as well not bother showing the goat

If shown – the contestant might just as well ignore it

The real offer is two doors together in exchange for one door

It always makes sense to swap

Monty’s antics and his subtle timing of door opening and then making his offer lead us astray – he is crafty devious and very smart – he sets out to confuse like a stage magician and he usually succeeds.

It’s two for one: 1/3 + 1/3 =2/3 and at least one of those 1/3 chances is a goat.

“Monty might just as well not bother showing the goat”

But if Monty doesn’t show a goat then you only have a 1/3 chance of winning whether you swap or not as Freddie said.

It’s imperative that Monty does actually show a goat for your chances to increase to 2/3.

No – although it’s unstated the Swap Offer is to exchange one door for the other two together – each having a 1/3 chance of the car for a combined 2/3 chance – and at least one of those 1/3 chance doors is absolutely guaranteed to have a goat behind it.

When we see a goat we see see what we knew in advance to be the case – a goat behind a door.

Chances of the car do not leap about from door to door just because we don’t approve of what we see – it takes two doors together to give a 2/3 chance.

1/3 + 1/3 = 2/3 and at least one of those 1/3 chances is a goat.

For a single door to have a 2/3 chance of the car there needs to be three doors and two cars – there is only one car and that is why each door has a 1/3 chance of hiding the car – two doors together have the 2/3 chance – and at least one of those 1/3 chances has a goat to show.

The Swap offer is your door for the unopened door, not for both doors. It’s “unstated” because it doesn’t exist. You can’t just change the problem conditions to suit you argument.

“For a single door to have a 2/3 chance of the car there needs to be three doors and two cars”. Where do you get this idea from? I ask because it’s not true. As I’ve explained the the unopened door has a 2/3 chance by itself.

Dear Plamer…

It’s unstated because Monty wants to fool you – if he said it was a two for one deal then there would be no Monty Hall Problem.

I know of no mechanism that causes the movement of a 1/3 chance of the car from one door to another just because a goat comes into view – why does the 1/3 chance move? How does it move? Why does it move to one particular door rather than another? Why does the 1/3 chance not get shared equally between the two unopened doors?

Do not try to answer these questions because there is no answer. Chances do not move about – chances have no knowledge of goats being revealed chances are fixed by the numbers.

If there are three doors and one of them has a 1/3 chance of the car – then the other two together have – perforce – a 2/3 chance and it is this – and this alone that causes the swap away from the selected door to have a better chance of success. The swap is away from one door in exchange for two

Because a goat comes into view most people assume that it’s not part of the mix and in their minds they move – as if by magic – the 1/3 chance from the goat door to the the last door – they arrive at a 2/3 chance but ascribe it to one door rather than two – they are unable to accept that an open door with a goat on view has a 1/3 chance of the car – yet accept without question that a closed door with a goat behind it unquestionably has a 1/3 chance of the car. if it’s true when the door is closed no matter what lies behind then it must also be true when it’s open – no matter what lies behind.

The 2/3 chance comes equally from two doors – it’s a one for two swap – but crafty Monty reveals a goat before making his offer – the naughty fellow.

Now – Three doors two goats and one car – these numbers in proportion provide denominators and numerators that make the 1/3 versus 2/3 probability.

Suppose there were two cars and only one goat? The probability for a car is now 2/3 – because there are two cars in the mix – similarly 1/3 the goat.

Probability is fixed by the numbers 3-2-1 Doors-Goats-Car – change one or more of the components and the probabilities change – leave the numbers as they are and the probabilities remain fixed.

Many people confuse a probability – what might happen in the future – with what has happened – probability allows for either result in proportion – what has happened is a result and has no bearing on the original – or indeed the current – probability. Probability stays fixed forever – unless something changes – a goat appearing where a goat was certain sure to be is not a change.

It really is childishly easy – Group of One for a 1/3 chance – Group of two for a 2/3 chance.

Any group of 2 doors will always have a least one goat to show – the revealed goat is of no consequence.

@richard buxton

What if there 100 doors? You have a 1% chance initially, 98 goats are shown.

Do you believe that there is still a 1% chance that the door Monty left has the car?

There’s no way you can manipulate the statistics to say that the blind 1/100 pick you made initially has the same odds as the 1 left from the 99 other doors.

Still not convinced? 1000 doors, Monty removes 998 doors, you feel confident in your initial pick?

1,000,000 doors? You would still stay with your initial pick?

With 1,000,000 doors, the only way that the door Monty leaves is not the one containing the car is if your 0.0001% chance initial guess was correct. How can that one initial blind 1/1,000,000 guess contain the same odds as the one door that Monty leaves?

The principal behind this mental exercise is that your initial guess has poor odds. Albeit it is possible that your 1/1,000,000 pick was correct it most likely was not. It’s much easier to see this shift in odds when the number of doors is amplified. If your initial pick was wrong (which based on the odds is most likely the case) then Monty must be showing you the correct door.

So, in reality you’re not betting on your ability to pick the correct door. You’re really betting on your inability to pick the correct door initially.

You state your arguments eloquently and you seem well spoken, but we obviously disagree.

Dear Ryan,

I tend to reject the 100 door – the 1000 door and the 1,000,000 problem scenarios because they just waste time with different numbers but the same problem – they take longer

The odds of a successful pick the first time are always lower because there are more unpicked doors that might hold the prize – we agree this.

Monty sets our one way or another showing goats until only two doors remain – the original selection and one other – he then makes his offer to swap – most people think this is a one for one swap but it isn’t – even though Monty does not actually say so – the offer is – in reality – the chosen door against all the other doors together – all those goats mean nothing – if there are N doors then Monty will reveal N-2 Goats.

On one side a chance of 1/N and on the other side a chance of N-1/N i.e 2/3 or 99/100 or 999/1000 or 999,999/1,000,000

I bet on the basis that 99 doors together are better than 1 door on its own

Or that 2 doors together are better than one

You get my drift I think.

I don’t let the numbers influence me – just Monty’s clever trick of trying to get me to believe it’s a one-for-one exchange – it isn’t – it’s a swap of one door versus all the others combined – I always go after the bigger number.

Simplicity is the key to understanding this thing. Two doors are better than one.

For whatever reason I thought you had been arguing against what we had all been saying, until I just read your last reply.

You’re essentially making the same argument that the rest of us are.

Dear Anonymous,

I think I am arguing against what most other people are saying -

I am saying:

1) That the swap is one door for two doors together

2) That a door can never have a 2/3 chance of the car

3) That a goat-door is not eliminated

4) That a goat-door still has a 1/3 chance of the car – very difficult to accept

5) That we learn nothing when a goat is revealed

6) That Monty might just as well not bother showing a goat

7) That a contestant should disregard the goat when Monty reveals it

8) That the 1/3 versus 2/3 probabilities remain fixed no matter what

Apart from these I’m in complete agreement with everybody else.

Again, without the reveal the chances remain 1/3 (stick) or 1/3 (switch to the first of the other doors) or 1/3 (switch to the second of the other doors). It’s a TV show, not a theoretical puzzle; you’ve got to have the final denoument and complete the game. So the contestant has to know which door to pick.

… a bit like the chances of me spelling ‘denouement’ correctly – I had to click ‘post’ to reveal the contents.

@Richard

I think you’re confusing the mathematical/logical puzzle that is the Monty Hall Problem with the 1960′s game-show called ‘Let’s Make a Deal’.

In the former Monty is constrained in his actions by the very precise rules governing how the “game” is played – he is an automaton, devoid of free will, guileless. In the latter Monty the game-show host is jokey, crafty, maybe even devious, free to change the rules on a whim (although I don’t think he ever offered a contestant a choice of 2 doors instead of one). As Monty Hall himself said “My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home.”

And so on to your post…..

“It’s unstated because Monty wants to fool you”. No he doesn’t, he just follows the rules of the game. He doesn’t offer “a two for one deal” because that is not in the rules. He offers a one for one deal because that is how the game has to be played.

“Chances do not move about”. I can assure that chances (probabilities) can and do change. Perhaps an example will explain:

Let’s suppose Monty does indeed offer “a two for one deal”, and he makes this offer to 90 contestants say. It’s clear that 60 of those 90 contestants will win the car and 30 won’t. Let’s also suppose Monty asks each contestant to initially open just 1 of their 2 doors… to prolong the suspense.

(we know for certain that behind each contestant’s 2 doors there is at least 1 goat)

Now, 30 of those 90 contestants will open a door and reveal a car straight off… Hey Presto! They become instant winners. The other 60 contestants will open a door with a goat. Since we know there are 60 winners in total, only 30 (or 50%) of those 60 contestants who open a goat door become winners, and the other 30 end up losers.

Therefore, if a contestant opens a goat door then their 2nd door only has a 1/2 chance of being a car, and the single door not picked also has a 1/2 chance. In other words after a goat door (which we knew was there anyway) has been opened, what was a 2/3 chance has become a 1/2 chance, and what was a 1/3 chance has also become a 1/2 chance . So clearly chances can change as new information becomes available.

As I said in my previous post the unopened door has a 2/3 chance once Monty reveals a goat.

“Chances do not move about”. I can assure that chances (probabilities) can and do change.

Quite so; they do this in relation to time and the (current) observer’s frame of reference. Consider the toss of a coin before a sports match. If that particular coin has landed heads down the last nine times out of ten in prior circulation it is 9/10 likely to be tails. But the captain who is asked to call by the referee is faced with just one random coin of the set available – he gets 50/50 from the same coin (the others in the set containing the complete range of probabilities).

There’s a video of Monty Hall – the man himself on Youtube – he talks of the Monty Hall Problem (MHP) and explains that he never ran it the way it gets discussed – he use to offer to buy the selected door from the contestant – he never made any swap offers.

The MHP – as we know it – was described by the Parade journalist Marilyn vos Savant an American magazine columnist, author, lecturer, and playwright who rose to fame through her former listing in the Guinness Book of World Records under Highest IQ.

It is she who sets the rules and describes the puzzle – not Monty Hall. It’s easier however to pretend that it is a game show and that Mr Hall is running things…

Well – here’s my challenge to those people who claim that chances move about – specifically how a door with a 1/3 chance of the car suddenly has a 2/3 chance of the car because a goat was revealed in another door.

What is going on that makes the transfer happen? Why does the 1/3 chance move – what causes it to move? Why does it move to that specific door and not the other? Why is the movement not split equally to both of the other doors?

Tell me in words…. make it simple and easy to understand.

I claim that switching is advantageous and gives a 2/3 chance of the car because the switch is from one door to two doors with each door contributing a 1/3 chance – that’s not 2/3 + 0/3 but 1/3 + 1/3 – YES the goat door is a 1/3 chance of the car! How can this be so?

You will have no trouble in accepting that a closed door – all closed doors – have a 1/3 chance of the car – and yet two of those doors have a goat hidden behind them – what? A door with a goat behind it has a 1/3 chance of the car – what lunacy is this? Well it’snot lunacy – it must be so – you know absolutely that there are two goat-doors which have the attribute of 1/3 chance the car.

So I ask – what difference if the goat door is open or closed? You might respond – well it’s no longer a door you could swap to – and I say – oh yes it is because if you reject the selected door then the alternative is two doors together – and surprise surprise – that’s why swapping gives you a 2/3 chance because the swap option – unstated I accept – is from one door to two.

But what about the goat you may ask? Well consider what might b e behind those two doors….

Goat – Gaot

Goat – Car

Car – Goat

Notice in each of those three possibilities there is always a goat – we know there’s a goat to see – not because the game’s definition obliges Monty to show a goat – but because logic and common sense says that they’ll always be a goat – 100% certain guaranteed.

Now – as for my claim that probabilities and some (several) people asserting that they can and do move about – well they may change if the situation changes but in the MHP nothing changes so in the MHP probabilities and the 1/3 versus 2/3 proportion of each door remains constant – forever.

The probability is set by there being Three doors – Two goats and One car – 3-2-1 these numbers set the probabilities and these numbers do not change – an open door with a goat or a car on view does not alter the numbers – if the arrangement were 2 doors 1 car and 1 goat it would be 50/50 guess – but it’s 3-2-1 so the probability is 1/3 the car and 2/3 a goat – fixed.

Probabilities may certainly change if new information comes along – a goat is not new information.

Vos Savant’s Parade question simplifies to “Having been shown which of the doors you didn’t choose has the goat, is it to your advantage to switch?” to which the answer is yes. Richard, you appear to be asking the question “Given the opportunity to choose a different unopened door, is it to your advantage to switch?” to which the answer is no. Seeing the goat is material to the solution.

“If that particular coin has landed heads down the last nine times out of ten in prior circulation it is 9/10 likely to be tails.”

I don’t toss many coins, but I do know I should have said “… heads UP …”

Freddie – why and how is the goat material to the solution – how does it hep us? What does it change?

I remind you… the two doors may hide…

Goat Goat

Goat Car

Car Goat

Notice that there’s always a goat – a 100% dead cert there’s at least one goat behind those two doors – how can it be of assistance – what difference does it make if it gets shown? – if we see it or not? It’s always there…

What do we learn when we’re shown something we knew to have been there? We do not need to see the goat to know about it.

I fully admit that the goat-reveal tells us which specific door it was behind – Door 2 or Door 3? – The Middle One or the One on the Left? The Blue or Red?

Not particularly thrilling intelligence I suspect.

How is the goat material to the solution?

Simply this – the contestant has to know which door to pick. Otherwise their chances don’t change.

Ok Richard, you and I are on a game show. I’m out with the host, and you’re out back and unaware of the happenings on stage.

I choose door A, Monty opens door B and leaves it open to show no car.

You walk in having no clue which door I selected, but have been informed that monty revealed an empty door B. You are left with two choices door A and door C, but have no clue about what I did, Are the odds still 1/3 per door, or have the probabilities in fact changed to 50/50?

That’s where you lose me, when you say probability does not change, because it always changes as populations change, or new information is made available.

Richard, I’ve never actually seen the problem anywhere. I’ve just stumbled across this while looking for math logics problems. Please enlighten me as to where I’m stumbling during the problem.

If I pick door A, that leaves doors B/C.

Which is 1/3 vs the field or 2/3.

Based on your replies I believe we still agree up until this point.

Now Monty shows door C and alas no car.

We still agree that door A maintains the 1/3 chance of holding the car. My reasoning is that the 1/3 is based on early information of there being 3 unknown doors, which I think you agree with too ( correct me if I’m wrong).

Where we disagree is in the next bit of logicial assessment (I think).

It appears that most of us claim that the 2/3 odds now rests on door B. You claim that door C being shown not to have a car is irrelevant because you’re picking two doors vs your 1 initial pick. Unless the car moved I’m not picking both doors, I’m picking the best of those doors which was shown by Monty to be door B.

We all agree that the reason for the switch is the 2/3 probability that initially that the car was in the field even before Monty opened a door.

But the point of the game is not to choose two doors, it’s to choose one. Therefore because the field has a 2/3 chance of being correct and now the field has been reduced to door B, door B now contains that edge over door A.

Another interesting take on it:

My initial pick is blind i.e. 1/3. Monty is left to select a door from the field 2/3.

The difference between me and Monty’s pick is that his is not a blind selection, he knows exactly where the car lies.

There is 2/3 probability that I picked the wrong car initially ipso facto there is also a 2/3 chance Monty is leaving the correct door. If I unknowingly choose an empty door (which happens 2/3 of the time), and Monty must reveal an empty door, he must be leaving the car door.

Monty only makes a random guess 1/3 of the time, which only occurs when you were correct initially.

Freddie dear chap…

My contention is that the swap offer away from the selected 1/3 chance door must be to the alternative of TWO doors each with a 1/3 chance – the host’s messing around and showing a goat causes observers to think it’s a one for one swap.

Furthermore people claim that the remaining closed door carries a 2/3 chance of the car – I can think of no mechanism other than magic that will cause a 1/3 chance to transfer to an adjacent door – people just say it does because they need and want it to be the case – they do not show how it happens.

Most people do not accept that the swap offer is from one door to two 1/3 chance doors together – the host doesn’t say it so it can’t be true.

If the swap offer is to two 1/3 chance doors then it’s a simple matter to arrive at a 2/3 chance the swap – If the swap offer is to a single door nobody is yet to explain the change that happens – the how and why the door’s chance probability doubles – why the 1/3 chance moves to one door rather than another.

The contestant exchanges one door for two – hence the 2/3 chance of the car – it’s no more difficult than that – a group of one versus a group of two.

The goat is immaterial – it’s always there – Monty need not show it.

How many doors can the contestant open?

Richard I had posted to you directly right before your last reply.

Ryan – your question – I walk in late and see two doors…

This is my attempt to make you smile…

I select the open door with no prize

Monty then opens the other door with no prize

I swap to the remaining closed door for a 100% guaranteed win…

But you won’t let me do that

For the contestant facing three doors the probability of the car is clearly and obviously one in three for each and every door – he selects from a field of three .

The late arriving contestant (me) faced with two closed doors and a dead loss door has a realistic choice of two.

The first set-up has three doors to choose from hence 1/3 – the second scenario two doors and an even money 50/50 bet.

For the first guy the probabilities are set and never change

For the second guy the probabilities are set and never change

They may be the same doors but the probabilities are different

It’s the same pack of cards but they’ve be re-dealt

Now – to get closer to where I lose you…

1)You fully accept that all closed doors have the same 1/3 chance of the car?

I guess you do…

2) You fully accept that behind two of the doors there’s a goat

I’m sure you do…

3) The concept of a door with a goat behind it actually having a 1/3 chance of the car does not particularly disturb your equilibrium.

4) If you can cope with a closed door and goat and 1/3 the car then it’s a small step to accept an open door and a goat and 1/3 the car – I can.

5) So for me the numbers 3-2-1 Doors goats and Car set the probabilities.

6) To change the probabilities requires the numbers to change

7) In the standard Monty Hall problem the numbers do not change – we get to see a goat but so what? – the numbers do not change and the probabilities remain fixed – even if a bomb explodes and vaporizes all three doors the probabilities stay the same because of the numbers 3-2 and 1

Monty’s door opening is merely a stage trick intended to mislead – he shows something we all know about – knew about before he did it – something of no value to us.

Monty has to know where the car is because he is obliged to show a goat – the risk of his not knowing is obvious.

You lost me a bit with the interesting take… Too obscure for me to follow properly – when you say…

- a 2/3 chance Monty is leaving the correct door -

I have no idea how to interpret this – what does correct door mean?

” – we get to see a goat but so what?”

How many doors can the contestant open?

Freddie – you ask…

How many doors can the contestant open?

Well in the standard Monty Hall Problem – where Monty shows a goat to gasps of shock and surprise – and then there’s a swap offer – if the contestant swaps then there’s only one door he can open – Maybe Monty opens it for him – or a leggy blonde?

In my view of the world in which I claim the swap offer is to give up one door in exchange for two – and Monty has taken my advice and not bothered messing around revealing his useless goat – then the contestant has two doors to open – if he first opens the 1/3 door with the car behind it I guess he won’t bother opening the other door – if he opens a 1/3 goat door then he’s bound to take a peek inside the second – wouldn’t you?

Of course I’m being silly about who actually opens these doors – it really doesn’t matter who does it.

The swap offer is – in reality – swap that group of one for this group of two.

“… if [the contestant] opens a 1/3 goat door then he’s bound to take a peek inside the second -”

Isn’t the game that the contestant has to settle for his first pick? Otherwise they could keep opening doors until they found a car; that wouldn’t make great TV – ‘Hey honey that show’s on where they give them a car. Fancy a drive?’

Freddie – you asked…

Isn’t the game that the contestant has to settle for his first pick?

Monty makes you think that but he doesn’t say it – You select a door – he shows a goat – he offers the swap.

People think the swap is one-for-one but it isn’t – it’s one for two – the door opening confuses people something rotten – Monty’s door opening is just a meaningless trick – he opens one of our own doors for us the naughty man – he shows us our own goat.

The pick – as you put it – is two 1/3 doors together – it’s just that Monty doesn’t actually say so – he just offers an ill-defined swap.

I can see through his trickery but many people are taken in by it – there are too many trusting souls in the world. Monty is a deceitful knave.

“Another interesting take on it:

My initial pick is blind i.e. 1/3. Monty is left to select a door from the field 2/3.

The difference between me and Monty’s pick is that his is not a blind selection, he knows exactly where the car lies.

There is 2/3 probability that I picked the wrong car initially ipso facto there is also a 2/3 chance Monty is leaving the correct door. If I unknowingly choose an empty door (which happens 2/3 of the time), and Monty must reveal an empty door, he must be leaving the car door.

Monty only makes a random guess 1/3 of the time, which only occurs when you were correct initially”

The correct door that I mentioned would be the one containing the car.

I’ll try to more clearly state my argument now that I’ve had some coffee.

