This is a problem that turns up repeatedly in newsgroup discussions. It's all about why the “intuitively obvious” solution to a problem is not always the correct solution, at least where statistical calculations are concerned.
I have been told, by a usually reliable source, that Monty Hall is or was a game show host on American TV. I've never seen him myself, but his identity is irrelevant to the problem, which is really about why “intuitive” solutions to statistical problems don't always agree with the solution obtained by a careful analysis.
Let us assume that you are watching a TV game show hosted by Monty Hall. The contestant has reached the point of having to decide among three doors. Behind one of the doors if a valuable prize; a car, for example. Behind the other two is a goat. We are required to assume that the contestant does not have erotic fantasies about goats, so that choosing one of the “goat” doors would be considered a failure.
The contestant chooses a door. Then, Monty chooses one of the other two doors, and reveals a goat behind it. Monty then offers the contestant the opportunity to change his/her guess.
Our problem: should the contestant choose a different door?
Most people, when faced with this problem, will say that choosing a different door or sticking with the same door are equally good choices. That is, it doesn't matter whether the contestant chooses to switch or to stay with his/her original choice.
The actual answer, by mathematical analysis, is that the contestant should switch. This violates most people's intuition, which is why the Monty Hall problem is such an interesting problem.
To do any reasonable analysis we have to state what our assumptions are. For the purposes of the present article, we are assuming that:
The contestant will take advantage of whatever information is available. For example, if Monty accidentally reveals the prize then the contestant will choose the door that, until this slip, hid the prize.
There is exactly one prize, and the contestant has no inside information.
If the contestant has a choice between N possibilities, and no information as to which of those is the best decision, the contestant will randomly choose one of those possibilities, with probability 1/N. This rules out, for example, a contestant who consistently chooses the first door. (It probably doesn't matter – in fact, the analysis for that case gives the same conclusions – but we still need to state our assumptions.)
In the usual version of the Monty Hall problem, it is stated that there is a goat behind both of the non-prize doors. We assume that the contestant is not someone who is attracted to goats, and who would therefore consider a goat to be a desirable prize.
If Monty knows where the prize is, he will not reveal the prize until the game is over. (This assumption is, of course, irrelevant if he doesn't know where the prize is.) Furthermore, if Monty knows where the prize is, the rules of the game say that the contestant knows that Monty knows. This last point is important, because it is the factor that biases the probabilities.
The initial choice is clear. Lacking any other information, the contestant has a 1 in 3 chance of picking the winning door.
The next step depends on how much knowledge Monty has.
If he knows where the prize is, then of course he will open a non-prize door. That means that there are just two places where the prize might be. Lacking any further information, that means that there is a one in two probability that the prize is behind either of the remaining doors. The contestant now knows that either of the two remaining choices is equally good. Since that's all he knows, it doesn't seem to matter whether he switches. In fact it does matter; if we know that Monty will always open a non-prize door, that has given away some important information.
The other possible scenario is where Monty doesn't know where the prize is. He picks a door at random.
He, like the contestant, doesn't know whether he is about to reveal the prize. If he picks the door where the prize is, the contestant has won. (If we assume, as we reasonably should, that the contestant has the choice of changing his/her choice at this point.) If he picks a wrong door, the contestant again knows that the two remaining doors are equally good. Again, it doesn't really matter whether he switches.
Now, one very popular analysis says that
the initially chosen door has a 1 in 3 chance of being correct;
after one door has been opened, the remaining doors have a 1 in 2 chance of being correct. Thus, the door that the contestant has not chosen has a 1 in 2 chance of being correct.
Since 1 in 2 is better odds than 1 in 3, the contestant should switch.
This is a faulty and naïve analysis. When Monty opens a door, he changes the probability for both of the remaining doors. (Each time you get new information, you have to recalculate the probabilities.) In fact, switching is a good strategy, but not for that reason. The real reason is that the very fact of opening a door reveals something about what Monty knows.
I know quite well that most of you are not going to go through the calculations, so I'm going to put the conclusions ahead of the detailed analysis.
The answer turns out to be that
If Monty does not know where the prize is, there is no advantage or disadvantage to the contestant's decision to switch or not switch. The probability of winning is the same either way;
If Monty does know, and the contestant knows that Monty knows, then the contestant can double his/her chances of winning by consistently deciding to switch.
