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.