There is a 1/3 chance that I choose the car blindly. That is the same as saying that there is a 2/3 chance that I do not choose the car with my initial guess.

Since Monty must reveal a door without the car there are three possible outcomes based on initial door chosen. For simplicity I’m leaving the second pick option out and just stating where the car lies.

1. I choose door A which has no car, Monty opens door B. The car is in door C.

2. I choose door B which has no car, Monty opens door A. The car is in door C.

3. I choose door C which has the car, Monty opens either door B or C.

2/3 of the time the wrong door is chosen initially, meaning that 1/2 goats have been selected 2/3 of the time. Because Monty must show a goat, 2/3 of the time Monty will show you the second goat, leaving the car door. The only time that Monty makes a random pick is when you selected the car initially, which occurs 1/3 of the time.

Monty operates like a blackjack dealer, he does not think but instead follows a fixed set of rules. When I Swap doors, I’m not necessarily betting on him showing me the correct door because 1/3 of the time he flips a coin (this happens when I picked the correct door initially, situation 3 above).

Instead I’m swapping my pick based on the 2/3 chance that I picked a goat initially, knowing that if I did pick a goat, the car would always be left.

“The pick – as you put it – is two 1/3 doors together – “

Surely not. The general understanding, despite the lack of specific detail in original descriptions of the problem, is that the MHP contestant is offered the choice of one door out of three, not two. Their chance of winning if they switch is indeed that of two 1/3 doors together, but only if Monty reveals a goat. Without that they only have one 1/3 chance which is assigned to the door they pick (another 1/3 chance remaining with the ‘switch’ door they do not pick).

How many doors can the contestant choose from? Three.

How many doors can the contestant choose? One.

Dear Freddie -you say…

Surely not. The general understanding, despite the lack of specific detail in original descriptions of the problem, is that the MHP contestant is offered the choice of one door out of three, not two.

And you would be right – that is the general understanding – but it’s not the case – not in my world it isn’t.

Cheeky Monty prematurely opens one of the two doors and makes people terribly confused – they imagine the opened door eliminated – it’s 1/3 chance the car transferring by some bizarre mechanism to another door – they need to say the remaining unopened door has a 2/3 chance – so they declare the goat-door somehow void – its 1/3 chance transferred elsewhere.

If they accept my 2 door explanation then there’s no issue about swapping having a 2/3 chance – I have yet to see how the unopened door gets to have – all by itself – the double probability.

My contention is that a door can only have a 2/3 chance of the car if there are Three Doors and Two cars – but that’s too deep for most of my interlocutors even though they have no issue with three doors and one car having a 1/3 chance – very strange.

Similarly I try to get across that Monty’s goat revelation is merely his trick – he opens one of the contestant’s two doors – says nothing about it – and shows a goat that’s certain sure to be there.

For me it’s a piece of cake – I’m devious – suspicious and never trust authority figures – my default condition is to disbelieve everything.

Other’s who may be more trusting than me seem to have difficulty with what I say.

I can easily justify in my mind how the swap offer is a 2/3 opportunity – it has to be 2 doors together – I am too dim to work through how an individual door can possibly have a 2/3 chance of the car – and nobody has ever explained it to me – they just say that it’s so…

Richard did my last last reply help clarify my earlier statement for you?

Ryan —

I’m still a bit confused by this… I’ll take it lowly – piece by piece…

There is a 1/3 chance that I choose the car blindly. That is the same as saying that there is a 2/3 chance that I do not choose the car with my initial guess.

1) OK I’m with you so far

Since Monty must reveal a door without the car (a goat) there are three possible outcomes based on initial door chosen. For simplicity I’m leaving the second pick option out and just stating where the car lies.

2) Confused now

1. I choose door A which has no car, Monty opens door B. The car is in door C.

2. I choose door B which has no car, Monty opens door A. The car is in door C.

3. I choose door C which has the car, Monty opens either door B or C.

3) Well this makes some sense but only to Monty because he knows where the car is – meaningless to the contestant – he doesn’t know where the car is.

2/3 of the time the wrong door is chosen initially, meaning that 1/2 goats have been selected 2/3 of the time.

4) I dispute “wrong” – goat is better but…

5) Yes – yes – go on…

Because Monty must show a goat, 2/3 of the time Monty will show you the second goat, leaving the car door.

6) What do you mean “the second goat”?

The only time that Monty makes a random pick is when you selected the car initially, which occurs 1/3 of the time.

7) Well sort of – the door Monty selects may be a random selection but it’s a don’t care random selection – the outcome is not random – he knows he’s going to reveal a goat

Monty operates like a blackjack dealer, he does not think but instead follows a fixed set of rules.

8) I reluctantly agree – he might just as well not be there – he says pick a door – then he fiddles around like an automaton – then he asks do you want to swap?

When I Swap doors, I’m not necessarily betting on him showing me the correct door

9) What do you mean – showing you the correct door? All you know as a contestant is that Monty shows a goat – and for the wise contestant an utterly useless goat.

because 1/3 of the time he flips a coin (this happens when I picked the correct door initially, situation 3 above).

10) Rubbish – Monty doesn’t flip a mental coin – his only freedom of action is when you first pick the door with the car behind it – his decision making then is limited to door 2 or door 3 – random as discussed but the outcome beyond doubt – a goat.

Instead I’m swapping my pick based on the 2/3 chance that I picked a goat initially, knowing that if I did pick a goat, the car would always be left.

11) Yes naturally – if you picked a goat to begin with the car must be in the alternative group of two

12) what are you saying here? Do you agree that the swap is one door for two or not? If not – how do you explain how the remaining door miraculously gets to have a 2/3 chance of the car – are the doors conspiring with one another -secretly passing information between themselves – - –

pst! I have a goat here – when Monty opens me up I’ll slip you my 1/3 chance of the car – oh bother – I’ve got a goat too – I’ll gladly accept your 1/3 chance and become a goat door with a 2/3 chance of the car… Ho-Ho-Ho.

“I am too dim to work through how an individual door can possibly have a 2/3 chance of the car – and nobody has ever explained it to me – “

I thought you explained it very well yourself when you said:

“I remind you… the two doors may hide…

Goat Goat

Goat Car

Car Goat”

a total of two cars and four goats, making chance-of-car two in six (or 1/3) and a field of two doors. Now take one goat and one door away, Monty having revealed a goat behind one of the doors, and there are now two cars and three goats, making chance-of-car two in three (2/3) and a field of one door.

Sorry my maths don’t stack up but perhaps someone who can express it properly could make the argument more elegantly than me.

Freddie – you’re making my brain hurt…

a total of two cars and four goats, making chance-of-car two in six (or 1/3) and a field of two doors. Now take one goat and one door away, Monty having revealed a goat behind one of the doors, and there are now two cars and three goats, making chance-of-car two in three (2/3) and a field of one door.

No – there are only two goats and one car – the list I gave – as you well know – was the possible content combinations of any two doors in the MHP – if I show on one line the possibility of a car behind door two – and on the next line the possibility that it might be behind door 3 – that does not sum to two cars – it’s the same car – or rather it’s the same 1/3 chance of a car – and two 1/3 chances together are 2/3 chance the car – but NOT from one door – from TWO Doors – two doors together – you are such a naughty boy.

And if it’s not two doors together how do we ever get to one single door having a 2/3 chance of the car – my question is still hanging – your explanation of just now – rejected.

At least it’s not 50/50 …

Freddie – I have no maths – but this is all pretty basic stuff – a word explanation should be enough – a minimum of easily followed logic but I get nothing.

Many say that the alternative to the selected 1/3 door is the other unopened door that now has a 2/3 chance of the car – that sounds very odd to me – a door changing from a 1/3 chance to a 2/3 chance just because some other door has a goat behind it – it really doesn’t stack-up in my view of reality – I’ve never seen anything to explain it – not even come close to explaining it – except for one thing… If you’re swapping to two doors together – then it all works perfectly without hurting your brain – two 1/3 doors together = a 2/3 chance.

Bed time for me here in England – so good night…

I can’t show it with maths (as you’ve seen), but using your table again:

“I remind you… the two doors may hide…

Goat Goat

Goat Car

Car Goat”

each time it’s the first row Monty opens one goat leaving a goat, and each time its the second or third rows he opens a goat leaving a car. So one out of three times it’s a goat that’s left behind the unopened door, two out of three times it’s a car. The unopened door may not be the same door each time, but whichever one it is has 2/3 chance-of-car. So 2/3 chance in one door.

@Richard Buxton

“Since Monty must reveal a door without the car (a goat) there are three possible outcomes based on initial door chosen. For simplicity I’m leaving the second pick option out and just stating where the car lies.

2) Confused now”

-What I meant was that if there are only three doors, then you are left with only three initial choices (1/3). Of those three doors, only one will contain the car, while the other two possess goats.

I then listed the possible scenarios (which I’m re-quoting below). I intentionally left out the swapping option, because for now I’m not concerned with the probability involved in swapping choices. Instead I’m focused on the bare bones of what is actually behind the doors. In attempting to do so the first time I was apparently unable to clearly display the following:

A: 1/3 chance of getting it correct on the first try (which of course you wouldn’t know until the end).

B: 2/3 chance of getting it incorrect, or as you like to say “picking the goat”. For argument sake I consider picking the goat an incorrect guess, seeing as the point is to select the car which renders anything other than the car incorrect regardless of what it is actually behind the door.

C. Whenever an incorrect choice is made initially (see scenario 1&2 in my quote below) Monty will always leave the door containing the car.

D. The only time that Monty does not leave the door containing the car is when it was initially selected.

“1. I choose door A which has no car, Monty opens door B. The car is in door C.

2. I choose door B which has no car, Monty opens door A. The car is in door C.

3. I choose door C which has the car, Monty opens either door B or C.

3) Well this makes some sense but only to Monty because he knows where the car is – meaningless to the contestant – he doesn’t know where the car is.”

-I disagree, I think this does indeed benefit the contestant. You are correct in saying that only Monty knows where the car is, but knowing that the contestant does not know where the car is, can be used to the contestants betting advantage (worded confusingly I know).

Much like in blackjack, the fact that I am at a disadvantage initially (1/3 odds) combined with the fact that Monty (the house) is bound to obey a fixed set of rules can be used to the bettors advantage. This is explained in more detail below.

“2/3 of the time the wrong door is chosen initially, meaning that 1/2 goats have been selected 2/3 of the time. Because Monty must show a goat, 2/3 of the time Monty will show you the second goat, leaving the car door.

6) What do you mean “the second goat”?”

-There are two goats in the puzzle. Based on the odds you will inevitably select 1 of the 2 goats 2/3 of the time with your first guess.

Because Monty will always show a goat from the field (the remaining two doors), it also means that the same 2/3 odds that apply to you picking a goat initially apply to Monty showing the second goat, thus leaving the car as the other door in the field 2/3 of the time. This only 1/3 of the time Monty leaves another goat in the field, which only occurs when you unknowingly select the car initially.

Notice that I still have not mentioned swapping doors yet? That is because I want to make sure the odds that go into the initial door selection and Monty’s subsequent revelation are clearly stated, as they are the meat and potatoes of my logical argument.

“The only time that Monty makes a random pick is when you selected the car initially, which occurs 1/3 of the time.

7) Well sort of – the door Monty selects may be a random selection but it’s a don’t care random selection – the outcome is not random – he knows he’s going to reveal a goat”

-Exactly. That inconsequential random selection is why earlier I only stated three outcomes and not 4 in my quote above.

Now for the whole door swapping aspect of the problem:

“When I Swap doors, I’m not necessarily betting on him showing me the correct door

9) What do you mean – showing you the correct door?”

-What I meant was showing me the correct door by revealing a goat. i.e. I choose a, he shows goat in b, left with c. In this case I’m saying by swapping to C I’m not betting directly on him leaving C, but instead betting on me incorrectly picking A to begin with.

“All you know as a contestant is that Monty shows a goat – and for the wise contestant an utterly useless goat.”

-I disagree. Because of poor initial 1/3 odds, I use my first pick just to get Monty to show a goat. I’m assuming that I chose a goat initially, because 2/3 times this would be the case. I use my pick and Monty’s subsequent revelation as deductive reasoning to leave the car.

“because 1/3 of the time he flips a coin (this happens when I picked the correct door initially, situation 3 above).

10) Rubbish – Monty doesn’t flip a mental coin – his only freedom of action is when you first pick the door with the car behind it – his decision making then is limited to door 2 or door 3 – random as discussed but the outcome beyond doubt – a goat.”

-Exactly, the only time Monty’s revelation is random is when by chance my initial pick was correct. As a bettor using my poor odds to my advantage, I’m hoping I was not correct initially and choose to ignore this 1/3 chance outcome as well.

“Instead I’m swapping my pick based on the 2/3 chance that I picked a goat initially, knowing that if I did pick a goat, the car would always be left.

12) what are you saying here? Do you agree that the swap is one door for two or not? If not – how do you explain how the remaining door miraculously gets to have a 2/3 chance of the car – are the doors conspiring with one another -secretly passing information between themselves – – – “

-From “Another interesting take on it:” on I’ve moved on from the door taking on 2/3 odds, instead to a leveraging of poor odds as a way to hedge one’s bet.

We’ve all stated the same foundation but take it different directions to get to the same point. We all agree that 1/3 odds initially is worse than the 2/3 odds that the field possesses, we just interpret those odds differently to get to the same result.

@Richard

“It’s easier however to pretend that it is a game show and that Mr Hall is running things…” Why? What makes it’s easier? It seems to me it’s making it harder as you’re attributing actions to him that are outside the scope of the ‘game’

“Well – here’s my challenge to those people who claim that chances move about”. I’ve already shown you how chances can move about. How what was a 2/3 chance can become a 1/2 chance. Do you require further examples?

“I claim that switching is advantageous and gives a 2/3 chance of the car”. Correct

” because the switch is from one door to two doors with each door contributing a 1/3 chance – that’s not 2/3 + 0/3 but 1/3 + 1/3″. Incorrect. The switch is to ONE door with a 2/3 chance as I’ve already explained.

“… a closed door – all closed doors – have a 1/3 chance of the car – and yet two of those doors have a goat hidden behind them” Correct again, but it is unknown what is behind those doors. Probability is about “unknowns”

“A door with a goat behind it has a 1/3 chance of the car – what lunacy is this?” An utter lunacy. If Door A has a goat behind it, there is zero chance that Door A has a car behind it. Once it is known precisely what is behind a particular door there ceases to be a chance that there is “something else” behind it. A goat doesn’t have a 1/3 chance of being a car – as you say that’s lunacy.

“you know absolutely that there are two goat-doors which have the attribute of 1/3 chance the car.”. There are 2 doors, behind which is a goat and either another goat or a car, we don’t know which, it is unknown hence we can assign a probability to what is likely to be behind each door.

“So I ask – what difference if the goat door is open or closed?” If a door is

open we know what’s behind it.

“well it’s no longer a door you could swap to”. The rules of the game don’t allow you to swap to the open door, and why would you want to – it’s a goat.

“and I say – oh yes it is because if you reject the selected door then the alternative is two doors together” That doesn’t follow. If you “reject the selected door” what’s to stop you selecting ONE of the other 2 doors.

“that’s why swapping gives you a 2/3 chance because the swap option is from one door to two.” It is a 2/3 chance but not for those reasons, since you don’t get to swap to 2 doors

“but in the MHP nothing changes so in the MHP probabilities and the 1/3 versus 2/3 proportion of each door remains constant – forever.” More lunacy. When a door is opened the situation changes – you didn’t know what was behind Door C before, you do now, and THAT changes the probabilities

“Probabilities may certainly change if new information comes along – a goat is not new information.” A goat behind Door C IS new information, because you didn’t know what was behind Door C before it was opened. By definition that is NEW information.

Dear Palmer Eldrich…

Everybody discussing this interesting problem does it from the perspective of the Monty Hall game show – not as the problem published in Parade Magazine by Marilyn vos Savant – so that’s what makes it easier…. We don’t argue the presentation circumstances.

https://www.youtube.com/watch?v=c1BSkquWkDo

I do not recall you showing how chances move about – either I didn’t notice you doing it – don’t remember what you said or didn’t believe you.

I still maintain that chances do not move about – I genuinely do not know how a 1/3 chance of the car can transfer from one door to the other. Or why it does – or why one particular door rather than the other…

The next bit is very contentious – few will accept it…

Furthermore – at the outset all doors have a 1/3 chance of the car and yet two doors have a goat behind them – how can this be so? Well it is so because the probability of the car is not decided by what lies behind the door – the contents dictate nothing – the chance probability is determined solely by there being Three doors – Two goats and Only one car.

Three doors – One car – All doors 1/3 chance the car

Two doors 1/3 chance the car but a hidden goat behind.

A door is closed – it has a 1/3 chance of the car but there’s a goat behind.

The door is opened – the goat revealed and then….?

Well I maintain that the goat-door retains its 1/3 chance of the car because for me the 1/3 chance represents what MIGHT happen not what has happened.

Others have the 1/3 chance moving about but they’re a bit weak on the logic of it – the how and the whys and the where of it.

I accept that to swap gives a 2/3 chance of the car – and for me it’s very easy in my mind – the swap offer is to two doors together – each with a 1/3 chance of the car for a combined 2/3 chance – and at least one of those two doors will always have a goat behind it.

Now yet more controversial stuff…

I maintain absolutely that Monty’s revelation of the goat makes no difference and offers no help to the contestant.

Imagine that you are the contestant – you’ve selected a door – and there are two other doors – you think about the possible goat/car distributions of those doors – (1) Goat/Goat or (2) Goat/Car or (3) Car/Goat – three different possible distributions and each one – all three – a goat in there. 100% dead cert that there’s a goat – you’re a smart cookie – you do not need to see a goat to know there’s a goat – might even be two goats you just can’t tell – but you will not be astonished or surprised or better informed if a goat wanders into view.

Monty might just as well not bother showing you a goat and telling you what you already know – Monty’s door opening antics are a phantom – just his trickery intended to confuse people – he opens one of your doors and shows you a goat – so what?

The problem for me is that I keep having to repeat myself on here because I have only one story to tell….

Chances do not move about – even a goat door has a 1/3 chance of the car

The swap offer is to exchange one door for two

When Monty shows a goat we learn nothing – it changes nothing

And my justification for these outrageous statements are above.

@Richard,

You make no justification for your outrageous statements at all. You maintain, or assert or state with nothing to back it up.

Did you even read my last comment? If so, you have conveniently avoided everything I wrote, and failed to address any of the points I made refuting your more ‘lunatic’ claims. There’s really no point carrying on further discussion with you if you’re going to adopt an attitude of “I’m right because I say so, therefore any alternative argument is invalid and can be ignored”

However on a final note I will return to this statement which you repeated in your last comment:

“Well I maintain that the goat-door retains its 1/3 chance of the car because for me the 1/3 chance represents what MIGHT happen not what has happened.”

So what you are saying here is: a door that is KNOWN to conceal a goat has a 1/3 chance of concealing a car. Tell me “what MIGHT happen” in this situation means. There’s a 1/3 chance the goat will turn into a car?

Dear Plamer…

You said… Probability is about “unknowns”

No – probability is about the likeliness that an event will occur

What might happen – looking into the future – going-forward as they say

In the MHP probability is determined by the numbers – Three doors – Two goats and One car – if those numbers change then the probabilities will change.

In the MHP we begin and end with the same numbers 3-2-1 therefore probability does not change.

Now to the goat reveal and what we learn when we’re shown a goat – we learn almost nothing – nothing of any real value or interest. We learn a door number 2 or 3 – perhaps the color of the door Red or Green perhaps we learn if it’s the middle door or one of the others. We do not learn the goat – we already know about the goat – the logic of our mind tells us of the goat – Monty does not need to show the goat for us to know about it.

Now – for me but perhaps not for you – my Swap decision is to two doors – there is no question in my mind that that is the offer – for you I think you’re only able to swap from the selected door to the other closed door – in which case you need – in your mind – to have the alternative door become a 2/3 chance of the car – you may be able to do that mental exercise but it’s beyond me.

In my simplistic world I swap from one door to two – all the difficult stuff goes away – and my brain is not stressed.

Yes I know that nobody says that the swap offer is to two doors together – that’s the knavish trickery of it all – do not trust what people tell you until you’re absolutely certain – don’t take my word for it -work it through for yourself.

Begin with the mystery of a single door suddenly having a 2/3 chance of the car – how does that work? Don’t tell me – just tell yourself how it happens – be honest.

“The swap offer is to exchange one door for two”

The swap offer is to exchange one door for another door (one of the other two). The contestant has to choose one door. They cannot choose two doors and have the contents of both. They only gain advantage from switching from their original door to one of the other two if one of those doors has been revealed, otherwise they would be exchanging one 1/3 chance for another. If a car were revealed they would gain the advantage of 3/3 (1/1) chance of winning the car – pointless both as a TV show and a statistical puzzle. As a goat is revealed they gain the advantage of 2/3 chance of winning the car.

