Understanding the Monty Hall Problem

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 10 times, using a “pick and hold” strategy. Now reset and play it 10 times, using a “pick and switch” approach.

There’s a chance the stay-and-hold strategy does decent on a small number of trials. (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).

If you’re not convinced, try playing 50 or 100 times with a “pick and hold” strategy: Just keep rapidly clicking on a single door. You’ll see your win percentage settle around 1/3. I’ll be here when you’re done.

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.

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Posted September 4, 2009, under Math
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Comments

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.

uwe — September 4, 2009 @ 3:52 pm
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!

Aaron — September 4, 2009 @ 3:59 pm
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.

Gary — September 4, 2009 @ 4:11 pm
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.

CL — September 4, 2009 @ 4:53 pm
@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!

Kalid — September 4, 2009 @ 6:20 pm
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.

ktr — September 5, 2009 @ 9:58 am
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.

Mike — September 5, 2009 @ 10:56 am
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.

Srikanth — September 6, 2009 @ 3:30 am
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 — September 8, 2009 @ 2:20 pm
@lewikee: Awesome, I like that formulation! Monty is basically picking the best door he can out of the remaining choices. Very nice .

Kalid — September 8, 2009 @ 2:51 pm
@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.

Kalid — September 8, 2009 @ 6:13 pm
Nicelly analysed and explained.

3rojka — September 9, 2009 @ 11:27 pm
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!

Seldon — September 11, 2009 @ 1:58 am
Interesting and crazy. I’ve read it twice and still don’t get it.

geld — September 11, 2009 @ 1:46 pm
@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.

Kalid — September 11, 2009 @ 3:44 pm
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 — September 11, 2009 @ 3:47 pm
@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!

Kalid — September 11, 2009 @ 3:59 pm
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?

Frank — September 11, 2009 @ 9:43 pm
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.

Lizzy — September 11, 2009 @ 11:51 pm
I like the way you encourage visualizing the numbers with both shape and color. Great stuff.

rajanKazhmin — September 12, 2009 @ 12:03 am
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

Domogo — September 12, 2009 @ 1:03 am
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.

Edmund — September 12, 2009 @ 3:01 am
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.

Mike Pooposterous — September 12, 2009 @ 7:38 am
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.

Mike Pooposterous — September 12, 2009 @ 7:47 am
Why are both doors examined before they are opened? I find that part confusing. So did he open both doors or only one?

Sludgie — September 12, 2009 @ 8:19 am
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…

DerekSmith — September 12, 2009 @ 1:53 pm
Ya right, i got 5 wins 5 losses both ways, 50 50. your strategy fails.

Ace — September 12, 2009 @ 2:00 pm
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.

SteaveG — September 12, 2009 @ 10:07 pm
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.

sweetestsadist — September 13, 2009 @ 2:22 am
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.

sweetestsadist — September 13, 2009 @ 2:37 am
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?

Puzzled — September 13, 2009 @ 7:54 pm
@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.

Kalid — September 13, 2009 @ 10:29 pm
still completely do not understand this.

CB — September 14, 2009 @ 1:20 am
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.

Greg — September 14, 2009 @ 11:02 am
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!

Alex Wyler — September 14, 2009 @ 11:16 am
@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!

Kalid — September 15, 2009 @ 3:07 am
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 M — September 19, 2009 @ 9:47 pm
@Bruce: Awesome, thanks for sharing!

Kalid — September 20, 2009 @ 12:19 am
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!

Pedro Juan — September 29, 2009 @ 6:52 pm
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.

Colin — October 6, 2009 @ 2:01 pm
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 — October 7, 2009 @ 8:35 am
@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!

Kalid — October 12, 2009 @ 1:15 am
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!

Scott — December 29, 2009 @ 10:21 am
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.

Aupey — January 25, 2010 @ 11:40 am
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.

zenk — January 28, 2010 @ 11:53 am

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