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I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.

The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):

You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?

The character picks A, as the odds are all equally in his favor.

The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.

The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.

What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?

I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.

So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful.

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This is known as the Monty Hall problem. The point is that your odds of winning with the original door have not changed. Since the total odds have to add up to 1, the odds of $B$ being the correct door are now 2/3. In fact, "switching to B" is equivalent to "pick the best of whatever is behind doors B and C" (you know that what is behind B is no worse than what has been revealed behind C), which clearly gives you a 2/3rds odds of winning. The precise conditions of the game are very important, though. –  Arturo Magidin Jan 6 '12 at 3:27
I would expect that this has been asked before; I found two questions asking about variants (here and here), but not the plain question. I'm probably just not finding it? –  Arturo Magidin Jan 6 '12 at 3:33
@ArturoMagidin I looked as well and was shocked not to find one. –  Alex Becker Jan 6 '12 at 3:37
I'd suggest that the question is ambiguous as commonly stated. It doesn't specify whether the teacher selected a door at random which just happened to have a donkey behind it, or if the teacher deliberately selected a door with a donkey behind it. –  Winston Ewert Feb 6 '12 at 3:43
Of course if you lived in the mountains of Nepal a donkey would be preferred to a car... –  Bogatyr Sep 16 '12 at 19:02
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8 Answers

up vote 43 down vote accepted

This problem, known as the Monty Hall problem, is famous for being so bizarre and counter-intuitive. It is in fact best to switch doors, and this is not hard to prove either. In my opinion, the reason it seems so bizarre the first time one (including me) encounters it is that humans are simply bad at thinking about probability. What follows is essentially how I have justified switching doors to myself over the years.

At the start of the game, you are asked to pick a single door. There is a $1/3$ chance that you have picked correctly, and a $2/3$ chance that you are wrong. This does not change when one of the two doors you did not pick is opened. The second time is that you are choosing between whether your first guess was right (which has probability $1/3$) or wrong (probability $2/3$). Clearly it is more likely that your first guess was wrong, so you switch doors.

This didn't sit well with me when I first heard it. To me, it seemed that the situation of picking between two doors has a certain kind of symmetry-things are either behind one door or the other, with equal probability. Since this is not the case here, I was led to ask where the asymmetry comes from? What causes one door to be more likely to hold the prize than the other? The key is that the host knows which door has the prize, and opens a door that he knows does not have the prize behind it.

To clarify this, say you choose door $A$, and are then asked to choose between doors $A$ and $B$ (no doors have been opened yet). There is no advantage to switching in this situation. Say you are asked to choose between $A$ and $C$; again, there is no advantage in switching. However, what if you are asked to choose between a) the prize behind door $A$ and b) the better of the two prizes behind door $B$ and $C$. Clearly, in this case it is in your advantage to switch. But this is exactly the same problem as the one you've been confronted with! Why? Precisely because the host always opens (hence gets rid of) the door that you did not pick which has the worse prize behind it. This is what I mean when I say that the asymmetry in the situation comes from the knowledge of the host.

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Fantastic explanation! Thanks for the clarification :). –  Avicinnian Jan 6 '12 at 4:11
+1 what if you are asked to choose between a) the prize behind door A and b) the better of the two prizes behind door B and C really cements the idea. –  dj18 May 2 '12 at 17:34
Good explanation. But I would say that the fact that the host knows the door which not to pick, is more a kind of asymmetry than symmetry. –  Cris Stringfellow Dec 20 '12 at 0:39
@CrisStringfellow I wrote asymmetry. –  Alex Becker Dec 20 '12 at 18:40
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To understand why your odds increase by changing door let us take an extreme example first. Say there are $10000$ doors. Behind one of them is a car and behind the rest are donkeys. Now, the odds of choosing a car is $1\over10000$ and the odds of choosing a donkey are $9999\over10000$. Say you pick a random door which we call X for now. According to the rules of the game, the game show host now opens all the doors except for two, one of which contains the car. You now have the option to switch. Since the probability for not choosing the car initially was $9999\over10000$ it is very likely you didn't choose the car. So assuming now that door X is a goat and you switch you get the car. This means that as long as you pick the goat on your first try you will always get the car.

If we return to the original problem where there are only 3 doors we see that the exact same logic applies. The probability that you choose a goat on your first try is $2\over3$ while choosing a car is $1\over3$. If your choose a goat on your first try and switch you will get a car and if you choose the car on your first try and switch you will get a goat. Thus the probability that you will get a car if you switch is $2\over3$ (which is more than the initial $1\over3$).

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+1 This means that as long as you pick the goat on your first try you will always get the car. - really clarified the topic. –  dj18 May 2 '12 at 17:36
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Its simple, switching allows you to pick 2 out of the 3 doors. Choosing door number 1 and then always switching is the equivalent of saying "door number 2 or door number 3, but NOT door number 1". When you look at it that way, you should see that you have a 2/3 chance of being right, and that the reveal simply confirms which door it must be if you are right. Increase the number of doors and it should become even more obvious that saying "door 2 or 3 or 4 or 5 or ... but not 1" is the right way to bet. You have an $1-1/x$ chance of being right, and a $1/x$ chance of being wrong.