Richard, you have described a game show where the host offers the contestant a switch from their original choice but what then? The script runs out once they have chosen to stick or switch. What door(s) subsequently get opened and who decides?

@Richard

“no -probability is about the likeliness that an event will occur”. or an event that has already occurred where the outcome is unknown. When you pick a door in the MHP Monty already knows where the car is, you don’t.

“In the MHP we begin and end with the same numbers 3-2-1 therefore probability does not change”. The situation changes when a door is opened which causes the probability to change

“we learn almost nothing” You’ve changed your mind then, from “nothing” to “almost nothing”

“nothing of any real value or interest.” We learn enough for the probabilities to change as I HAVE EXPLAINED.

“Now – for me but perhaps not for you – my Swap decision is to two doors – there is no question in my mind that that is the offer” On what basis? Not by anything Monty says , nor by any possible interpretation of the rules of the game.

“for you I think you’re only able to swap from the selected door to the other closed door” Because that’s the offer, those are the rules.

” – in which case you need – in your mind – to have the alternative door become a 2/3 chance of the car” Yes, and I have explained how that is so. It is not a mental exercise (I can provide a mathematical proof if you prefer).

“In my simplistic world” you require a goat to have a 1/3 chance of turning into a car, which is beyond me (and every other sane person I would imagine)

“Yes I know that nobody says that the swap offer is to two doors together” – That’s because it isn’t, there’s no “knavish trickery” involved.

“work it through for yourself” I have.

“Begin with the mystery of a single door suddenly having a 2/3 chance of the car – how does that work? Don’t tell me – just tell yourself how it happens – be honest.” It’s not a mystery, it’s probability theory, and I have told you how that works. If you didn’t understand that explanation, if it was beyond you, I’m happy to provide you with another explanation or two (or three).

I gave an example in an earlier comment of how a 1/3 chance becomes a 1/2 chance after a goat is revealed, perhaps you should look at it again.

Hey Freddie – I have said quite enough on here – you will think me very boring but as I’ve said before – I have only one story to tell.

In my mind I can’t work out how a single door can possibly and suddenly have a 2/3 chance of the car based on a goat being shown – it means that the 1/3 chance of the car – that the goat-door enjoyed – moving itself to one particular door just because a door gets opened to show a goat – for me that’s more magic than logic – and what’s more – nobody is able to explain it.

If I try in my mind to make this 1/3 chance move about – I can’t make it happen – yet I know that swapping results in a 2/3 chance…

How can I reconcile this mystery and still stay sane?

If I allow in my mind that the swap offer is to exchange one door for two then it all suddenly makes sense to me. There is no single door carrying a 2/3 chance but two doors together each with a 1/3 chance. I swap one for two – Bingo!

If in your mind the contestant HAS to choose one door – then so be it – I have never seen this stated – in which case YOU have the problem of explaining to yourself how the 1/3 chance moves from one door to the other – don’t tell me how it works – just think it through for yourself – make it happen in your mind and be true to yourself as you do it.

In my mind the swap offer is to two doors together – a group of two each with a 1/3 chance and the whole having a combined 2/3 chance – yes I agree – nobody actually says the swap is one for two – and the open goat-door makes it look very much as if it is a one-for-one swap – but nobody says that it isn’t an exchange of one door for two – it’s pretty much unstated either way – I ask you – who in their right mind would want to swap to a goat-door? – well me for a start – because in my mind I swap to two doors.

You ask…

What door(s) subsequently get opened and who decides?

Well the sensible contestant always swaps for his 2/3 chance so the contestant decides – in the defined problem Monty opens the first of the contestant’s doors and shows him a goat – tell me something I didn’t know Monty – he might think to himself – after that – in the group of two only one door remains to be opened – it actually has a 1/3 chance of the car because of the 3-2-1 numbers but many (most) people will claim it’s got a 2/3 chance – so be it – no matter – I don’t agree with them.

It doesn’t really matter who opens the last door – Contestant – Monty – Passing midwife – gust of wind…

The whole point of the TV show (it was actually never like this) is to entertain – in much the same way as a stage magician might amaze and mesmerize his audience – the big trick is (a) to fail to describe the swap offer (b) to first reveal a meaningless goat and allow people to think that there’s something significant going on – and (c) make the swap offer after the goat reveal – further confusing people into thinking it’s a one-for-one swap.

Monty is merely an automaton – he adds little (no) value – he asks to pick a door – he shows a goat by obligation – he asks the swap question – he just facilitates according to a predetermined formula – he uses his knowledge of the car to be sure that he keeps it hidden.

” – in the defined problem Monty opens the first of the contestant’s doors and shows him a goat – “

No no no no no; in the defined problem Monty opens one of the doors the contestant did NOT choose in his first choice, revealing a goat, THEN he offers the opportunity to switch.

… the opportunity to switch from their originally-chosen door with 1/3 chance of a car to either (a) an open door with 0/3 chance of a car or (b) a closed door with 2/3 chance of a car.

Dear Freddie – you say…

I gave an example in an earlier comment of how a 1/3 chance becomes a 1/2 chance after a goat is revealed, perhaps you should look at it again.

Could you not find it and paste it in for me? I really can’t come to terms with the concept unless something changes – and the appearance of a goat is not a change – not a significant change.

Let me explain again (for the last time) – it’s all quite simple…

1) Any door has a 1/3 chance of the car

2) But two doors have a goat behind them and yet they still have a 1/3 chance of the car – how can this be so?

3) Because probability is merely a forecast of what might happen – it might be a car and it might be a goat in the proportion of 1:2

4) What actually happens does not retrospectively change the calculation of what might happen – the opening of a door does not alter the goat/car forecast

5) The probabilities 1/3 and 2/3 are set by there being 3 doors – 2 goats and one car – only if these numbers change can the probabilities change – and in the MHP the numbers stay fixed – nothing added & nothing removed.

6) If there were just two doors one goat and one car then there would be a 1/2 chance – but there isn’t – in the MHP the numbers 3-2-1 don’t change.

Where we differ – one area where we differ – is when you say that an opened door with a goat has a 0/3 chance of having the car and that’s TRUE from the perspective of a new observer – but from the outset that goat-door carried a 1/3 chance of the car – a probability of what might happen in the future – a probability set by the 3-2-1 numbers and probability is not altered by subsequent events.

You go your way and I’ll go mine – for me it’s all very simple.

Finally – you wrote…

” – in the defined problem Monty opens the first of the contestant’s doors and shows him a goat – “

For me the contestant OWNS both of the other two doors – so the door that Monty opens IS the contestant’s

The next paragraph I agree with entirely – but your No no no no no; is unnecessary

No no no no no; in the defined problem Monty opens one of the doors the contestant did NOT choose in his first choice, revealing a goat, THEN he offers the opportunity to switch.

“If there were just two doors one goat and one car …”

Two doors, one goat, one car. 50/50. One door opens to reveal a goat. What is the chance the other door conceals a car? Put another way, how many times in a series of opening that door would it contain a goat?

@Freddie,

Some people will never, ever admit they’re wrong. Richard is one of those people.

He considers it perfectly acceptable to change the rules of the game, ignore comments that prove him wrong, or simply make stuff up in support of his flawed and illogical arguments.

I’ve come to the conclusion it’s a waste of time continuing this discussion with him, he obviously has no interest in broadening his understanding of the MHP – I’d bet he thinks he already knows everything there is to know about it anyway.

Don’t be taken in by those self-effacing “in my mind” phrases, I suspect he’s actually quite arrogant.

Freddie dear chap – please stop this – you just posted…

“If there were just two doors one goat and one car …” My quote…truncated…

If there WERE two doors one goat and one car then there would be a 50/50 split – but there NEVER IS two doors one goat and a car. I was showing – trying to show – how a 1/2 chance could be achieved – by changing the numbers 3-2-1 – and I went on to say that in the MHP the numbers begin at 3-2-1 and end with 3-2-1 – therefore probabilities remain fixed throughout.

Your next paragraph…

Two doors, one goat, one car. 50/50. One door opens to reveal a goat. What is the chance the other door conceals a car? Put another way, how many times in a series of opening that door would it contain a goat?

Why do you ask such basic stuff?

If there are two doors then each has an equal chance of the car – a 50% chance – the revelation of a goat does not change that probability – something that might happen in the future.

When something does happen – as you say – a goat being shown – we do not need to discuss it further – we have a result – we know absolutely where the car is – it sits behind a door that has a 50% chance of a goat – whoopee!

Probabilities are set by the numbers and only the numbers – probabilities are not affected by subsequent events – It’s probable that Wicked Strong will win the Kentucky Derby – Oh! Wicked Strong finished 4th – the result does not affect the probability – they are different – one a forecast the other a result – a confirmation of the forecast perhaps but not a negation of it.

I got the math now for that table using the total of all probabilities:

“I remind you… the two doors may hide…

Goat Goat

Goat Car

Car Goat”

a total of two cars and four goats, making chance-of-car two in six (or 1/3) and a field of two doors. Now take away one goat from each line and one door, Monty having revealed a goat behind one of the doors, and there are now two cars and one goat, making chance-of-car two in three (2/3) and a field of one door.

“the result does not affect the probability”

Not the probability in the series, but it does affect the outcome in that particular game. And the contestant is trying to win that game, not improve their performance over the series. To complicate the argument, the outcome in an individual MHP game is a change in probability. However in this instance it is the probability in that particular game, after a goat has been revealed, rather than in the series of games both past and hypothetical.

Palmer – arrogant is my middle name – arrogance is my default condition.

Some years ago I used to manage a big and professional web development team –

When interviewing developers for a job I would often ask them this puzzle – I had no idea it was called Monty Hall in those days – around 1998 – one or two would get it right first time but most would argue with me – if they argued hard I would bet them a bottle of Scotch Whisky and leave it there – they always gave me my Whisky eventually – sometimes after several months – but they always paid up.

You may send my Whiskey to me here in Reading UK – I can wait…

“When something does happen – as you say – a goat being shown – we do not need to discuss it further – we have a result – we know absolutely where the car is – it sits behind a door that has a 50% chance of a goat – whoopee!”

Just to be clear about the 50/50: if we open the one on, say, the left of two closed doors each time a new game starts it will indeed average 50% chance of a car. But if the game is changed and the goat door is opened for us, every time we open the other door it will reveal a car. 100% chance of a car. seeing the goat changes the probability.

@Richard

Ignorant is your nickname, ignorance is your mental condition.

I’ve cut and paste the relevant example for you – if it seems “more magic than logic” I’ll try and explain it for you (of course I’m assuming you’ll actually read it in the first place)

Let’s suppose Monty does indeed offer “a two for one deal”, and he makes this offer to 90 contestants say. It’s clear that 60 of those 90 contestants will win the car and 30 won’t. Let’s also suppose Monty asks each contestant to initially open just 1 of their 2 doors… to prolong the suspense.

(and we know for certain that behind each contestant’s 2 doors there is at least 1 goat)

Now, 30 of those 90 contestants will open a door and reveal a car straight off… Hey Presto! They become instant winners. The other 60 contestants will open a door with a goat. Since we know there are 60 winners in total, only 30 (or 50%) of those 60 contestants who open a goat door become winners, and the other 30 end up losers.

Therefore, if a contestant opens a goat door then their 2nd door only has a 1/2 chance of being a car, and the unpicked single door also has a 1/2 chance. In other words after a goat door (which we knew was there anyway) has been opened, what was a 2/3 chance has become a 1/2 chance, and what was a 1/3 chance has also become a 1/2 chance

Freddie – this…

I agree this – what you say… in a 2 door game… Which is not the MHP

But if the game is changed and the goat door is opened for us, every time we open the other door it will reveal a car. 100% chance of a car.

Naturally – the appearance of the only goat is a result – because we know immediately where the car is. We do not have to open the car door – we know it’s there.

But I reject this…

….seeing the goat changes the probability.

Seeing the goat just gives a result – the probability for each door is 50% goat and 50% car – because – there ate two doors – one goat and one car.

Probability can ONLY change if the numbers change – the goat appearing does not change the numbers – there is still 2 doors – one goat and one car.

Palmer – so gracious of you to follow my theory of a two for one swap… thanks…

I followed and generally agreed up to and including this…

Now, 30 of those 90 contestants will open a door and reveal a car straight off… Hey Presto! They become instant winners. The other 60 contestants will open a door with a goat. Since we know there are 60 winners in total, only 30 (or 50%) of those 60 contestants who open a goat door become winners, and the other 30 end up losers.

Well that’s what we would all expect 2/3 of them winning the car and 1/3 of them not winning it – it’s not difficult to follow so far but I deplore scenarios of multiple iterations – they just take too long…

Now for the meat in your sandwich… where is falls apart…

You say…

Therefore, if a contestant opens a goat door then their 2nd door only has a 1/2 chance of being a car

No – each door has a 1/3 chance of the car – the originally selected door has a 1/3 chance – each of the other two doors has a 1/3 chance – the door that’s still closed has a 1/3 chance and – difficult for you to accept I know – the door with the goat on view also has a 1/3 chance – it’s a losing door I grant you – but it represents a 1/3 chance of the car – as do all the doors.

The chance probabilities are set at the beginning – Three doors – Two goats etc the 3-2-1 relationship – unchanging 1/3 the car and 2/3 a goat for each and every door no matter what lies behind it – open or closed.

Probabilities are not changed by subsequent events in this simple MHP – what has happened does not retrospectively affect what might happen – they don’t change the starting price of a horse if it comes in last.

So in those 60 games where a goat is revealed you’re saying the contestants win 40 of them? Are you saying that Richard?

How can that be? 40 wins + 30 wins (when a car is revealed) = 70 wins in total – out of 90 games in total. That’s not 2/3 it’s 7/9.

“I deplore scenarios of multiple iterations ” – I’m sure you do since their evidence disprove yours “theory” (and I use that term loosely) that the odds remain unchanged at 2/3.

“Probabilities are not changed by subsequent events in this simple MHP” It is apparent that you don’t know the 1st thing about probability. Where did you learn about it? You should ask for your money back.

Palmer – you treat me bad you naughty chap…

Sure I know in a 90 door iteration – if they all elect to swap – then 60 of them will be successful and of them all – 30 of them will just get 2 goats – about 30 of them will get a goat in the first door they open and the rest will see a goat in their second door – You chide me without justification – I know full well that 30 & 40 sum to 70 – where did you get 40 from?

As for asking for my money back – my education ended in 1962 – Probability was never on the school curriculum in those days – I picked it up as I went along – the school of life – and I know virtually nothing of the subject – for a while I taught maths in college – I was a poor teacher and very disrespectful to management – I played Contract Bridge until recently and needed to concentrate hard and think quickly for three hours at a time – I guess this is where my ideas about probability were formed – establish the probabilities of a layout and be prepared to change as new information presented itself – as the cards fell. If there was a 63% chance that a certain player had a certain card – and that player didn’t have that card it made no difference – I would try to play in accordance with the best chance – the same is true in poker so I’m told.

Nevertheless I was club champion for 9 consecutive years 2004 until 2012 – not bad for a person weak in probability. I’m confident in my approach to the subject at the basic level – nothing academic though – but OK with Monty.

I’m sure your probability credentials are more advanced than mine.

It seems probable (67% probable) that we shall never agree – thanks nevertheless for the chat and the challenges.

Best wishes…

P.S you might consider taking up Bridge – it requires a devious and suspicious mind.

“about 30 of them will get a goat in the first door they open” I think you meant to say “about 30 of them will get a CAR in the first door they open (60 of them get a goat in the first door they open)”

“where did you get 40 from?” 40 is 2/3 of 60 (the games where the contestant reveals a goat)

If you’re now saying that in those 60 games (where the contestant reveals a goat) the contestants only win 30 games (which they do) then 30 out of 60 represents a 1/2 chance not a 2/3 chance. In those 60 games the car will be behind the single door not picked by the contestant 1/2 the time, or to put it another way the single door has a 1/2 chance of being the winner.

Before any door was revealed, the single door (the one not picked by the contestant) only had a 1/3 chance, now (after a goat has been revealed by the contestant from their 2 doors) it has a 1/2 chance , Its probability has increased from 1/3 to 1/2 by the revealing of a goat.

“I know virtually nothing of the subject”. And I’m trying to teach you something about the subject. You seem extremely reluctant to believe what I’m saying. I don’t know why, I know what I’m talking about.

This all seems pretty straight-forward to me. The probability appears to “change” because you’re being asked at different times in the game to consider two different probabilities:

First, whether the car is behind any one of the three doors. You’re being asked to consider three sets of one door each. Each one has a 1/3rd chance.

Then, whether the car is behind the door you picked (1/3rd chance), or behind *one of the remaining doors* (2/3rd chance). You’re being asked to consider two sets, one set containing one door, the second set containing two doors.

I think what confuses people is that they’re assigning the probability to the doors themselves, not the sets.

You still have to get from the set containing two doors to the single door with 2/3 chance of a car (which is a subset of the set containing two and is the whole point of the question ‘whether to switch’). Plenty of confusion left there I think …

395. “You still have to get from the set containing two doors to the single door with 2/3 chance …” That was me, having cleared my browser cache and forgotten to re-enter my name.

Doesn’t your experience in Contract Bridge answer your question, Richard: 392. ” – establish the probabilities of a layout and be prepared to change as new information presented itself – ” such as a card being revealed … or a goat door being opened?

392. ” – establish the probabilities of a layout and be prepared to change as new information presented itself – ”

Probabilities of layout established after first choice but before goat is revealed:

1/3, 1/3, 1/3.

Probabilities of layout after goat is revealed (new information) changes to:

1/3, 2/3, 0/3.

Freddie – I will discuss this with you no more – because all I do is repeat my single truth…

You say…

Probabilities of layout established after first choice but before goat is revealed:

1/3, 1/3, 1/3.

And I agree with that…

But not Your next…

Probabilities of layout after goat is revealed (new information) changes to:

1/3, 2/3, 0/3.

Here you have a 1/3 chance the car moving from one door to another – nobody explains why or how this can happen – it’s a figment of a confused mind – it moves for you only because it needs to move – no other reason or explanation – why does it not move to the selected door? why not in equal proportion to the two unopened doors?

What is the force – the cause of its movement? Seeing a goat – surely not?

It doesn’t move at all… It stays just where it started – It’s a horse that began the race at odds of 1/3 – and it comes in last – you would think it very silly if they reported the odds next day – Skewball – last – 0/3 and then added Skewball’s odds onto another horse – it’s betting Jim but not as we know it…

Of course probabilities can change – a wise player makes adjustments as new information is presented – in card games it happens all the time – card players become very familiar – they recall the drop of each card and can very soon work out the distribution of the concealed cards in each hand.

I maintain – here we go again Richard for the umpteenth time – that when Monty shows a goat we learn no new information because we were well aware of the goat – we may learn which specific door was hiding the goat but that’s hardly earth shattering news. Door 2? Door 3? So what?

Play this game just once in your mind – imagine that when Monty shows the goat you’re looking the other way – you close your eyes and play the obstinate fool (it’s very easy for me to do that) – and then you elect – as you should – to swap – There’s a goat to be seen but you’re not looking at it – you know it’s there – well you knew that before the door was opened.

Is this silly variation of mine any different to the model of the game you used before? – is the result any different? Can you get – in your mind – to a 2/3 chance of the car without seeing the goat?

Do you really need to see the goat?

Do not answer me directly – I shall reply no more…

Thanks & goodbye…

@Richard

“Here you have a 1/3 chance the car moving from one door to another – nobody explains why or how this can happen ” . I thought I had in my last comment; either you didn’t read it (again), you’re simply too stupid to understand it, you’re quite happy in your ignorance, or you’re too arrogant to admit you’re wrong

“Do not answer me directly – I shall reply no more…”

Thank God for that.

A contributor here said…

…nobody explains why or how this can happen ” . I thought I had in my last comment;

No you didn’t explain it – you just think you did.

When a horse comes in last they do not take the starting odds (probability) and assign that chance to another horse – that would be very silly don’t you think?

One more thing – if somebody doesn’t understand what you say it may be because they’re stupid but it might also be because your explanation was poorly framed – or – dare I say it? Wrong..

“When a horse comes in last they do not take the starting odds (probability) and assign that chance to another horse – that would be very silly don’t you think?”

And what has that got to do with the MHP? You ever hear of “in race betting”? The odds change all the time, a bit like in the MHP

If you didn’t understand what I wrote, then I’ll explain it. What exactly didn’t you understand? Do you disagree that only 30 of the 60 (or 50%) contestants who reveal a goat win the car? If so, why?

““I know virtually nothing of the subject (probability)”. Indeed.

I don’t think anyone is saying that each door doesn’t always, throughout the game, have a 1/3rd chance of containing the car.

The question is whether you stick with the door you opened (1/3rd chance), or choose ALL THE OTHER DOORS (2/3rd chance). You *can* choose all the other doors, because Monty has ensure that the remaining set of two doors contains only one door to which you can switch.

Again, don’t think of the odds being on the doors. They’re on the choice sets.

@Dave

“I don’t think anyone is saying that each door doesn’t always, throughout the game, have a 1/3rd chance of containing the car.”

That’s exactly what I’m saying: that a door that is KNOWN to contain a goat (the door Monty has opened) has a zero chance of containing the car …..because we can actually see the goat, it’s not simultaneously 66.7% goat AND 33.3% car – it’s a GOAT.

@Palmer:

“That’s exactly what I’m saying: that a door that is KNOWN to contain a goat (the door Monty has opened) has a zero chance of containing the car”

Understood. I guess I was, in a poor way, trying to get at the crux of the understanding problem: Those who say every door always has a 1/3rd chance of containing the car are saying “‘chance’ means the odds of the car being behind any door, regardless of any new information being revealed.”

“Here you have a 1/3 chance the car moving from one door to another – nobody explains why or how this can happen – “

Might you find during a card game that, for example, you calculate one player has 70% chance of holding the Ace of Spades. Then you are dealt the Ace of Spades and there is now 0% chance the other player has it.

In a series of MHP games each door starts with a 1/3 chance-of-car. But in any one individual MHP game when one door has been dealt a goat you have new information that makes a difference. Richard, consider one game, not the series.

Greetings Dave…

Are you saying that a closed door with a goat behind it does have a 1/3 chance of the car – but when the door is opened the 1/3 chance no longer exists – that a door’s chance of car or goat is altered by what lies behind it – but only when we get to see what’s behind the door – is this what you’re saying?

Granted that an open door with a goat has no chance whatsoever of having the car but that door still represents a 1/3 chance of the car. The contents do not dictate the chance probability – chance probability is due solely to there being Three Doors – Two Goats – and One Car – chance probability allows for either eventuality – a goat does not change the 1/3 chance of a car – a car does not change the 2/3 chance of a goat – the door has the probability – not the contents.

” an open door with a goat has no chance whatsoever of having the car “

” a goat does not change the 1/3 chance of a car “

Make up your mind Richard, either it has “no chance” or “1/3 chance”. Which is it?

Hi Richard;

“Are you saying that a closed door with a goat behind it does have a 1/3 chance of the car – but when the door is opened the 1/3 chance no longer exists – that a door’s chance of car or goat is altered by what lies behind it – but only when we get to see what’s behind the door – is this what you’re saying?”

A door’s chance of car or goat is altered by new information on what lies behind it.

We may just be arguing semantics. Consider this. If a door is closed, is there a chance that an object is behind it? If the door is opened and you can see there’s no object behind it, is there then a chance that there’s an object behind it?

@Richard

Replace the doors, goats and car with a bag containing 2 red balls and 1 white ball.

You pick a ball out of the bag but don’t look at it.

Monty deliberately removes a red ball from the bag and shows it to you.

What is the probability that the remaining ball in the bag is the white ball?

(Hint: the answer is 2/3)

Richard, here are some examples of shifting and changing chances. Note that each event may have its own unique result; probability is measuring the likelihood of particular results over a series of events, not the result we observe in one individual event.

Two equal cars, Red and Blue, repeatedly have a race. They win four each; Blue has 50% chance in the next race. Red wins nine in a row; Blue has 90% chance in the next race – a punter arrives to bet on Blue; he accepts 50/50. Eventually Red blows a tyre (about once in 100 races) and cannot catch up which gives Blue 99% chance of winning that one. In one race we watch Red pass the finish line ahead of Blue who now has 0% chance in that particular race. Over many races they win 50% each.

Come off it Freddie – you speak with forked tongue…

…probability is measuring the likelihood of particular results over a series of events, not the result we observe in one individual event.

You could not be more wrong – probability is the likelihood of an event occurring – and if you feel the need to confirm it you can run a series of multiple tests if you wish – the test series has nothing to do with it – just the numbers 3-2-1.

The standard MHP has Just Three Doors – Two Goats and One Car

The probabilities are 2/3 a goat and 1/3 the car

If you swap away from the 1/3 single-door then by definition you swap into the 2/3 double door

Swapping – as we (nearly) all accept – provides a 2/3 chance…

Multiple iterations are just a waste of time – proof of the 2/3 is for babies

In the standard MHP there is only one go – you don’t get multiple games.

The MHP is so kindergarten-simple it does not require multiple simulations to confirm the 1/3 versus 2/3 probabilities – so why bother?

Your car race analogy is very interesting but it has no connection with the MHP – If the two cars had an equal chance then they would cross the line at the same time surely?

It would be better as a coin-toss

They each – according to you and common sense – have a 50% chance – and one of them wins 9 in a row – and then you make the MOST OUTRAGEOUS statement that in race 10 the car that has yet to win has a 90% chance of first place – do you really mean this? You you actually believe it?

And you have the Gall – the bare-faced cheek to lecture me about probability ?- go to the back of the class and sit on the naughty seat until you get your mind synchronized with reality.

@Richard

Replace the doors, goats and car with a bag containing 2 red balls and 1 white ball.

You pick a ball out of the bag but don’t look at it.

Monty deliberately removes a red ball from the bag and shows it to you.

He then says “Do you you want to keep your ball or swap it for the one left in the bag?”

What is the probability that the remaining ball in the bag is the white ball?

I might have asked this question before

411. “If the two cars had an equal chance then they would cross the line at the same time surely?”

Richard, we are discussing probability not Utopia. All this means is that the circumstances which might give one car an advantage in one race (eg the other has a tyre blowout, driver distracted, wheel-slip on an oil stain) happen as often to one as the other so they both average the same number of wins. And an occasional draw might well be declared. Heads and tails have an equal chance in a toss of a fair coin; do they ever both come up at once?

Somebody – an anonymous person -said…

Richard, we are discussing probability not Utopia.

And if in reality two card DID have an equal chance they would always cross the line in unison – my point is that two cars can not possibly have an equal chance – I was making a silly joke to show the ridiculousness of the Two-Car scenario – it has nothing whatsoever to do with the Mont Hall Problem.

I am not discussing Utopia – just making fun of Freddie – who seems to deserve it when he says a certain car – with an equal chance to another – has a 90% chance of winning because the other car has just won 9 in a row.

Before the TV show begins – before the audience file in to take their seats – a stage hand will set up the three doors – somebody – it doesn’t matter who arranges the car and goats. The probabilities for Goat or Car are set at this point and will never change – BECAUSE – it begins and ends with Three Doors – Two Goats and One Car – The doors do not go away – The goats do not go away – The car does not go away – the 1/3 versus 2/3 probability remains constant forever and ever until the end of time no matter what happens.

The MHP is not affected by oil on the track or by tyres blowing out – no siree.

Heads and tails have an equal chance in a toss of a fair coin; do they ever both come up at once?

Of course not – it was a JOKE! To point out the ridiculous statement about two CARS having an equal chance – idiotic – if they really had an equal chance there would be no point in having a race would there?

” – just making fun of Freddie – “

Here’s a fun question Richard: you flip a fair coin and get five heads in a row. The chance of a random ‘tails’ is 50% while the chance of six heads in a row is 1%. Is chance of ‘tails’ 50% or 99% next throw?

“The probabilities for Goat or Car are set at this point and will never change – BECAUSE – it begins and ends with Three Doors – Two Goats and One Car – The doors do not go away – The goats do not go away – The car does not go away – the 1/3 versus 2/3 probability remains constant forever and ever until the end of time no matter what happens.”

An ignorant comment made by an arrogant fool who admits he knows nothing of the subject he’s talking about.

What is the probability that the remaining ball in the bag is the white ball Richard?

Freddie – I do not need your Fun Question thank you very much – I am not here to sit your juvenile test – your questions – if you have a point then make it – you (we) get nowhere if you ask questions – to give answers and explanations is best – don’t you think?

Now – with a fair coin the probability of one side or the other is 50% – why do we need to go here?

The chances of having the same result – Heads – Heads – Heads – Heads – Heads (five in a row) is 0.5 to the fifth power – I can not be arsed to work it out – a very small number – why do you need to lead me down this dead end?

And on the next throw the chance of heads or tails is 50% but the likelihood of the sixth Head in a row is an even smaller number than it was before – half the size.

This has sod-all to do with the MHP as far as I can see – it’s about tossing a coin – six independent events – independent – you know – not connected with one another.

The MHP is just one event – they don’t do it several times over to see what happens – the contestant just gets the one go.

“The MHP is just one event – they don’t do it several times over to see what happens – the contestant just gets the one go.”

They do it several times over with a different contestant each time. Probability does the same but in theoretical terms. Each game is one event in a series. The series has probability, each event has results.

@Richard

“The MHP is just one event – they don’t do it several times over to see what happens – the contestant just gets the one go.”

The MHP is a probability puzzle. It can be solved logically or mathematically, and can be empirically tested via computer simulations (and has been millions of times).

Richard, what is the probability that the one remaining ball in the bag is the white ball? Are you avoiding the question for some reason?

No Freddie – each game has the same probability as every other game – most people accept it’s a 2/3 chance to swap – they do not place a condition on swapping and say – for instance – it’s a 2/3 chance to swap except on the Nth game.

If it’s done several times over the results will no doubt average out to 2/3 the swap and one third for not swapping – but the MHP is a single event each time – the question is…

Is it best to swap or best to stick or does it make no difference?

The question is not – what happens if we do it N times?

Doing it over and over is for those sad souls who can’t come to terms with the 2/3 better to swap concept – the doubters who can’t work it through for themselves. And I know that you’re not one of them – THEREFORE – there is no need to mess around with multiple iterations of Cars or Coins or Goats – it just wastes time and energy – don’t you think?

Plamer – please return to planet Earth

Richard, what is the probability that the one remaining ball in the bag is the white ball? Are you avoiding the question for some reason?

Yes – I’m avoiding the question because I can’t be arsed to answer it – for me it’s a non-question and I am not here to answer juvenile questions – but if there are 2 red balls and one white one then there’s a greater chance for the double colour to appear more often and the single colour less often – Lord give me strength! If you know the answer to your little problem why do you need me to help you with an explanation? Is it lack of confidence?

If you need a computer simulation to solve Monty Hall then there’s no help for you – fools may wish to try it several times over – but they end with the same result as if they’d just worked it through – ONCE – in their head – it pays to swap.

Please – Please – You and Freddie both – stop asking questions – put up a proposition of your own – or critique my views (in grown up language) but drop the questions – I can’t be bothered to answer them and they NEVER have any bearing on the standard MHP.

@Richard

“Yes – I’m avoiding the question because I can’t be arsed to answer it “

More like because it proves your “theory” (that the probabilities in the MHP can never ever change) for the pile of crap that it is.

Just as you avoided the question why the 60 contestants who reveal a goat only win 1/2 of games and not 2/3.

As you said “I know virtually nothing of the subject”.

Ignorant, arrogant and stubborn, that’s a real winning combination you’ve got going for you there Richard.

420. “… the question is…

Is it best to swap or best to stick or does it make no difference?”

Richard, the MHP is “having been shown a goat, is it to your advantage to switch?”. If no goat is revealed, the answer to the question “Is it best to swap or best to stick or does it make no difference?” is that it makes no difference.

In the aptly-named post 321. you claim “When Monty reveals a goat it is no help whatsoever”. Without re-reading every post since, my impression is you have stuck to this. However, revealing the goat changes the chance-of-car of the remaining unchosen door from 1/3 to 2/3. Not revealing a goat leaves it at 1/3.

While my math is not always accurate and I am no expert (but hey, those were the people left with egg on their faces in the Vos Savant affair), I have tried to show through examples how chances change as information becomes known.

Why not try answering Palmer’s question – it looks to me like another example of how information changes chance.

@Freddie

“However, revealing the goat changes the chance-of-car of the remaining unchosen door from 1/3 to 2/3″

Not according to ‘Richard The Arrogant’ (or is that Ignorant?) it doesn’t. When Monty picks a RED ball out of the bag and shows it to you, Richard thinks that it still has a 1/3 chance of being the WHITE ball.

Dear Freddie

I am not motivated to even look at Palmer’s question let alone answer it – and it is not your position to tell me which third-party questions to answer – if it’s that important to you then be my guest – go ahead and answer it – I am in a don’t give a care situation.

Now – fro the last time I hope – I designate the selected door as Door 1 – we forget about it and concentrate on the other two doors – watch closely now….

The doors 2 & 3 may be hiding in combination…

Goat – Goat

Goat – Car

Car – Goat

See? a 2/3 chance of there being the car behind those two doors but….

Absolutely 100% guaranteed dead-cert never fails – there’s always a goat behind at least one of them – we can work this out BEFORE Monty opens a door – Yes – yes I know that he’s obliged to show a goat – and we know in advance that the rules oblige him to do that.

BUT WE DON’T NEED TO SEE THE GOAT TO KNOW THAT THERE’S ALWAYS A GOAT TO SEE – sorry for shouting but it’s important – we don’t need to see a goat to know that one’s there – there is always one there – ALWAYS!

When Monty opens a door and shows a goat we do learn something – we learn that behind either door 2 or perhaps door 3 there’s a goat. This IS new information I admit but it’s pretty useless information – not the sort of information that could possibly change any thing – could not possibly alter probabilities – goat behind Door 2? – goat behind Door 3? – So what? What does it matter the door number?

Just pretend you’re not looking – you get to the point where Monty shows the goat but you refuse to acknowledge that he’s done so – you ignore his goat – what do you do next? Why you swap of course – you shoot for the 2/3 chance – you do not need to see the goat – THEREFORE the revealed goat changes nothing.

And why is this? Because the swap offer is to TWO Doors not one.

But – if it was to just one door – I don’t accept that it is – but suppose it was to one door – then you can only swap to one door – the one that’s still closed – so I ask again – how does the revealed goat affect the actions of a contestant? – he swaps to one door either number 2 or number 3 – or he swaps to both doors 2 AND 3 – the revealed goat does not guide his hand or inform his decision making because – between any two doors there will ALWAYS be at least one goat.

At the time Monty makes his swap offer the choice is either – two doors one open and the other closed – or – One closed door.

You say – Not revealing a goat leaves it at 1/3 – well certainly if the swap is one-for-one – But if the swap is – as I maintain – one-for-two then the goat makes no difference.

Now why do I say it’s one for two? Because by accepting a one for two swap I do not have to invent the fantastical movement of a 1/3 chance from door 3 to door 2 – or vice versa – I just sum 1/3 and 1/3 and arrive at 2/3.

“Because the swap offer is to TWO Doors not one.”

The swap offer is to the single unopened door – read the problem definition. Tediously repeating the same factually incorrect statement, doesn’t make you right. What do you call this fixation “Proof by Assertion”?

“by accepting a one for two swap I do not have to invent the fantastical movement of a 1/3 chance from door 3 to door 2″ But you do have to accept the lunancy that a RED ball retains a 1/3 chance of being the WHITE ball

“BUT WE DON’T NEED TO SEE THE GOAT TO KNOW THAT THERE’S ALWAYS A GOAT TO SEE”

We know that.

The contestant can only switch to one door. The MHP does not allow them to choose two doors and select the best of them. Their chance of a car increases if they see a goat BEFORE CHOOSING TO SWITCH and remains the same if they do not.

The contestant can only switch to one door. The MHP does not allow them to choose two doors and select the best of them.

It’s not defined – but no matter – if one door is open and showing a goat the effective move in terms of new discovery is the door not showing a goat – if you want to think it’s one for one – no matter – if you want to think it’s one for two – as I do – also no matter – there’s really not much difference is there?

“Their chance of a car increases if they see a goat BEFORE CHOOSING TO SWITCH and remains the same if they do not.”

Well the last part we can only agree – don’t switch = 1/3 chance the car – you did not need to say…

As for the decision to swap – is it not generally accepted that it always pays to swap ? How does the revealed goat inform our decision to swap or not swap?

Goat behind door 2 – definitely swap – Goat behind door 3 – also definitely swap.

Goat not shown – definitely swap. Goat shown but not looked at – definitely swap.

Under what circumstances might a wise contestant not swap? Does the wise contestant actually need to see the revealed goat? If a goat door is opened – say door 2 – should the contestant swap to door 3? What other swap option does he have? If a goat door is opened – say door 3 – should the contestant swap to door 2? What other swap option does he have.

Just how is that revealed goat helping with all this decision making?

Answers on a postcard should be sent to me – here in Reading UK – a goat is offered to the person sending in the most amusing and creative suggestion.

Without … seeing … the … goat … which … of … the … two … doors … would … you … switch … to ???

Without … seeing … the … goat … which … of … the … two … doors … would … you … switch … to ???

Why do you ask?

If the goat door had been opened – and in the Standard MHP it always is opened – but suppose I had my eyes tight shut – then I would Swap to the unopened door – I would just say – Swap Please That is my Final Answer – but in my own private world I would really be swapping to both doors.

If on the other hand a goat door had not been opened – I would just say – Swap Please That is my Final Answer – because in my own private world I would really be swapping to both doors.

A few moments thought should tell you that the offer to swap to either one or the other of the two unopened doors does not arise.

You should also recognise that the swap offer is not defined as either one door or two.

In my view of the world – I swap – I always swap – I take no notice of Monty’s blessed goat and and I swap to two doors because nobody says I can’t – I do this to achieve a chance of 1/3 + 1/3 – the sum of which equals a 2/3 chance – I do not have to worry my simple brain with messing around with moving a 1/3 chance from the goat-door to another place – it requires no effort. If you wish to do it another way – be my guest – but allow me my little foibles – we each get to the same place – my way might tax your brain but it keeps my brain cool.

You may be interested – or you may not be interested – to know that I’ve enrolled in the Adult Education Course here in Reading – starting January…

https://www.conted.ox.ac.uk/G100-52

“You may be interested – or you may not be interested – to know that I’ve enrolled in the Adult Education Course here in Reading – starting January…”

Heaven help them.

Heaven help them.

Freddie – when I get there I shall tell them about the friends I’ve made here on Better Explained…

Thanks for the entertainment – I elect to Swap – That is my final answer.

“I elect to Swap – That is my final answer.”

But without goat being revealed you have one more answer to give – which door do you wish to swap to?

Perhaps you should take a class in a place called ‘Math’ instead of ‘Reading’.

Freddie – I thought I had made myself clear on this – I swap to both of them – both doors – because I can – It’s not your place to dictate here – I don’t think you’re the king of swapping – I decline to play by your made-up restrictive rules – I am an anarchist – I reject all authority but I do pay my taxes – taxes are a good idea – the foundation of society – I swap to both weather you like it or not – Monty doesn’t mind – he privately winks as I do it – he understands you see.

“But without goat being revealed you have one more answer to give – which door do you wish to swap to?”

I thought we were talking about the Monty Hall problem?

“I thought we were talking about the Monty Hall problem?”

We are. Richard is arguing that revealing the goat is not material to the solution of the MHP; I am arguing that it is. Richard is suggesting that it is possible to interpret the solution of the MHP as ‘swapping to both doors’ while I am suggesting it is necessary to nominate a single door. Both of us are talking about the MHP.

” – I thought I had made myself clear on this – I swap to both of them – both doors – because I can – “

In that case may I offer a new solution to increase the chances to 100%: the contestant simply chooses all three doors – because they can – and opens them until they find a car.

Re Freddie

“We are. Richard is arguing that revealing the goat is not material to the solution of the MHP; I am arguing that it is. Richard is suggesting that it is possible to interpret the solution of the MHP as ‘swapping to both doors’ while I am suggesting it is necessary to nominate a single door. Both of us are talking about the MHP.”

I thought Richard was just saying that you don’t need to see that there was a goat behind the door Monty opened, because you know there was, if you know that Monty will not open the door containing the car.

“I thought Richard was just saying that you don’t need to see that there was a goat behind the door Monty opened, because you know there was, if you know that Monty will not open the door containing the car.”

The difference between seeing a goat and knowing it is there because Monty will not reveal a car seems too inconsequential to discuss – in both cases it’s a goat behind an open door. Hanging an honest sign saying ‘goat’ on the door would get round the need to even open it.

However Richard asserts that he can make an advantageous choice without any door being opened, and that the chance of the ‘final’ door does not change to 2/3 from its original 1/3. He is making fundamental points about the way probability behaves in the MHP.