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Let us use some theory here. Lets call the doors $0, 1, 2,$ and the right door $D$. $$P(D=0)=P(D=1)=P(D=2)=\frac13$$ $D$ is random. Now let us call the door we choose $C$. $C$ is not random. Without loss of generality, let $C=0$. Also we have $R$, the revealed door, which is random. Since the person won't reveal the right door or the one you choose, $R \neq C$ and $R \neq D$. Since we know $C$, without knowledge of $D$ we have: $$P(R=1)=P(R=2)=\frac12$$ Let us say, without loss of generality, we get the information $R=2$. Then what we are looking for is $P(D=1|R=2)$. $$\frac{P(D=1 \wedge R=2)}{P(R=2)}$$ $$\frac{P(R=2|D=1)P(D=1)}{\frac12}$$ Now, $P(R=2|D=1)=1$, since $R$ can't be 0 or 1, since those are already taken by $C$ and $D$. $$\frac{1 \cdot \frac13}{\frac12}$$ So the answer is: $\frac23=66.\overline6 \%$.

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The person who changes his choice will win if and only if his first choice was wrong, and there is a probability of $\frac{2}{3}$ on that.

The person who does not change his choice will win if and only if his first choice was right. There is a probability of $\frac{1}{3}$ on that.

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Rather than looking at the player, I prefer to explain the paradox from the host's standpoint, as this only involves one step.

As the player gets one door, the host gets two. There are 3 possibilities with the same probability:

  • donkey-donkey => leaves a donkey after a door is open
  • car-donkey => leaves the car
  • donkey-car => leaves the car

So in two cases out of three the door that the host leaves closed hides the car.

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Merely knowing that the teacher showed a losing door does not provide any information unless one knows how the correctness of one's initial answer would influence the likelihood of the teacher showing the losing door. Consider the following four possible "strategies" for the teacher:

  1. The host knows where the prize is, wants the contestant to lose, and will show an empty door only if the contestant had picked the one with the prize [if the contest already picked a wrong door, the host would reveal either the contestant's door or the one with the prize].

  2. The host knows where the prize is, wants the contestant to win, and will show an empty door only if the contestant had picked the other empty door [if the contestant had already picked the right door, the host would simply show it].

  3. The host knows where the prize is, and will always show an empty door [the empty door if the contestant's initial guess was wrong, or an arbitrarily-selected empty door if it was right].

  4. The host picks a door at random; if it contains the prize, the contestant loses; otherwise, the contestant is allowed to switch to the other unseen door.

In the first two scenarios, the host's decision to show or not show an empty door will indicate to anyone who knows the host's strategy whether the player's guess was right or not. In the third scenario, the host's decision to offer a switch provides no information about whether the contestant's initial guess was right, but converts the 2/3 probability that the contestant's initial guess was wrong into a 2/3 probability that the prize is under the remaining door.

To evaluate the last scenario intuitively, imagine that the host draws an "X" on the player's door, flips a coin to pick a door at random and draws an "Y" on it, and finally draws a "Z" on the remaining door. If neither the host nor player has any clue as to where the prize is, doors 1, 2, and 3 will each have an equal probability of holding the prize, and the marking of letters X, Y, or Z by people who have no idea where the prize is don't change that. If the host asks the player if he'd like to switch to Z before anyone knows what's under Y, the decision will be helpful 1/3 of the time, harmful 1/3 of the time, and irrelevant 1/3 of the time. If door Y is shown to be empty, the irrelevant case will be eliminated so of the cases that remain, the other two will have 1/2 probability each.

Note: Many discussions of the "Monty Hall Paradox" assume that the host uses strategy #3, but fail to explicitly state that fact. That assumption is critically important to assessing the probability that a switch will be a winning move, since without it (depending upon the host's strategy) the probability of the prize being under the remaining door could be anything from 0% to 100%. I don't know the strategy used by the real-life game-show host for whom the "paradox" is named, but am pretty certain I've seen players revealed as winners or losers without being given a chance to switch, implying that while Monty Hall might sometimes have used strategy #3, he did not do so consistently [the normal arguments/proofs would hold if the host's decision of whether or not the player would be shown an empty door and allowed to switch was made before the player selected his door, but I have no particular reason to believe Monty Hall did things that way].

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Look The Monte Hall problem is corrupted b/c everyone is influenced by previous answers given in movies and an article in a magazine years ago.

You are all solving a false equation. Door #1, #2, or #3. there is 1 door of value and 2 doors of non value.

The Host always removes a non value door, single time. This gives you a new choice, a new equation. Which is never accounted for in the given explanations.

The real choice is A, B or C where C = B = $0 and A=$ After you choose, regardless of that choice B or C is removed from the equation and you are given another choice. A or B (or C)

The first equation is thrown out and a new one put in its place. essentially A or 0.

The Host was always going to make the real choice between A and 0, so you never really had a choice of A, B and C.

The fact that you don't know that doesn't change the real equation.

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Your explanation is confusing. Please clean it up, and explain how other approaches go wrong (better yet, what yur assumptions are, and why they are more realistic). –  vonbrand Feb 11 '13 at 2:01
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