Dave – at least you don;t ca;; me stupid and ignorant – thanks for that…

Richard’s problem is an inability to understand – accept – believe – that a 1/3 chance of the car can transfer from the goat-door to it’s friend – the other un-chosen door. What makes it move? Why does it move? Why does it not move to the chosen door? Why not in equal 1/6 part to each of the closed doors?

I have yet to be convinced that this can happen – sure many people say that the other unopened door now has a 2/3 chance of the car when the goat-door is opened but they sort of fudge the mechanism of the transfer – as George Bush was given to say – fuzzy math.

Now this transfer is a pretty basic thing – it should be easy to explain in words but I’ve never seen it done – people say that they’ve just shown how it but they have yet to convince me of the feasibility of it or the truth of it or the mechanism.

This phantom transfer troubles me – but I have a way round it.

If I have – in my mind – the swap offer being to both of the other two doors together – each with a 1/3 chance of the car then the sum chance is 2/3 and I need not worry about this fanciful magic of a single door suddenly being worth a 2/3 chance of the car.

Another advantage is that – if the scenario is as I describe – we swap to two doors rather than one – the goat is of no consequence – we as a contestant own the goat – the goat contributes to our success by being behind a door that carries a 1/3 chance of the car.

Clearly what lies behind a door does not govern that door’s chance of a goat – a closed door with a 2/3 chance of a goat – you can’t in all honesty say – oh no – it only has a 2/3 chance of a goat if its got a goat behind it – it has a 2/3 chance of a goat irrespective of what lies behind it surely – the 1/3 and 2/3 relationship being due to the combination of Doors – Goats and the single car.

As the Monty Hall game progresses nothing changes – there are still Three doors – Two Goats and a car from start to finish – so I maintain the probabilities of 1/3 the car and 2/3 a goat remain fixed permanently to each door – unchanging. Many find it very difficult to accept that an open goat-door has a 2/3 chance of a goat and consequently a 1/3 chance of the car – they say -oh no – it has a 0% chance of the car if it’s got goat behind it – they confuse the chance probability – what might happen – with a result – what has happened.

I try to illustrate this with the horse race analogy – when a horse comes in last – they don’t set the horse’s starting price to zero and assign it to another horse.

Rocks get thrown at me about swapping to two doors – nobody says you can have two doors Richard – the swap offer is to one door. Well in my world it’s a one for two swap and I get the sum of two 1/3 doors and I know absolutely that at least one of those doors will hide a goat – so if I get to see it – that’s fine – and if I don’t get to see it that’s also fine – I do not need to see something I know to be there to know that it’s there – I trust my judgement well enough not to bother about the goat.

Much is made about the revealed got bringing in important new information – vital in decision making – what decision making? If you stick it’s a 1/3 chance – where’s the difficult decision?

Swap to two doors – always swap no matter what – ignore goats.

The network has unearthed some early footage from the show. Here are the transcripts.

Episode 1

Monty: Before the break, Freddie chose Door 1. Now I’m going to open Door 3 – hey it’s a goat! Freddie, do you want to give up your original choice and switch?

Freddie: Yes please Monty, I choose Door 2.

The Voiceover of Fate: And Freddie now has Door 2 with 2/3 chance of winning the car.

Episode 2

Monty: Welcome back viewers, and Richard you chose Door 1. Now I’m going to open …

Richard (interrupting): No need Monty – leave them all closed. I don’t need to see anything.

Monty: Er, OK Buddy, if you’re sure? You, er, wanna stick with Door 1 or switch?

Richard: Switch Monty, switch every time.

Monty: So … which door?

Richard: Both. I choose both doors Monty.

Monty: Sorry Buddy, it’s one door or nothing. You gotta choose one door.

Richard: Door 3 Monty. No wait, Door 2. Or 3. Please can I have both Monty?

The Voiceover of Fate: And Richard now has 1/3 chance of winning the car whether he chooses Door 2 or 3 or goes back to his original choice, Door 1.

And now a word from our sponsors …

(music) “Bet-ter Ex-plained – we ex-plain things bet-ter”

@Richard

“Richard’s problem is an inability to understand – accept – believe – that a 1/3 chance of the car can transfer from the goat-door to it’s friend – the other un-chosen door”

A) You are told that there is a 66.7% chance that there’s a car behind EITHER DoorA OR DoorB

B) You are subsequently told that there is definitely NOT a car behind DoorA.

Now, most people would deduce from A) and B) above that:

C) There’s a 66.7% chance that there’s a car behind DoorB

But not you Richard: you seem incapable of making that simple logical leap from Step B) to Step C). Why is that?

For someone who purports to be something of a ‘whizz’ at Bridge (a game replete with logical inferences and deductions) I find that quite astounding.

PS. The question was rhetorical Richard, I don’t expect an answer from you, I’m sure I’d be disappointed if I did

Dear Palmer – You said this…

A) You are told that there is a 66.7% chance that there’s a car behind EITHER DoorA OR DoorB

1) Nobody tells us this – the wise contestant might deduce it but nobody says it…

2) The capitalized word OR is incorrect – the word AND is more accurate – the 2/3 chance applies to both doors together – 1/3+1/3 = 2/3.

It’s not 66.7% on one door alone but 66.7% on two doors together.

B) You are subsequently told (shown) that there is definitely NOT a car behind DoorA.

The opening of the door to reveal a goat – I’m OK with that.

Now, most people would deduce from A) and B) above that:

C) There’s a 66.7% chance that there’s a car behind DoorB

They might deduce this but the 66.7% chance applies to two doors together – not one door on its own.

I contend that the opened door’s 1/3 car and 2/3 goat chances stay permanently with the door – only in an observer’s mind might they move about – the 1/3 chance moves off to the door next door? How can this be so? And what happens to that door’s 2/3 chance of the goat?

No – these probabilities are fixed – the numbers that show what might happen in the future – the numbers that allow for either eventuality in proportion – they stay the same because they’re determined by the 3-2-1 Door Goat Car combination – and the 3-2-1 door combination does not change . The contents of a door do not determine that door’s probabilities – it’s just the numbers that determine probabilities – not what we see.

It’s a nice try Palmer but I still don’t see how any door can have a 2/3 chance of the car- I’m not convinced.

Now – if you allow in your mind the swap offer being from one door to two doors then all the problems of the flying 1/3 chance the car evaporate – the 1/3 chance does not need to be moved about – you don’t even need to see the goat – or even consider it if it does get shown.

Yes I know that Monty doesn’t actually say it’s to two doors – but he doesn’t say that it isn’t either.

If you reject one door with a 1/3 chance what are you left with ?

Yes – two doors – each with a 1/3 chance.

One minus 1/3 = 2/3 It works for me…

Freddie – your excerpts from the historical shows made me smile – but neither historical or hysterical – jut smile-worthy.

The MHP we discuss here is vos Savant’s invention – on his show Monty used to open a door to reveal a goat and then make an offer to buy back the originally selected door from the contestant – swapping never entered into it.

I think you invented those conversations…

https://www.youtube.com/watch?v=VuuTc_9kpYE

At the 58 second point Monty himself talks about buying the door rather than making a swap offer. He later says that he never did make any offer to swap.

“I think you invented those conversations… “

… and I based them on the hypothetical Vos Savant question, not on ‘Let’s Make a Deal’.

I think you invented the contestant’s entitlement to choose two doors – which is not permitted in ‘Let’s Make a Deal’, the Vos Savant version or any reasonable interpretation of the MHP rules. Perhaps if you re-wrote ‘Episode 2′ from post 441. as you would see it happening it would enlighten me.

I think you invented the contestant’s entitlement to choose two doors – which is not permitted in ‘Let’s Make a Deal.

Well I DID invent the two door swap – to solve a problem I have.

Just as others invent the single door swap.

Just because an authority figure (Monty) fails to clearly define the swap offer and leaves it open to interpretation – then I interpret it in the way I want.

And when I do consider it as a one for two swap then all doubts and uncertainties fly away – Two doors together are a 2/3 chance – No issue with the 1/3 chance magically moving to another door.

How do the doors know that the appearance of the goat is significant? They don’t of course – they have no power to move their probabilities around – nobody even mentions what happens to the 2/3 chance of a goat when a goat gets shown.

And the other advantage in the 2 door swap is that you don’t need to consider the goat – you don’t even need to see it.

And lastly – when he shows the goat and then asks do you want to swap?

What options are there?

Well you could be swapping to a single closed door – that’s the only other unopened door – so not much choice really – or you could be swapping to that unopened door and the opened goat-door – no difference there either.

It seems to me that the revealed goat can only inform the swap or stick decision – the revealed goat does not point the contestants attention in any specific direction – at this point it’s swap or stick nothing more – what difference that you swap to my version of two doors – or – the goat’s next door neighbor?

I’ll tell you – no difference whatsoever…

The man himself…

https://www.youtube.com/watch?v=c1BSkquWkDo

Monty tells us that there’s no Swap offer – he tells us at the 54 second point and again at 1:41 that there was no swapping in his show.

The swap offer is open to personal interpretation.

“Monty tells us that there’s no Swap offer – “

Richard, we are discussing the MHP (as posed by Vos Savant in Parade), not ‘Let’s Make a Deal’. In the MHP there is an explicit and material swap offer. It is not open to interpretation given the other facets of the question and the clarity with which Vos Savant subsequently defended her position. It is not reasonable to claim it offers a two-door-for-one swap without the contestant nominating a single door of the three as their final choice, though I believe this is what I believe you are suggesting. I illustrated that in my ‘Episode 2′ scenario and have invited you to rewrite ‘Episode 2′ your way as a way of clarifying any misconceptions I might have about your argument.

@Richard

I said in my previous comment that “You are told that there is a 66.7% chance that there’s a car behind EITHER DoorA OR DoorB”

You countered with “The capitalized word OR is incorrect – the word AND is more accurate”

So if we substitute OR with AND in my comment we get “You are told that there is a 66.7% chance that there’s a car behind DoorA AND DoorB” That would seem to imply there’s 2 cars, but that’s just me I guess.

“I contend that the opened door’s 1/3 car and 2/3 goat chances stay permanently with the door” .

If we restate this comment in the context of the 2 RED balls and 1 WHITE ball in a bag problem (identical in all respects to the standard MHP) we get “I contend that the RED ball Monty shows us has a 1/3 chance of being the WHITE ball and (only) a 2/3 chance of being a RED ball ” That may make sense to you Richard, but not to anybody else.

“then all the problems of the flying 1/3 chance the car evaporate”. The ‘problems’ exist only in your mind Richard, but then you’re the only person I’ve ever seen assign probabilities to events that already happened and the outcome of which are already known.

“Yes I know that Monty doesn’t actually say it’s to two doors – but he doesn’t say that it isn’t either.?

(From Wiki) As posted in Marilyn vos Savant’s “Ask Marilyn” column in Parade magazine in 1990:

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?”

Mmm ….. He doesn’t say “Do you want to pick door No. 2 AND door No. 3?” does he? Still make up your own rules Richard if it makes you happy, it still won’t make your maths anything other than fabricated nonsense.

OK Freddie – here’s my version…

Monty opens a door to reveal a goat – I’m looking in the other direction with my eyes shut tight – I have my back to the doors…

Monty asks, “Richard – you ignorant and arrogant worthless contestant – do you want to give up your selected door and swap?”

Richard responds – “Yes Monty – I’ll swap thank you very much – and you can stick that goat where the sun don’t shine for all I care.”

Freddie – when Monty show a goat and makes his offer to swap – what are the realistic options?

Firstly: Not to swap

Secondly: To swap to the other closed door

Lastly: To swap to BOTH the closed door and the opened goat-door together

I see no difference between the second option and the last – they both get the same door opened – do they not?

And – we come back to it again – IF the swap offer is to both doors then we need not bother ourselves with the goat.

And IF the swap offer is to just one door then there’s no choice to be made if we elect to swap – there’s only one place to go.

“I see no difference between the second option and the last “

Therefore:

a) If you swap “to the other closed door” you win 66.7% of the time,

b) If you swap “to BOTH the closed door and the opened goat-door together” you win 66.7% of the time.

So the “opened goat-door” adds precisely zero to your chances of winning

… which we knew all along, since it has 0/3 chance of being a car (with it actually being a goat )

“Lastly: To swap to BOTH the closed door and the opened goat-door together”

I still don’t understand what you think will happen after you have chosen ‘both doors’ Richard. There must be another rule that says what happens. Is it:

(a) Both doors are opened and all their contents are given to the contestant,

(b) The contestant is asked to nominate one of the doors and is given its contents,

(c) Monty nominates one of the doors and the contestant is given its contents,

(d) Both doors are revealed and the contestant is invited to choose the contents of one of them,

(e) Some other rule?

I still don’t understand what you think will happen after you have chosen ‘both doors’ Richard. There must be another rule that says what happens.

What do you mean – “there must be another rule that says what happens” ?

There is no rule that says there must be a rule – there is no must in it – in my world I set the rules – if it’s not explicitly stated then it’s not explicitly disallowed.

And I don’t mention the two doors – I just say -

“Swap please – that is my final answer.”

Then I reach out and begin door opening expecting all the time to encounter at least one goat.

So – I do this…

(a) Both doors are opened and all their contents offered to the contestant,

If you reject one door then you take the contents of the other two – two goats perhaps – or a goat and a car.

“Both doors are opened and all their contents offered to the contestant,”

Thank you for that clarification Richard. It explains your repeated contention that there is no need for, or point in, Monty revealing a goat. However I think you have answered the RBP (in which doors retain the same chance and two may be chosen), not the MHP (in which two of the doors’ chances change and only one can be chosen).

Dear Palmer… You wrote…

Therefore:

a) If you swap “to the other closed door” you win 66.7% of the time,

b) If you swap “to BOTH the closed door and the opened goat-door together” you win 66.7% of the time.

So the “opened goat-door” adds precisely zero to your chances of winning

EXACTLY – I knew that with enough time and patience the penny would drop in your mind – the goat is a phantom – a figment when it comes to decision making – the goat does not inform the Swap / Stick decision – and if there’s a goat on view – I ask you in all seriousness – if you swap what can you actually swap to?

One door or both – it really doesn’t matter. The goat’s useless in our search for the car…

Two doors both closed – No difference (me)

Two doors one showing a goat the other closed – No difference (also me)

One closed door on its own – No difference (you)

Well done you… for recognising that the goat makes no difference…

And here’s the nub of it Freddie…

You say – you claim this…

the MHP (in which two of the doors’ chances change and only one can be chosen).

Here we disagree – First the statement – only one can be chosen – what? when you swap from one door to the other – the Stick or Swap decision – when you see a goat and a closed door – and you say only one can be chosen – seems like stating the blindingly obvious to me – can’t imagine the revealed goat being much help in deciding which door to swap to – can you?

As for the probability chances of a door being able to change – that’s a magic trick far beyond my imagining . Shifting the starting odds to another door just because the door didn’t win? – most unusual.You must explain how this can be so – the triggers – the mechanism – the prime mover. Do the doors arrange it or is it down to the observer? And why is the 1/3 chance not split equally between both of the other doors? You know – the 50/50 thing?

” – the Stick or Swap decision – when you see a goat and a closed door – “

Richard, you have frequently argued against seeing a goat:

“428. Goat not shown – definitely swap.

374. Monty does not need to show the goat for us to know about it.

372. I maintain absolutely that Monty’s revelation of the goat makes no difference and offers no help to the contestant …

372. Monty might just as well not bother showing you a goat and telling you what you already know – Monty’s door opening antics are a phantom

361. … Monty’s goat revelation is merely his trick -

355. – and Monty has taken my advice and not bothered messing around revealing his useless goat –

350. The goat is immaterial – it’s always there – Monty need not show it.

339. 6) That Monty might just as well not bother showing a goat

334. – the revealed goat is of no consequence.

330. Monty might just as well not bother showing the goat

325. The goat might just as well not show up at all …

321. If for instance Monty does not bother to reveal a goat – nothing changes – “

and you have claimed that the rules allow the contestant to open two doors. Richard, you are not talking about the MHP. Your question about shifting chances can be answered within the MHP but cannot be addressed within a sea of moving goalposts.

“the goat does not inform the Swap / Stick decision”. Without the revealed goat you only have a 1/3 chance whether you stay or swap. so your statement is plainly incorrect

“if there’s a goat on view ……. what can you actually swap to?” Only the closed door, which is a 2/3 chance.

“The goat’s useless in our search for the car…”. On the contrary, once the goat is revealed the search is narrowed down to 2 doors, one goat, and one car (since one door and one goat have been eliminated from the equation)

Nice try Richard ….. but a complete load of bollocks as usual.

Since you’ve previously acknowledged “I have no maths” and “I know virtually nothing of the subject” you really should stop pretending you do.

Stick to Bridge – perhaps you can win a 10th (?) championship and maybe the rest of us will receive fewer of your posts containing the same repetitive drivel.

Palmer – I have to repeat myself because I have only one story – now if you were to stop challenging me I might not feel the need to be repetitive….

And you say… “Without the revealed goat you only have a 1/3 chance whether you stay or swap” – well not if you’re swapping to two doors – you get a 2/3 chance with two doors – no dispute that 1/3 + 1/3 = 2/3…

And then you go on – over the edge of rationality on the brink of the precipice that could drop you into the mire of confusion – you go on to say this…

“The goat’s useless in our search for the car…”. On the contrary, once the goat is revealed the search is narrowed down to 2 doors, one goat, and one car –

That sounds very much like the 50/50 scenario to me – Two doors – One Goat – One car – very dodgy – very old-school…

As for bridge – I only watch it online these past 2 years – BBO – I recommend it.

Freddie – you said…

Richard, you have frequently argued against seeing a goat:

Perhaps – but not actually ARGUED against seeing a goat – I claim the goat is of no help whatsoever – frankly I don”t give a care if I see the goat or not – I ignore the goat – I can play the game without seeing it.

So you must admit that I’m consistent in my attitude about the useless goat – I’ve mentioned it 11 times I think – and this makes 12!

And this…

and you have claimed that the rules allow the contestant to open two doors

My claim is that two doors is a valid option because it’s not forbidden – I also claim that if you reject one door then the alternative is – perforce – 2 doors.

And lastly this…

Your question about shifting chances can be answered within the MHP but cannot be addressed within a sea of moving goalposts.

I maintain the goal posts remain constant until the end of time but this bit particularly…

Your question about shifting chances can be answered within the MHP

Go on then – answer it – How? Why? Trigger? Why just one particular door? What happens to the 2/3 chance of the goat?

I answer this by swapping to two doors – I don’t even need to tell Monty I’m having two doors because he’s already opened one of them for me – our little secret eh?

“if you were to stop challenging me I might not feel the need to be repetitive”

I keep challenging you because you’re wrong, and being repetitive doesn’t make you right – just tiresome.

“well not if you’re swapping to two doors”. But you’re not.

“That sounds very much like the 50/50 scenario to me – Two doors – One Goat – One car ” To a fool maybe it does, and since you “have no maths” you’re not really qualified to comment on it’s veracity anyway are you?

Watch more BBO…. please !

462. “Your question about shifting chances can be answered within the MHP

Go on then – answer it – “

OK, I’ll try, emphasising the difference between an event (which has results) and a series (which has probability). For example, imagine these results over a series of three MHP games (chances of car in brackets):

Game A – Door 1 = First choice (1/3), Door 2 = Monty’s goat (0/3), Door 3 = switch choice (2/3).

Game B – Door 1 = Monty’s goat (0/3), Door 2 = switch choice (2/3), Door 3 = first choice (1/3).

Game C – Door 1 = switch choice (2/3), Door 2 = first choice (1/3), Door 3 = Monty’s goat (0/3).

Adding up gives each door 3/9 chances over the series (1/3) while the results in each game need not be exactly 1/3. This is the first step in the answer; I’ll let it lie before I continue.

Freddie – You begin with this…

OK, I’ll try, emphasising the difference between an event (which has results) and a series (which has probability)

We begin on the wrong foot – your wrong foot.

It sounds very rum (odd) to me – you say a single iteration of the MHP only has results but it takes several goes to determine probability?

If there’s Three Doors – Two Goats and One Car – before anything happens – even before the contestant walks on stage – each door has a 1/3 chance of the car and a 2/3 chance of a goat – fixed by the numbers 3-2-1. Looks very much like a probability with the result still yet to happen- you know – a probability is what MIGHT happen – it allows for either eventuality. A result is what has happened – in something as simple as the MHP we don’t need to make multiple test runs – a contestant only gets to play one game.

If a person feels the need – lacks the confidence – they may wish to confirm the 1/3 versus 2/3 probability – a sensible person however would be content with the simplicity of the 3-2-1 combination – multiple tests would be time wasting and something of an irrelevance.

Now to the meat of your sandwich…

Oh! – I can’t go on – you have a fraction with a numerator of zero – what’s happened?

The door remains but the car has disappeared? Is that it? Remember the numerator is the count of how many cars there are – as in One Car between three doors 1/3 – suddenly you have it as no cars behind three doors just because you’ve seen a ghost – sorry – a goat.

I can’t be doing with this creative accounting – I freely admit it’s too deep for me – I can’t go on…

Better by far to swap to two doors but say nothing about it – let Monty open one of your 1/3 chance doors and reveal a goat – and then when he offers the chance to swap say “yes please” – there will only one door to swap to because dear old Monty has already opened the other door – no need for him to know you were swapping from one door to two – don’t tell him – let him think it’s a one for one deal – let him think the revealed goat has made all the difference to you.

0/3 chance the car – three doors & no car – whatever next?

Re Richard

“I know absolutely that at least one of those doors will hide a goat – so if I get to see it – that’s fine – and if I don’t get to see it that’s also fine – I do not need to see something I know to be there to know that it’s there”

Yep. That’s what I was saying in my previous post. The rules of the “problem” are that Monty opens a door that he knows has the goat. Probabilities are always based on information. The information changes during the game.

I would never call you stupid or ignorant. In keeping with Internet discourse, I would more likely say “that sounds suspiciously like something Hitler would have said.”

Probabilities are always based on information.

Well if that’s not stating the obvious I don’t know what is….

The Information that defines the probabilities of 1/3 and 2/3 in the MHP is – wait for it…. Three Doors – Two Goats and One car – where have you heard this before?

If it’s too difficult for you I’d be happy to put it in capital letters.

Then your next preposterous assertion…

The information changes during the game.

How does the information change ? Does the car go away? A goat? A door? A wink and a nod from Monty. None of these surely…

Grant you that a door is opened and a goat revealed but you know very well – you know it in your bones – in the depth of your soul – that the MHP starts with 3-2-1 and ends with 3-2-1.

So the chance probabilities of each door remain constant – because the numbers remain constant – the probability of Goat and Car are not affected by the contents – and certainly not affected by somebody getting a glimpse of the contents. As far as I know Maths is not like that – “oh! somebody looked at me – I need to change.” Ludicrous – don’t you think?

So – this gives a lie to your statement…

The information changes during the game.

I don’t think it does…

Richard;

While I still wouldn’t call you stupid and ignorant, I would say that based on your last post, I would call you arrogant, condescending, and thoroughly without people skills. Evidently one or both of your parents didn’t love you enough. So I will render unto you my final post, then disassociate myself from this discussion which has regressed to childish insults.

“The information changes during the game.

How does the information change ? Does the car go away? A goat? A door? A wink and a nod from Monty. None of these surely…”

Before Monty opens the door, you are unaware of which of the two remaining doors is more likely to contain the car. After he does, you are aware. Before he opens the door, you have no reason to switch your choice to a different door than the one you originally chose. After he opens the door, you do have a reason: it has been revealed that the remaining, third door has a 2/3rds chance of containing the car, rather than a 1/3rd chance.

Good bye, frustrated one.

By the way, Richard, you may want to re-read the Summary at the top:

“The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after. But the goal isn’t to understand this puzzle — it’s to realize how subsequent actions & information challenge previous decisions.”

Dave – yes I read this…

“The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after.

…some time ago and it caused me to make my original comment here – I was outraged by the silliness of it…

Just because it’s written does not make it correct.

Monty does not “filter” he just opens a goat-door because he has an obligation to do so – please consider the possible contents of two doors just before Monty shows a Goat…..

1) Goat / Goat

2) Goat / Car

3) Car / Goat.

It does not take wisdom, insight or imagination to notice that all the possible combinations have a goat that he can show.

Does the revealed goat point to any of those three different possibilities being true? – The revealed goat might be in any of 1,2 or 3 – seeing the goat does not tell us which – I can see no filtering going on.

….. Monty’s actions are pretty much dictated by the rules – he is mostly an automaton – a robotic compare.

The fatal flaw is blindly believing what people say rather than working it out for yourself.

The chances ARE the same before and after the goat-reveal – if you can show me how the chances change when a goat comes into view I will supply the alcohol at your wake – which is another way of saying – you’re having a laugh.

Monty just shows a goat from behind door 2 or perhaps from behind door 3 – that’s not filtering – he shows a goat 100% certain to be there – that’s not new information – not in my world it isn’t. However some may think a door number of vital importance but they will lack the ability to explain the significance of it.

Dave – you said this…

Before Monty opens the door, you are unaware of which of the two remaining doors is more likely to contain the car.

TRUE they each have a 1/3 chance the car – Two of them together a combined 2/3 chance…

This is what I do…. Without telling Monty – and before he opens a door to show a goat – I reason that there’ll always be a goat to show – I also conspire with myself that when the swap offer comes I’ll swap to BOTH doors – but I tell nobody. At this point I own the selected door and at the same time I secretly know that I’ll swap to the other two doors.

Then Monty opens one of my two doors and shows me a goat – my own goat – well I knew that was coming so no surprise there. Then he asks about the swap – what swap? There’s only one realistic swap option the remaining unopened door – naughty Monty has already opened one of my doors – but no matter – I swap to 2 doors for a combined 1/3 + 1/3 chance of 2/3 the car. Quite simple really and the goat was no help at all – I didn’t need to think about it.

You know – when people feel themselves on dodgy ground they often change the subject to my arrogant and condescending style – it’s not my fault – my parents didn’t understand me – others call me stupid and ignorant – it’s their way of trying to save face in a public forum.

“The door remains but the car has disappeared? Is that it? Remember the numerator is the count of how many cars there are – as in One Car between three doors 1/3 – suddenly you have it as no cars behind three doors just because you’ve seen a ghost – sorry – a goat.”

No Richard, the car hasn’t disappeared; it’s behind one of the other doors. The numerator is how many times the car appears in a series (not “the count of how many cars there are”): in the case of the goat door, zero.

“However some may think a door number of vital importance but they will lack the ability to explain the significance of it.”

The significance of door numbering is that it is necessary to identify doors according to their attributes and numbering is one way of achieving this. For example, some people have assigned number 1 to the door the contestant first chose and thereafter referred to it as Door 1 in their explanation.

You do pick up some inconsequential tosh to discuss Richard.

471. “Then Monty opens one of my two doors and shows me a goat – “

Richard, does this mean you accept that the MHP requires a goat to be revealed at some point AND requires the contestant (before they choose whether to switch) to be aware of this?

Not sure what has happened to this thread, but fairly broad consensus seems to have given way to barely coherent arguments of the 50/50 variety. Perhaps their proponents have not read some of the clearer explanations in this thread and elsewhere online.

Whatever the case, rather than arguing the theory, perhaps if any “50/50s” could consider what would happen in reality if the MHP were run 100 times simultaneously, say, in a huge hall for a TV audience. If they accept that, maybe then they can get their heads around the theory.

Bob, the TV station manager, wants to know how many Buicks he is likely to have to give away during this 100 game extravaganza so he asks Walter, the show’s producer.

“Each contestant gets to pick one of three doors. There’s a car behind one of them.”

“So all things being equal, I’m gonna be up for 33 Buicks” reasons Bob.

“That’s right Bob, tomorrow night, from the gantry when you look on the studio floor, statistically, only 33 contestants should have chosen a door with a car behind it.”

“Excellent, Walter, I’ll let procurement know”, says a relieved Bob.

“There’s one more thing you should know about, Bob.”

“What’s that, Walt…”

“After everyone has chosen a door, Monty is going to reveal to each contestant one of their three doors that conceals a goat and offer them the opportunity to switch.”

Bob, who has an IQ of 165, quickly says “Holy Shit! This could be an unmitigated disaster!! If every contestant switches doors, the 33 who first chose a car will end up with a goat, and the 67 who first chose a goat will end up with a car!”

“Don’t worry, Bob, these contestants are average Americans, not Rhodes scholars like you. I’ll be amazed if even a handful knows what as Rhodes scholar is let alone has the brains to switch. We’ll lose two, maybe three, extra Buicks.”

“Yeah, you’re right, Walt, most people will be wedded by superstition to their first pick and think two doors means they have a 50/50 chance of winning [chuckles.] I love schmucks. Does Buick do a 3-door?”

24 hours later….

“What the fuck happened, Walt! You told me I was dealing with schmucks but everyone last one of those cocksuckers switched. I’m down over half a million bucks more than I budgeted.”

“I can’t explain it, Bob. Except before the show someone circulated a printout of a web page called “Betterexplained.com” that showed switching doubled your chances of winning. All the contestants got together and agreed to switch en masse and share the proceeds equally.”

“That’s the last time we recruit contestants from my alma mater!”

@Jonathon,

If the tone of this thread has degenerated into a farce over the past week or so you can thank a certain Richard Buxton.

An elderly man (with obviously too much time on his hands) who freely admits he has no math knowledge and knows nothing about probability theory yet feels qualified to lecture all and sundry on …… the mathematics and probabilities of the MHP, whilst at the same time making nonsensical remarks such as ‘there is a 66.7% chance there’s a car behind DoorA AND DoorB’ (because saying OR would invalidate his entire argument) , and who then takes umbrage when I call him a fool.

Oh, and he’s immeasurably arrogant too – which he seems quite proud of.

@PalmerEldritch, isn’t Richard Buxton correctly saying there is a 2/3 chance of winning if you switch and a 1/3 chance of winning if you stick?

I have tried but I cannot understand what most people’s positions are any more .

It seems more about semantics than logic now.

Is everyone violently agreeing with Ms vos Savant’s conclusion but arguing about the explanation?

I just cannot tell anymore.

@Jonathon

Yes he is, and no one is arguing with him about that.

The disagreement is with his assertion that the (prior) probabilities of 1/3, 1/3, 1/3 for the car remain unchanged after Monty opens a door with a goat behind it – that it is (in his opinion) IMPOSSIBLE for the probabilities to be ever anything other than 1/3, 1/3, 1/3 no matter what happens (For example: if a gust of wind accidently blows a door open to reveal a goat then you still have a 2/3 chance if you swap doors).

And I forgot to mention he’s also annoyingly condescending.

As you would all know, when a door is first opened to reveal a goat, that door ceases to be an option and so ceases to be part of the game. Indeed, one could consider it to belong to Stage 2 of a three stage Game, with Stage 3 comprising a choice among the two remaining doors (and Stage 1 comprising the contestant’s initial selection of one door from three).

In all stages, the door first chosen will be found, on average, to conceal the car 1/3 times. In Stage 3, the other remaining door will be found to conceal the car 2/3 times, or have a 67% chance of concealing the car.

So in Stage 3, the door first opened has 1/3 chance, the door no longer in the game because it was opened to reveal a goat has 0/3 chance, and the other remaining door has 2/3 chance.

As these two probabilities add up to 1 and generally reflect what happens in reality, this theory would seem to be sound.

It is axiomatic that a door that concealed a goat and that has been eliminated from consideration has zero chance of ever being the location of the car.

@Jonathon

And good luck getting Richard to agree with any of that

Johnathan – here’s my challenge to you…

Yes I accept swapping carries a 2/3 chance of success – the issue for me is how we determine the 2/3 probability…

Here we go… you said… (I think it’s a typo)

…because it was opened to reveal a goat has 0/3 chance, and the other remaining door has 2/3 chance…..As these two probabilities add up to 1 and generally reflect what happens in reality, this theory would seem to be sound.

Next…You said…

As you would all know, when a door is first opened to reveal a goat, that door ceases to be an option and so ceases to be part of the game.

Well – clearly it may not be a realistic option to be hiding the car but it’s still a component part of the game – it began as a 1/3 chance of hiding the car and a 2/3 chance of a goat – got opened and revealed a goat. If you notionally remove it from the game just because it showed itself to have the most likely outcome seems quite bizarre. The doors and their contents are benign – they cannot act – mathematics does not know the desire of the contestant is to win a car – mathematics does not care what lies behind the door – it provides probabilities of the ratio of what might happen i.e 2/3 and 1/3 – it allows for either eventuality.

A contestant or an observer naturally looks at what’s happening with a mind sharply focused towards the car – just the car – and adjusts accordingly. They say that a goat is eliminated – but that’s just an emotional response – only in their mind can a door – and or a goat – be eliminated – logic, common sense and mathematics do not eliminate things just because the observer sees something they don’t approve of – I don’t think it works this way.

My analogy of the horse race helps – if a horse comes in last they do not take the starting price (1/3 the car) – declare it to be zero (0/3 the car) and assign the odds to an alternative horse.

This mental juggling arises because people want the alternative of a 1/3 chance the car – the single selected door – to be a 2/3 chance – which it is – there is no dispute about that – and yet there’s an opened door with a goat on view! So they need to have the remaining unopened door to have a 2/3 chance – else it doesn’t work for them – they invent and fudge the transfer of the 1/3 chance from the goat door to its neighbor – nobody ever explains the trigger (seeing a goat?) – the mechanism (opening a door?) or the distribution – why is the transfer not 1/6 each to the other 2 doors?

They try to explain it by inventing multiple iterations of the MHP – as if multiple iterations make any difference – and they fail – they fail to convince me – and this and this alone is why I make my condescending challenges – I do not set out to be condescending – any character flaws are just my default condition – somebody said it was a deficiency in parental love – what do they know?

I can easily overcome the difficulty of the flying 1/3 chance – for me it requires no mental effort whatsoever. I swap to two doors each with a 1/3 chance – and I expect at least one of those 1/3 chance doors to have a goat. I make the swap in my mind even before Monty shows the goat. I keep it a secret – I tell nobody.

Monty opens a door and shows a goat – my goat. This comes and no surprise and it gives me no help whatsoever because I knew in advance that there’d be a goat in there somewhere – door 2 or door 3 – what does it matter which?

Then Monty makes his switch offer – not much of an offer really – the only realistic place to swap to is the only unopened door – so I accept his offer and elect to swap – I know I’m Swapping to two 1/3 chance doors together – I know I’m swapping from a 1/3 chance door to a combined 2/3 chance – no mental gymnastics required. I might tell Monty about my ruse later on – I doubt he will care… he’s just a compare – a facilitator – an automaton – he does what’s required of him – he doesn’t really help at all.

@Jonathon,

And now you can see the unremitting bullshit the rest of us have had to put with since Richard unfortunately discovered this heretofore excellent blog.

Even though I am required on planet Earth where they have “comperes” running game shows, and even though your logic is without compare, “challenge” accepted.

What you “mentally” DECIDE to do BEFORE the first goat is revealed is neither here nor there.

The only relevant question is, “what is the relevant probability for each of the two remaining doors?”

They are, a 33% chance the door you first chose conceals a car and a 67% chance the door you did not choose conceals a car.

This is borne out theoretically and sufficiently in practice.

Your obsession with an element, the first door opened, that has no bearing on the last stage of the game, is quite misplaced and the reasons for it are quite beyond me.

As for your attempt at analogous explanations, they are irrelevant or, at best, imperfect analogies. Whatever they are, they are unnecessary and too hard to understand.

I mean, how can the losing horse in a race that has been run have anything to do with the final stage of the Monty Hall problem which is still in progress? And you appear to be confusing odds with probability.

On the other hand, if the horse were scratched before the race, as happens to the door first opened in the MHP, when the goat is revealed and that door “scratched”, the bookies most certainly recalculate the odds of the remaining horses winning, more so the shorted the odds of the scratched horse. Opening that first door is somewhat analogous with scratching a horse but what you did in taking into account a horse that lost a race is bunkum.

Furthermore, there is no “transfer” of probability from the goat door to the other door. There is simply a new set of facts for which a new set of probabilities must be calculated (just as the bookmakers recalculate the odds when the field gets smaller):

There are two doors remaining in Stage 3 of the Game. The total of all probabilities must be 1. Stay with me here, Richard. We know the probability of the door first selected is 1/3. Therefore the probability of the only remaining door must be 1 minus 1/3, or 2/3.

No transfer of probabilities, just a simple deduction using basic arithmetic.

And that is how the probability of the other remaining door comes to be 2/3. 1 minus probability of only other possible outcome. Say it to yourself a few times, Richard.

As I said, planet Earth beckons. I must away.

And I trust we can all go home now, as there is nothing to see here.

And it’s Jonathan.

Richard, meet your long lost identical twin, Jonathan. Good evening to both of you. Please try and shorten your posts – for conciseness and focus you can’t beat Palmer.

Richard is correct in 481. that the goat door is not ‘eliminated’. A goat-loving contestant could choose it, and it is important to Monty’s role that he is revealing its chance of a car, not prohibiting its being chosen. It contributes to the distribution of chances by its 100% chance-of-goat.

Richard, it would be really helpful to have an answer to my question at 474., many thanks. At 477. Jonathan points out “It seems more about semantics than logic now.”, a point already made by Dave earlier, and it would be useful to try and reverse that position.

Johnathan – sorry about spelling your name incorrectly…

You said…

Your obsession with an element, the first door opened, that has no bearing on the last stage of the game, is quite misplaced and the reasons for it are quite beyond me.

I’ll try to explain…

There are three options for those two doors… Doors 2 & 3

A) Goat – Goat

B) Goat – Car

C) Car – Goat

It’s quite clear that each of the three possible layouts has a goat in it so – if a goat is shown – our discovery is restricted to a particular door number – the goat itself comes as no surprise.

The MHP definition tells us that a goat is shown but notwithstanding that information – we can see for ourselves that there’s a goat in there – we do not need to be shown the goat to know of the goat.

The goat reveal is just a goat reveal – the goat reveal does not make the 1/3 versus 2/3 probability ratio change – it does not and it can not make the probability of an adjoining door grow from 1/3 the car to 2/3 the car.

The probabilities are determined solely by there being Three doors – Two goats and One car – probabilities are not adjusted because one of the doors is shown to have a goat behind it – there are two such doors – the arrangement of three doors and two goats makes for the 1/3 versus 2/3 ratio – what we see – what gets shown does not alter the 3-2-1 arrangement.

So – when Monty shows the goat – and offers the chance to swap – what option does the contestant have? In your view of things there is only one door he can swap to – but in my view of the world he can swap to both doors – the closed door and the useless goat-door – there’s no difference really.

I’m sorry that my horse race analogy was too difficult – I thought it quite good – clear – short and appropriate.

Still – I still await somebody to adequately explain how a 1/3 chance of the car becomes a 2/3 chance just because a neighboring door is shown to have the most probable contents – a goat!

It’s probability Jim – but not as we know it. Enjoy your interplanetary trip – I’ll look out for you.

Freddie #474…

“Then Monty opens one of my two doors and shows me a goat – “

Richard, does this mean you accept that the MHP requires a goat to be revealed at some point AND requires the contestant (before they choose whether to switch) to be aware of this?

Now – please don’t blame me for this contribution – I was asked to answer…

Yes I accept that the standard MHP requires Monty to reveal a goat – not necessarily that the MHP requires the revelation – Monty is obliged to show a goat and he does so – he follows his script – a subtle difference I accept – a matter of semantics perhaps – it matters to me but may not matter to you – but no matter… When Monty or some other actor – gust of wind perhaps – reveals a goat I am unconcerned – unfazed – completely uncaring – the goat has no relevance for me – I see what I already knew about – a goat.

Now to the next bit…

…requires a goat to be revealed at some point…. AND requires the contestant (before they choose whether to switch) to be aware of this?

I do not accept that there is any requirement on a contestant to be aware of anything – not even their own name – prior awareness is not a condition of entry or decision making.

You and I know that Monty is obliged to show a goat – you may think this of vital importance in decision making – I do not.

You may think that the switch-decision typically (always) comes after Monty’s offer to switch – but – knowing what I know about how the whole thing operates I will make my switch decision in the bus on the way to the TV studio – I just won’t tell anybody – it would make me look even more like a foolish condescending smart-arse if I let my secret out – my little plan.

Then I select a door – Then he shows a goat (my goat) – Then he makes his offer – and I think WTF is going on here? – does he think me mad? I switch to what appears to the casual observer to be the only logical place – the remaining closed door – but in reality I’m switching to that door and the goat-door together – away from a 1/3 chance into two doors and a 2/3 chance. How could anybody not switch? That’s what I think…

Now – why did you ask this question in 474? – is it possible for you to please – stop asking questions ? What can you possibly hope to gain?

Just say how a door can double its chance of the car just because we get shown a goat – the most probable contents of all three doors – we see a goat and BANG! a 1/3 chance of the car moves by magic to join its friend – is it magnetism that does it or some other force?

Please tell me without multiple simulations – or racing cars – or balls in bags – or playing cards just how this 1/3 chance manages to move about – in the standard MHP – in one game – please tell me how it happens.

@Richard

” I still await somebody to adequately explain how a 1/3 chance of the car becomes a 2/3 chance just because a neighboring door is shown to have the most probable contents – a goat!”

No you don’t ….. as your preposterous statement that “there’s a 2/3 chance the car is behind BOTH doorA AND doorB” demonstrates. The logically correct phrasing “there’s a 2/3 chance the car is behind EITHER doorA OR doorB” immediately exposes your crackpot “theory of probability” for the piece of garbage it is.

Palmer- people continue to rudely shoot arrows at me and I give return fire – many complain about my return fire but I feel justified…

Now we should not bother ourselves arguing about AND versus OR – we surely agree that each door is 33% and that two of them together represent double that.

Now to your most important piece…

(You quote me) I still await somebody to adequately explain how a 1/3 chance of the car becomes a 2/3 chance just because a neighboring door is shown to have the most probable contents – a goat!

(You) No you don’t …..

(Me again) Oh yes I do – so far nobody has managed to do this – they change the subject – they make criticism of my style and character – make no allowance for my position on the Autism Spectrum – but they DO NOT…

…adequately explain how a 1/3 chance of the car becomes a 2/3 chance just because a neighboring door is shown to have the most probable contents – a goat!

Might I beg your indulgence to do this – no multiple iterations – no Racing Cars – No balls in bags – no playing cards – no multiple doors – just the one go with 3 doors – keep it simple if you can…

“Now – why did you ask this question in 474? – is it possible for you to please – stop asking questions ? What can you possibly hope to gain?”

Thank you Richard. I hoped to establish whether we were discussing logic or semantics. I am still unclear – do you believe the decision to switch could be advantageous if a goat were never revealed before the contestant’s final choice of door?

Please answer for the following two cases:

(a) the contestant is allowed two doors, and the contents of both, as their final choice (as you appear to believe), and

(b) the contestant is allowed only one door as their final choice (as the rest of us appear to believe).

“we should not bother ourselves arguing about AND versus OR”. Yes, let’s just ignore the fact they have completely different meanings.

Logically illiterate too – why am I not surprised?

“make no allowance for my position on the Autism Spectrum”, Believe me ….. I’m making a HUGE allowance.

Freddie – don’t you get fed up asking all these questions and then getting overly long answers from me?

Do not answer that… it was rhetorical

(you) I hoped to establish whether we were discussing logic or semantics. I am still unclear –

(me) Both – lack of accuracy in the use of language can cause confusion – you say semantics – I prefer clarity.

(you)…do you believe the decision to switch could be advantageous if a goat were never revealed before the contestant’s final choice of door?

(me) either deliberately – or by accident – you place your own condition on this – you say DOOR – singular – I only ever swap to DOORS plural. And to the gist of your question – it’s not a case of COULD be advantageous – the revealed goat is neither advantageous or disadvantageous – if I’m swapping to TWO DOORS – which I am in my mind – the goat makes no difference as I’ve explained too many times.

Now to your two cases…

Case the first…

(a) the contestant is allowed two doors, and the contents of both, as their final choice (as you appear to believe), and

Case the second…

(b) the contestant is allowed only one door as their final choice (as the rest of us appear to believe).

You appear to be offering me two cases and seek my opinion on which is correct – I see no question – I hope I have it right – your intention…

It is not a case of what a person is allowed – as an anarchist I am allowed whatever I choose – if I want to swap to two doors I shall do so – because it’s so contentious to take the two-door swap I shall keep my counsel – remain silent about it – tell nobody – I shall let Monty open one of my doors and reveal my goat – thanks Monty – most helpful (my irony – don’t think for a moment that I believe that Monty is helping in any way).

Your (b) the contestant is allowed only one choice? Where do you get this from? The choice conditions are not stated as far as I know – and even if there was a single door restriction – then so what?

1) The selected door

2) An opened door with a goat

3) A closed door

The switch offer is from door # 1 to ?????

The switch offer can only be to door 3 – what does it matter if we include door 2 as well? Are you denying me my goat?

The switch offer is undefined – it’s purely academic if the switch is to one door or to two – helps keep my brain cool to think of it as a combined choice – 1/3 the goat door and 1/3 the closed door – that gives me the 2/3 probability – others are not so devious – they need to invent the Manhattan Transfer – the piece of illogical magic that they know should be the case but can’t explain – how the goat-door’s 1/3 chance moves about and joins with another 1/3 chance and creates a 2/3 chance of the car – in my world that needs Three doors and Two cars – but no matter -many accept it but I do not.

“Freddie – don’t you get fed up asking all these questions and then getting overly long answers from me?”

No. May I rephrase the questions as follows:

(1) Richard, let us assume that the contestant is allowed two doors, and the contents of both, as their final choice. If a goat were never revealed before the contestant’s final choice, do you believe it is to their advantage to switch from their first choice (that being a single door)?

(2) Richard, let us assume that the contestant is allowed only one door as their final choice. If a goat were never revealed before the contestant’s final choice, do you believe it is to their advantage to switch from their first choice (that being a single door)?

I believe the answers to these questions will help make clearer the discussion about the transfer of chance.

Freddie – the questions are literally coming in thick and fast today are they not?

Your rephrasing…

(1) Richard, let us assume that the contestant is allowed two doors, and the contents of both, as their final choice. If a goat were never revealed before the contestant’s final choice, do you believe it is to their advantage to switch from their first choice (that being a single door)?

(Me) Well of course – if I read you right – Two doors – as I’ve boringly said before – are better than one – and as I’ve also boringly repeated – behind any two doors there will always be at least one goat. Therefore- seeing a goat is no advantage in decision making – we see a known quantity.

I dispute your use of the word “allowed” – insert TAKE if you wish but I do not accept allowed – and if I keep my swap to two doors private then nobody will know – what could it matter?

Now your second…

(2) Richard, let us assume that the contestant is allowed only one door as their final choice. If a goat were never revealed before the contestant’s final choice, do you believe it is to their advantage to switch from their first choice (that being a single door)?

(Me)

This is cunning – you pose a problem that’s somewhat not the MHP but no matter…

The arrangement of doors is one of (I’ve told you this before) Doors 2 & 3

Goat-Goat

Goat-Car

Car-Goat

And Monty – for whatever reason – fails to show a goat

We currently have possession of door #1 with a 1/3 chance of the car

Your rule – we can only open one door gives the choice of all the options on the left column

Goat- Goat -Car (2/3 a goat)

Or all the options in the right hand column

Goat-Car-Goat (also 2/3 a goat)

Remarkably door #1 has the same 2/3 a goat chance.

So there’s no difference what You do – Door 1 – 2 or 3 all the same chances – 1/3 the car and 2/3 a goat

BUT don’t make anything out of this – You set the conditions – Swap to one individual door only and no goat reveal.

In my world – where I swap to two doors – the goat reveal is immaterial – the swap creates the 2/3 chance because I swap to two doors each of which may have a goat lurking behind it and similarly – each of which has a 1/3 chance of the car.

Thank you Richard. I believe you are answering:

(1) Without a goat ever being revealed, it is to the contestant’s advantage to switch from one door to two doors.

(2) Without a goat ever being revealed, it is not to the contestant’s advantage to switch from one door to another (single) door.

I agree with you on both counts.

Now we have a problem. To answer the question of whether chance can transfer from one door to another, we need to all be discussing the same scenario.

As you point out, (2) is somewhat not the MHP. Perhaps you will agree that an additional step, the revealing of a goat at some point in the game, would change things and switching would then have an advantage. Would you agree?

Freddie – we talk together so often – it’s as if we were brothers!

You say…

Perhaps you will agree that an additional step, the revealing of a goat at some point in the game, would change things and switching would then have an advantage. Would you agree?

Well the goat-reveal happens after the original selection and before the swap-offer so I pedantically define your “some-point”.

And no I would not agree. Monty’s messing around is nothing more than a stage trick – he shows a goat where a goat ought to be.

Now for the people who think the swap offer is from one door to just one other door this might make a difference to them – it allows them to make the fantasy transfer of that door’s 1/3 chance the car to the neighboring door – it allows them to think that the 1/3 chance moves about and that the still-closed door has a 2/3 chance the car.

Of course you have seen me insist before that this transfer is beyond reason – for a door to have a 2/3 chance of the car requires there to be Three doors and Two cars – quite impossible I’m sure you’ll eventually agree.

So when a door gets opened and a goat gets shown they feel in their bones that Monty has given them some advantage . They know the goat-door is not an option – they imagine it to be 0/3 the car – they assign the 1/3 chance the car to the next door neighbor and when the swap offer arrives they swap to that closed door in the belief that the door has a 2/3 chance of the car.

This does them no harm – the result is the same no matter what they think – to swap gives a 2/3 chance – not to swap is still just a 1/3 chance of the car.

If – like many – you think the swap offer is to a single door – then this works but as I keep saying – for it to work there has to be a magic transfer of the 1/3 chance and this is where most of the disagreements arise – there also has to be a goat reveal – so that the goat door is not an option when swapping – people say that Monty has eliminated a door by showing a goat – this may be true but for it to be true the (impossible) 1/3 transfer must happen.

So- for a one door swap we need a goat reveal AND – what I call – the Magic Transfer.

And I can’t get to grips with this Magic Transfer – so I overcome the conundrum by having a two-door swap – then I don’t need to mess around looking at goats – revealed or not -I just swap to two doors together -if Monty opens a door and shows me that one of my 1/3 chance doors has a goat behind it I will take it in my stride -I’m swapping to two doors so I fully expect to see a goat behind at least one of them.

I can’t make the magic transfer work and nobody can satisfactorily explain it – I get to the same end point by – in my mind – swapping to two doors – I need not tell anybody if Monty shows my goat I can do this – he’s opened one of my doors for me – prematurely – I then say swap and get what’s coming to me-

So – I stand corrected – when Monty shows a goat I do not have to reveal to the world that my swap is from one door to two doors.

But if Monty does not show a goat then I have to come clean and assert my right to open both of the closed doors .

People make it up I think – this thing about the swap being to just one door – but because the rules oblige Monty to always show a goat I can keep my dirty little secret and innocently allow Monty to open one of my two doors – nobody will ever know.

“So – I stand corrected – when Monty shows a goat I do not have to reveal to the world that my swap is from one door to two doors.

But if Monty does not show a goat then I have to come clean and assert my right to open both of the closed doors .”

To keep things clearer then, may we concentrate on the first version of the game you describe? So, if you were in the shoes of the contestant, when Monty shows a goat and offers the option to switch you would do so (without revealing your tactic of swapping from one to two doors). And if I were in the shoes of the contestant, when Monty shows a goat and offers the option to switch I would do so (and state my wish to swap to the unopened door).

Can we agree on that version, Richard?

So – even before Monty shows the goat I consider that the two non-selected doors are mine – they’re my two Swap doors – the Swap has yet to be offered – I keep a discrete silence – I could tell about my scheme but perhaps people might come along and object – so silence is probably best.

Monty prematurely opens one of my two doors (each 1/3 the car) and shows me my own goat – then he asks if I wish to swap – well there’s only one door to swap to – a person would have to be a bit daft to swap to the opened door that clearly has no car on offer – so I elect to swap – well who wouldn’t? This tactic of having two doors provides the 2/3 chance of the car – two doors each with a 1/3 chance – I do not have to bother moving the 1/3 chance from one door to another.

You – I suspect – would do the same – wait for Monty to do his messing around and then when the offer arrives – swap to the only other available door worthy of swapping to – the decision is Swap or Stick.

However – I can’t agree what you might do – if you were sensible that’s what you’d do – but it’s your choice – I agree that’s what you said you’d do. I don’t know your motivations for your alleged actions – you don’t say.

This is leading somewhere – your planning a trap I’m sure of it…

Why do you keep asking questions? Have you ever offered your own clearly stated and cohesive explanation of this stuff -or do you just pick away at the opinions of harmless fools like me?

I think you understand my views on this – I don’t recall seeing your views – why is that? Could it be that you only feel 67% confident?

No trap – just taking it slowly. So if I might check that I have it right, I have based the following steps on the version we have just discussed, which I have suggested is one we could agree on:

The contestant makes their first choice of door.

Monty opens one of the other doors and reveals a goat.

Monty offers the switch and the contestant opts to switch.

The contestant wins the contents of the unopened door.

Does that version work for you as well as for me?

Freddie – could we move on a bit quicker – I’m taking a short break in Malta very soon and will be off line.

Yes – your summary is correct and matches my view of the sensible game.

Get a move on – please hurry up… Get to the point…

Freddie – are you messing around just to get post number 500?

You pipped us all to the post! I hadn’t even spotted 500 coming up.

“The contestant makes their first choice of door.

Monty opens one of the other doors and reveals a goat.

Monty offers the switch and the contestant opts to switch.

The contestant wins the contents of the unopened door.”

This is our version of the game. Now we have completed the steps, what are the possible contents of those doors?

Starting with the first choice, we know it is either one of the two goats (call them Goat A and Goat B) or the car. Let me copy the steps and give the possible contents in brackets.

The contestant makes their first choice of door (say it’s Goat A).

Monty opens one of the other doors and reveals a goat (this must now be Goat B).

Monty offers the switch and the contestant opts to switch.

The contestant wins the contents of the unopened door (the only thing left is the car).

The contestant makes their first choice of door (Goat B this time).

Monty opens one of the other doors and reveals a goat (must be Goat A).

Monty offers the switch and the contestant opts to switch.

The contestant wins the contents of the unopened door (the only thing left is the car).

The contestant makes their first choice of door (the car).

Monty opens one of the other doors and reveals a goat (say it’s Goat A).

Monty offers the switch and the contestant opts to switch.

The contestant wins the contents of the unopened door (the only thing left is Goat B).

So for possible contents we had:

First choice door: Goat A, Goat B, the car.

Monty’s reveal door: Goat B, Goat A, Goat A.

The unopened door: the car, the car, Goat B.

“So for possible contents we had:

First choice door: Goat A, Goat B, the car.

Monty’s reveal door: Goat B, Goat A, Goat A.

The unopened door: the car, the car, Goat B.”

… and if those are the possible contents, the unopened door has 2/3 chances of the car. As for why, Monty’s reveal has indeed altered the chances for two of the doors and left the first choice untouched.

Freddie – You make a giant leap – and your way of describing things is a bit difficult for me – but I’ll try…

When there’s a goat behind the selected door (door #1) – which is a 2/3 chance -then the car sits behind either door 2 or 3 with the same 2/3 chance. Monty shows a goat 100% of the time – the goat’s name rank, number or letter doesn’t matter – If Goat Norman is lurking behind the selected door then Monty will show goat Billy. If the car is behind door 1 Monty can show either Norman or Billy it really doesn’t matter which.

Now what you wrote…

“So for possible contents we had:

First choice door: Goat A, Goat B, the car.

Monty’s reveal door: Goat B, Goat A, Goat A.

The unopened door: the car, the car, Goat B.”

Yes I see it – in the bottom line you have the car appearing twice – there’s only one car and there are two possible ways it gets to be behind an unopened door.

And then your giant leap for mankind….

… and if those are the possible contents, the unopened door has 2/3 chances of the car.

You show two ways for the car to end up in the group of doors 2 & 3 and say that the door has a 2/3 chance of the car? They’re different doors Freddie each with their own 1/3 chance of the car – you’re havin’ a larf…

I give up.

Come come Freddie – You put the same car behind one door – then you have it behind a different door – then you put one and one together and claim that a single door has a 2/3 chance of the car ! ? !

As I said – nobody is able to show that a single door has a 2/3 chance of the car – A door has a 2/3 chance of a goat because there are Three doors and Two goats – surely you can see the connection?

For a door to have a 2/3 chance of a car requires Three doors and Two cars – that’s how it works Numerator over Denominator. There’s One car and three doors – that’s how we get the 1/3 chance the car.

I’m away now to the Med – think it over and come back here in a week or so – tell me if the fog of understanding has cleared for you. Admit that it’s impossible for a single door in the Standard MHP to have a 2/3 chance of anything – except for a goat.

Tell me what sets these initial probabilities of 1/3 versus 2/3 – tell me what can change those probabilities.

Well, Palmer, Ryan, Dave and Jonathan – I fear there is no hope. I painstakingly created a scenario where Richard could visualise the actual MHP while still playing his precious ‘secret swap’. Then I enumerated the door probabilities without recourse to the ‘p’ word and avoided the use of door-position identifiers. And I omitted any suggestion of repeated games over a series. But Richard has thrown himself into the trap he perhaps castigated others for in post 470: “However some may think a door number of vital importance but they will lack the ability to explain the significance of it”. He says in 503: “You show two ways for the car to end up in the group of doors 2 & 3 and say that the door has a 2/3 chance of the car? They’re different doors Freddie each with their own 1/3 chance of the car – “. He has mistaken the position of the door as its attribute governing probability. So once he has imagined the unopened door as in position 2, he thinks it is a different door when in position 3. Of course it is the same door in terms of probability, the unopened door. As fast as he recovers from contradicting himself or reality in one matter, he does it again in another. It’s been a long night.

Of course it is the same door in terms of probability

No – EACH door has the same probability – there are Three Doors – each with a 1/3 probability of the car . In one scenario you have the car behind door 2 – and in another scenario behind door 3 – in each scenario the individual doors have their own 1/3 chance of the car.

You employ sophistry to achieve the 2/3 chance in a single door – it doesn’t work

For a door to have a 2/3 chance of the car requires Three doors and Two cars – two actual cars – not the same single car being in two different places.

“In one scenario you have the car behind door 2 – and in another scenario behind door 3 – “

I know I said I’d given up Richard, but have a look at 501 again. I did not mention door numbers – it is you who have done that and you are mistaken to do so. Rename the doors ‘First choice’, ‘Monty’s Goat’ and ‘Unopened Door’ if you must – without using numbers – and re-run the scenario.

Enjoy your trip.

Freddie, I gather you have already met, Richard, your co-inhabitant of Planet Wacko.

Freddie, your statement that, “Richard is correct in 481. that the goat door is not ‘eliminated’. A goat-loving contestant could choose it, and it is important to Monty’s role that he is revealing its chance of a car, not prohibiting its being chosen. It contributes to the distribution of chances by its 100% chance-of-goat.”

is as non-sensical as it is irrelevant to the question as to the probabilities of each of the two remaining doors concealing the car. That is the question posed in the MHP. An open door with a goat in the doorway has nothing whatsoever to do with calculating the probability of each of the two closed doors concealing a car.

Just because another inhabitant of planet Wacko might prefer a goat to a car has as much to do with calculating the relevant probabilities as the intentions of a punter to back a 500-1 outsider in a horse race have on the probabilities of the favourite winning the race. That is, None.

The intentions of actors/punters have no bearing on the probabilities being calculated. To think otherwise is madness unless you are trying to predict how actors will behave or punters punt. But I can tell you, on Earth, the chances of anyone, especially a typical sane American, choosing the goat over a chance at the car is NIL.

But it does not matter either way because the open door is not an option within the rules of the MHP as proposed and even if it were, it’s availability as an option has no bearing on the probability of either door concealing the goat, which is 1 or 100%.

Second, whether they choose the open door with the goat in the doorway or not has no bearing on the probability of each of the two remaining doors concealing a car, which are the relevant probabilities.

Further, whether the problem started with three doors or three million doors, the probability of the two remaining doors concealing a car is always calculated as follows (with NO reference to which doors have been revealed to conceal a goat or what a contestant MIGHT do):

Probability of first door chosen concealing a car = 1/n,

where n = initial number of doors at the start of the game (Stage 1).

The probability of the only other remaining door concealing a car = n-1/n

The fact that some goat-loving nut job might choose an open door with a goat in the doorway does not alter these probabilities one iota.

So whether or not you consider the door “eliminated” is neither here nor there, the door is irrelevant to calculating probabilities among the two remaining doors. Only the number of doors in Stage 1 of the MHP is relevant to calculating the probabilities.

PS – it is possible that a new immigrant from Afghanistan might choose the certainty of a goat over the possibility of an Impala…

I think you have rather missed some subtlety of interaction here Jonathan. Identifying a point of agreement on an inconsequential element which is not material to the central argument was my attempt at establishing rapport with Richard, while dampening his tendency to take an automatically contrary position by arguing semantic or irrelevant issues. While my carefully-crafted layering of areas of agreement ultimately failed, I did make inroads into some of his wilder meanderings.

Clearly.

Hello again Jonathan – I’m up in the night with indigestion – I should really be asleep… I thought you might come back…

You say this…

The probability of the only other remaining door concealing a car = n-1/n

Well this looks semi-official – it looks and is quite mathematical – in the MHP it translates as 2/3 but how is it arrives at – you correctly say that the probabilities for goat and car are to do with the starting numbers and then you adopt some other standard – quote n-1/n but do not say how you arrive there.

Sure 1-1/n translates to 2/3 – but you show no logical route to get there – you talk of “remaining doors” as if a door has gone away – you talk from the perspective of car-loving mathematics. Mathematics and probabilities have no interest in the car – or in goats – the numbers dictate the chance of one over the other – they allow for either eventuality – if the most likely event happens – a goat -the door does not go away – the chance probability of the least likely event does not jump over to another door and neither you or Freddie have shown that it does – you only assert that it must be so.

And while you make your assertion you resort to name calling which tends to weaken your case – and you lose any moral authority you thought you had.

Freddie has a car behind this door and goats behind the remaining doors -

he has a car behind that door and goats behind the others

(I must not assign an identifier to any of the doors)

He has a car behind another door and goats elsewhere

Three different places that a car might be – two situations where a switching contestant might win the car – Freddie offers this as evidence that a door can carry a 2/3 chance of the car – you offer N-1/N – merely a statement – not an explanation.

And I get called names

For a door to carry a 1/3 chance of the car requires Three Doors and One Car

For a door to carry a 2/3 chance of a car requires Three Doors and Two Cars.

Are you – either of you – arguing differently?

And Monty offers no help – he acts as an automaton – his actions do not change the 3-2-1 numbers so chance probability remains constant…

This statement from the opening…

The fatal flaw of the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after he filters the other doors.

… is tosh –

Monty does not filter anything – he just reveals a goat from among two doors – where a goat was certain to be – Is that what filtering means? Showing something known to be certain.

I’m not seeking “moral authority”, Richard, just to explain the logic.

(n-1)/n for the “other” door comes from the simple proposition that the total of probabilities for all possible outcomes must equal 1. I alluded to that when I said the probability of either of the two remaining doors concealing a car was 1 or 100%.

In other words, the total of all probabilities = n/n, where n is the number of doors.

The probability that the first door chosen conceals the car, be it from three doors or three billion doors, is 1/n, ie 1/3, or 1/3000000000, respectively.

It is not a great leap from those axioms that the probability of the only other remaining closed door concealing a car equals the total of all probabilities minus the probability of the only other possible outcome, ie the car being behind the door the contestant first chose.

the probability of the “other” non-contestant door concealing the car =

Total of all probabilities minus probability of all other possible outcomes, ergo, where probability of contestant’s door concealing car equals 1/n and total of all probabilities equals n/n:

Probability of other door concealing car = n/n – 1/n = (n-1)/n, or

3/3 – 1/3 = 2/3, or in the case of a 3000000000 door MHP:

3000000000/3000000000 – 1/3000000000 = 2999999999/3000000000

No reference to any other doors is necessary as we are dealing with the probability of each of the two remaining closed doors, and no others, and the total of their probabilities must, and naturally enough, equals 1.

QED.

Maaaaaaaaaaaate, Richard, what are you going on about? Fair dinkum, stone the crows, do you really think this:

“For a door to carry a 1/3 chance of the car requires Three Doors and One Car

For a door to carry a 2/3 chance of a car requires Three Doors and Two Cars.”

Clearly that is not the case, as has been shown when, after the first goat is revealed, the odds of the non-contestant door increase from 1/3 to 2/3 but the number of cars still equals 1.

If the host opens a second goat door there is a 3/3, or 100/100, or 100%, however you want to express it, chance the last door conceals a car. There is no need for the numerator or denominator to be referable to the number of doors or cars.

Saw some more of your post and I do not simply assert that a probability “jumps” from one door to the next. It is simply a process of elimination, like in a murder mystery where only one of two suspects could have committed the crime. You might say there is a roughly 50% chance each one committed the crime and a 100% chance either of them did it. After it is revealed that one of the suspects was dead at the time of the murder, through a process of elimination, Poirot concludes that it is 100% probable that the only other suspect must have committed the crime.

The probability of the other suspect does not “jump” over to the remaining suspect, merely a process of deduction where Poirot takes the total of all probabilities (ie 1) and when he deducts the zero probability of the dead suspect having committed the crime, ie 0%, is left with a 100% probability that the only other suspect committed the crime.

It is deduction, not transfer, that causes the probability of one possibility to increase, simply because the total of all possibilities must equal 1 and when the probability of a possibility reduces to zero it is eliminated and so causes the probability of other possibilities to increase. A process or elimination.

If you add a Ferrari to a race of 9 Volkswagen Beetles, the probability of the Ferrari winning do not “juymp” from the Beetles, they are simply recalculated in light of the new information, a much faster car in the field, and vice versa when the Ferrari is scratched from the race. The Ferrari’s 90% probability does not jump to the other cars when it is scratched, but the probabilities are recalculated so they equal 1, so naturally the probability for each Beetle winning increases.

Now, for God’s sake, give it a rest, there is truly nothing to see he except someone clutching at straws.

Streuth, mate, you’ve had a fair shake of the sauce bottle and ended up with the rough end of the pineapple every time. Just give it a rest or at least spend some time thinking about the cogent analyses before writing another response.

@Freddie and @Jonathon

If you ignore him he might go away.

Jonathan…You said this…

No reference to any other doors is necessary as we are dealing with the probability of each of the two remaining closed doors, and no others, and the total of their probabilities must, and naturally enough, equals 1.

(Me) Here you talk only of two doors – Surely the MHP has Three Doors – not two.

The three doors give rise to the ratio of 1/3 versus 2/3 for each door irrespective of what the door conceals – i.e –

The two closed doors hiding a goat each carry a 1/3 chance of the car

The single door hiding the Car has a 2/3 chance of a goat

The chance probabilities are FIXED by there being Three doors – Two goats – and One car – they are not altered by our discovery as a door is opened.

You conflate the future situation – what might happen in the proportion 1/3 versus 2/3 – with what has happened – what’s eventually seen – you retrospectively adjust the earlier prediction – not only do I not see why this happens – I can’t conceive how it possibly could happen. You discard doors on a whim because the predicted contents displease you.

You reduce the game to two doors just because you’ve seen a goat and risk the 50/50 outcome when you say…

…we are dealing with the probability of each of the two remaining closed doors, and no others…

And you think me misguided?

I shall continue to uphold the premise…

For a door to carry a 1/3 chance of the car requires Three Doors and One Car

For a door to carry a 2/3 chance of a car requires Three Doors and Two Cars.

…and am intrigued to know why you think it wrong – looks quite simple to me.

@Richard

“The chance probabilities are FIXED by there being Three doors – Two goats – and One car – they are not altered by our discovery as a door is opened.”

And that is incorrect. Since you admit to being a mathematical ignoramus you’re in no position to make pronouncements on a subject you know absolutely nothing about.

Look Richard, the question only relates to the closed doors as the question the contestant faces is, “Do I stick with my first choice or switch the other closed door?” Not, do I change to one of the two other doors.

So it is a problem concerning just TWO doors, not three, after one of the three doors has been revealed to contain a goat so any sane / logical contestant ceases it to consider it as an option because they know it has 0% chance of concealing a car.

So:

There is a 0% chance the open doors conceals a goat, so we discount that as an option, it can have no impact on our consideration of the chances of the other doors concealing a car.

However, there is still a 1/3 chance the door first chosen by the contestant contains a goat.

Agreed?

The probability the only other door concealing a car must, by deduction, equal 2/3.

This is arrived at by deducting the probability of the only other door concealing a car from 1 (the total of all probabilities).

1 minus 1/3 equals 2/3.

So the total for the two remaining doors is 1, 1/3 plus 2/3 equals 1. Ok?

The probability door the contestant first chose conceals a car is 1/3, so the probability of the only other closed door concealing a car must be 2/3. So logica says switch choices at this point in the game when there are only two doors to consider.

The opened door is irrelevant and no needing and consideration because it has been shown to conceal a goat and so has 0% chance of concealing a car. So ignore it.

If you cannot understand this I suggest you give it some time until you do or else everyone here will probably think your simply having a lend.

And the moderator should seriously consider deleting your posts.

Jonathan – You

So the total for the two remaining doors is 1, 1/3 plus 2/3 equals 1. Ok?

Me

No – not OK – each door has a 2/3 chance a goat and a 1/3 chance the car – the actual contents – what lies behind any door – do not – and can not – retrospectively change the original probabilities – they stay fixed forever.

The 2/3 chance of a car is arrived at from the sum of two doors i.e. two lots of one third added together – not 2/3+0/3 – A door can not possibly have a 2/3 chance of the car because that will require three doors and two cars – and as we know there’s only the one car.

Before you can make progress here you will need to accept that an open door with a goat on view has a predicted 2/3 chance of a goat and a predicted 1/3 chance of the car – those predictions are not erased just because of what we see – it’s a difficult concept to hold in your mind but when working through the chances of success a goat door is not 100% a goat and 0% the car – the ratio of probability is just as it started 1/3 versus 2/3.

So 1/3 the car from the closed door PLUS 1/3 the car from the goat-door add together to make the 2/3 chance of the car – if you can accept this then there’s no need to indulge in the mathematical magic of a single door having a 2/3 chance the car.

And if it’s too difficult a concept then the need arises for you to somehow have a single door to have a 2/3 chance of the car – preposterous and impossible.

Much easier to accept that a door always has a 1/3 chance the car no matter what the door conceals – you accept this when the door is closed – but not when the door is opened – when a door gets opened you have its 1/3 chance the car set to zero and an adjacent door uplifted to 2/3 the car – very creative but not logical – not possible.

Palmer…

This from you…

“The chance probabilities are FIXED by there being Three doors – Two goats – and One car – they are not altered by our discovery as a door is opened.”

And that is incorrect.

And from me…

What law then allows them to change – don’t tell me it’s the new information about seeing a goat – everybody knows there’s always at least one goat behind any two doors – the goat’s appearance changes nothing.

Do you not recall my most excellent Horse Race analogy – where a horse with a 1/3 chance of winning comes in last – and how when it gets reported the starting price is shown – fixed forever – they do not set it to zero and give the starting odds to some other horse. The starting price is reported – for all to see – as long as the record exists until long after we’re dead.

If you need to challenge this quite superb horse race analogy by saying how the odds can change during the race – sure they do – but the papers report the starting price not the half way round betting – the starting odds are fixed forever.

@Richard

“I’m taking a short break in Malta very soon and will be off line.”

We’re ALL looking forward to that. Malta’s holding the “Boring Old Fart Of The Year” award for 2014 – I see you’re odds on favourite to win it for a record 10th time.

Jonathan…

…after one of the three doors has been revealed to contain a goat so any sane / logical contestant ceases it to consider it as an option because they know it has 0% chance of concealing a car.

Yes – Yes – Yes – nobody puts a bet on the horse that comes in last – not even me – I would not select the Goat-Door – But I maintain the starting odds of that door were 1/3 the car and 2/3 a goat – and that those odds remain attached to the door for all time – they represent a component part of the whole – they do not get reset just because somebody doesn’t like what gets shown – after all the Goat is the most likely result.

Sorry guys, I think I made him worse. I tried really hard to help.

You did well Freddie – you worked hard and I admire you for it.

One day when you’re old and crusty like me you will not be so worried about swimming against the flow – of being thought a fool – and the name calling.

Staying close to the lavatory may well be more important to you….

My truth is simple – in the MHP the probabilities never change – it’s a swap to two doors not one – Monty can be relied on to open one of my doors – the Goat he shows when he opens my door is of no consequence.

And because it’s a swap to two doors I do not need to bother my head and struggle to get that 1/3 chance to move to another door.

For reasons unclear – when I state my truth – some people get quite agitated – and even I bet bored repeating myself – but I only have the one simple story.

“and even I bet bored repeating myself “

I find that very hard to believe.

“I only have the one simple story.” Yes you do, and a very poorly written piece of fiction it is too.

When are you off to Malta (and more importantly for how long)?

Palmer – it’s as if you find my comments here somehow threatening to you – the way you wish I would shut up and go away – the criticism of me rather than any rational discussion about the MHP – like me with the constant repetition of my story there’s a repetitive element to your comments too.

I fly out Sunday and back on Thursday – no doubt during this short break from me you will have a chance to reflect on my wisdom. I relish the opportunity for me to reflect on yours.

@Richards

Why would I feel threatened by the comments of somebody whose understanding of probability is on par with that of a goldfish?

“you wish I would shut up and go away”. Yes, because your asinine and repetitious comments are boring the tits off the rest of us.

“any rational discussion about the MHP”. Everyone’s tried that, it’s been a dismal failure. Explain the difference between ‘AND’ and ‘OR’ again Richard.

Only 5 days …… that’s a shame, I was hoping it might be longer – much longer

Richard, are you shittin’ me?

You say:

“Yes – Yes – Yes – nobody puts a bet on the horse that comes in last – not even me – I would not select the Goat-Door – “

But a horse that has run last in a race that has been run is not analogous with the “Goat” door in MHP. A horse that has been scratched BEFORE the race has started is closer.

Or look at it this way, a 100m sprint. 12 runners, two heats to find the 8 fastest runners. The fastest 8 go into the final. The slowest four runners can be forgotten abut for the purposes of calculating the probability of each of the 8 finalists winning. as they have ZERO chance of winning and so do not affect the chances of the 8 finalists.

Same with the goat door, if chance of revealing a car equals zero, impact on probabilities of possibile outcomes is also zero

Jonathan

Richard, are you shittin’ me?

Certainly not dear chap – my horse race analogy clearly failed to register with you – my point was to do with the starting odds – they get determined at the start- and the Starting Odds are fixed at that moment when the flag drops – the subsequent reports of the race show the starting odds – they don’t reset them to zero and move the odds about based on what happened.

In the MHP when one of the joint favorites – a goat – appears from behind a door some observers seek to reset the starting price (odds – chance probabilities) solely on the basis that they see a goat – why? A goat is the most likely occupier of a door – the door had a starting price -1/3 the car and 2/3 a goat – they see a goat and – in their mind – the 1/3 chance of the car becomes 0/3 = you see this all over MHP discussions and it doesn’t get challenged.

I try to challenge this concept by making the horse race comparison – it works for me but clearly not for you – my sincere apologies – here in the UK on the sports pages they report starting prices – they report them exactly as they were at the start – they do not get adjusted depending on the outcome – seems a reasonable analogy to me – but no matter.

My argument is that the 1/3 versus 2/3 probability of each door is set at the beginning and – like the odds in a horse race – does not vary depending on the outcome.

There is no issue with the concept that all three doors will each have the same probability ratio of 1/3 the car – and yet two of those doors will have a goat behind them – people cope with this conundrum – 1/3 the car but most probably a goat behind it.

The 1/3 versus 2/3 chance represents what might happen in the future – the appearance of the goat does not retrospectively cancel out that calculation – it’s valid before opening and just as valid after opening – what we see does not change the very simple calculations we made at the beginning – there are still Three doors – Two goats and one car.

A door with a goat on view is a 2/3 a goat door – the prediction not the result

A door with a car on view is also a 2/3 a goat door – a prediction not a result

A door with goat on view is a 1/3 the car door – merely the prediction

A door with a car on view is a 1/3 the car door – again just the prediction

The 1/3 versus 2/3 probabilities allow for either eventuality – the probable future outcome – they should not be discarded or altered just because one particular event occurs – they’re a prediction of what might happen – not a record of what has happened.

No Jonathan I’m not shittin’ you – why should I ?

Somebody may be shittin’ you – you may even be shittin’ yourself – but I’m not shittin’ you – I have no reason to shit you – or anybody else.

In the standard MHP – probability predictions do not change – they stay fixed for ever – BECAUSE they’re predictions – a calculation of what MIGHT happen.

What has happened is a result – the result does not alter the prediction of the likely outcome – if you think it does then you’re shittin’ yourself.

“In the standard MHP – probability predictions do not change – they stay fixed for ever – BECAUSE they’re predictions – a calculation of what MIGHT happen.”

Thus speaks the man who thinks that logical ‘AND’ is the same as logical ‘OR’ – clearly his opinions on the probabilities of the MHP are fit only for rubbish bin.

Isn’t it time you got on your plane Richard?

Hi Jonathan – this is me shittin’ you – or at least trying to…

Scene – Interior – TV Studio – Jonathan is called forward…

We see the action – people speak but we only hear Jonathan’s thoughts…..

– That door there – the one he’s reaching out to open – I forecast that it’s probably got a goat behind it but it may be hiding a car – It’ll be 2/3 for a goat and 1/3 the car – that’s my prediction of the future chances.

Jeez! – it’s a goat so I was totally wrong – and double jeez – I can’t be giving up the precious 1/3 chance of the car – I’ll shift that 1/3 chance to the next door along. I’ll abandon my earlier prediction and start over… I’ll make a new forecast that gives me a better chance of the car…

That was my juvenile attempt at shittin’ you…

You can reflect on this while I’m away and ask yourself just who is doing the shittin’ here?

@Richard

The fish disagrees with you.

http://www.candypegram.com/wp-content/uploads/2013/06/goldfish-bowl.jpg

Jonathan – before leaving for the airport I looked again at your recent comment…

You wrote…

Or look at it this way, a 100m sprint. 12 runners, two heats to find the 8 fastest runners. The fastest 8 go into the final. The slowest four runners can be forgotten abut for the purposes of calculating the probability of each of the 8 finalists winning. as they have ZERO chance of winning and so do not affect the chances of the 8 finalists.

Same with the goat door, if chance of revealing a car equals zero, impact on probabilities of possibile outcomes is also zero.

That’s what you wrote -

It’s as if some of your cylinders are not firing properly….

But I secretly suspect you’re doing it on purpose – just to get a discussion going…

Your race – when the final is run between the 8 fastest those eliminated from the final do not enter the calculation.

In the MHP there are no qualifying heats – just the one game – unless you’re prepared to believe of course – that the revelation of the goat is meaningful.

In that case you might have a point – but as I claim – the revealed goat is about as useful as as a newspaper horoscope.

So – this may be something you could dissect and explain over the next 5 days – what value is the revealed goat?

The revelation of the goat is meaningful. Just like your plane might have a 0.001% chance of being involved in a mishap, once it has landed safely you plane has no chance.

But there are two “races” in MHP. A “three-horse” race and a “two-horse” race

In the first, with 3 doors, the chances of a door concealing a goat are equally split, ie total of all probabilities (1) divided by number of possible outcomes (3), ie 1/3

After the goat is revealed, another “race” is held with just two possible “winners”. The door the contestant chose and the other remaining closed door.

Some people, even trained mathematicians, have said that the chances of these two door concealing a car are equally split, ie 1/2, however, Marilyn von Savant showed these and others that the chances were different.

The chances of the contestant having picked a winning door are 1/3, that much is incontrovertible.

According to the rules of statistics, one of which says the total of probabilities for all possible outcomes must equal 1, or 100%, the chances of the only other “horse” or door in the second race winning equals the total of all probabilities minus the probability of the only other “horse” or door winning, ie

Total of all probabilities Probability of only remaining possibility

1 – 1/3 = 2/3.

In considering the probability for each of the two remaining doors concealing a car, or horse winning a race, we need only consider the doors or horses in the contest.

As running this experiment more or less confirms the predictions in the theorem, you either understand it or accept it, but there is no point challenging it ad nauseum, or at all for that matter.

Claiming that the odds of a horse at the start of race 1 are relevant to race 2 in which that horse is not participating is beyond illogical, on planet Earth at least.

“Once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.” – Arthur Conan Doyle.

The value of a revealed goat is that it tells the contestant where the car isn’t. So, if there were only two doors in the MHP, and the goat was shown to be behind door 2, there would be a 100% chance of the goat being behind door 1.

If you want to call that “odds” jumping or whatever, do so. however, it is just simple deduction, as shown above.

If it is impossible for door two to conceal the car, the truth, with 100% certainty, if the rules of the game have been observed (as they are assumed to be in a proof of a theorem) door one must conceal the car.

That is the obvious value of the revealed goat in a two-door game of MHP.

In a three-door game of MHP, the revelation of the goat is similarly useful, so long as one knows how to use the information.

Interestingly, the predictive value of the goat goes from 100% in the case of a two-door game of MHP, to 67% in a three door game, to 75% in a four-door game, to 80% in a five door game, to 83.333% in a 6 door game…99% in a 100 door game and so on, never quite reaching, but getting ever closer to, 1 as the number of doors increases.

If you do not understand that, the revelation of the goat is as useful to you as tits on a bull, an ashtray on a motorbike or pockets in a singlet.

To those who do understand that, the value of the information is in proportion to the number of doors in any game involving three doors or more…

Richard, the meaningfulness, or predictive value, of the revelation of the goat, or goats in a game involving more than 3 doors, equals:

n-1/n, where n = the number of doors.

So, (3-1)/3, or 2/3 in the standard game or

(100-1)/100, or 99/100, in a 100-door game.

Jonathon – you do realise you’re completely wasting your time with Richard?

If you produced a mathematical proof (with Bayes Theory) he still wouldn’t believe you.

Just be thankful he’s gone away – and let’s hope the old fart is too senile to remember to return.