You can prove this by a theoretical analysis, and that analysis is contained in the next two sections. Of course, it's possible to make mistakes in the analysis. To check my conclusions, I ran a simulation of all scenarios. If you doubt my views, I'm making the source code available to everyone. You can fetch it here. For those programmers who prefer low-level languages like Java, C, and D-, I've even done (and debugged, at great personal cost) a translation into C.
Since the numbering of the doors is arbitrary, let us define door A as the one that hides the prize. The contestant does not know this, of course.
One possible outcome in this case is that Monty will accidentally reveal the prize. (Something that would never happen if he knew where the prize was.) In that case, of course, the contestant will certainly not switch doors, because he/she has already won. For the purpose of the following discussion, the words "the contestant switches" should be taken to mean "the contestant has a policy of switching provided that the game is not yet over".
There are three possibilities for the first round:
The contestant chooses door A. (Probability 1/3, because the contestant does not yet have any information.)
In this case the contestant has not yet won, because Monty has not yet revealed the prize. Monty must open one of the other two doors, revealing that there is no prize behind that door. Now we have two scenarios, depending on whether the contestant decides to switch.
The contestant changes his/her choice.
This gives a certainty of being wrong.
The contestant remains with door A.
This gives a certainty of being correct.
Thus the contestant's overall probability of winning, in case 1, is
if the contestant switches: 0
if the contestant doesn't switch: 1
The contestant chooses one of the other two doors (probability 2/3).
Now the contestant has chosen the wrong door, but he/she does not yet know it. Let us call the door the contestant has chosen door B, and the remaining door door C. In this scenario, doors A, B, and C are necessarily all different. Now we have the subcases:
Monty opens door A. (Probability ½, because Monty must choose between doors A and C, and he has no information to bias him one way or the other.)
In this case the contestant wins, because the correct door has been revealed.
Monty opens door C. (Probability ½, as above.)
If the contestant switches, he/she will switch to door A and win. Otherwise, he/she loses.
Thus the contestant's overall probability of winning, in case 2.2, is
if the contestant switches: 1
Putting these results together, the overall probability of winning in case 2 is:
Now we have to put cases 1 and 2 together. Recall that case 1 has a probability of 1/3 of happening, and case 2 has a probability of 2/3. This gives us the answers:
if the contestant switches: 1/3 x 0 + 2/3 x 1 = 2/3
if the contestant doesn't switch: 1/3 x 1 + 2/3 x 1/2 = 2/3
This says, of course, that it doesn't matter whether the contestant changes his/her mind. Switching neither improves nor worsens the odds.
Notice that the contestant's chances of winning, in either case, are a lot better than random. This is because there is a moderately high probability that Monty will accidentally reveal the prize.
This is a more plausible scenario. Most of us, I think, believe that a game show host has more information than the contestants.
Again, let us define door A as the one that hides the prize. The contestant does not know which door this is.
There are two possibilities for the first round:
The contestant chooses door A. (Probability 1/3, because the contestant does not yet have any information.)
In this case the contestant has not yet won, because Monty has not yet revealed the prize. Monty must open one of the other two doors, revealing that there is no prize behind that door. Monty does not care which door he opens, because he knows that neither of the other doors hides a prize; and in this case the outcome is independent of which door Monty chooses. Now we have two scenarios.
The contestant changes his/her choice.
This gives a certainty of being wrong.
The contestant remains with door A.
This gives a certainty of being correct.
Thus the contestant's overall probability of winning, in case 1, is
if the contestant switches: 0
The contestant chooses one of the other two doors (probability 2/3).
Now the contestant has chosen the wrong door, but he/she does not yet know it. Let us call the door the contestant has chosen door B, and the remaining door door C. In this scenario, doors A, B, and C are necessarily all different. The prize is definitely behind one of A and C. Monty knows which one it is, and will of course open the door that does not hide the prize. The remaining door does hide the prize. The contestant doesn't know, so can only decide that the two remaining doors are equally good. The contestant's probability of winning in case 2 is therefore:
if the contestant switches: 1
Putting cases 1 and 2 together, we get the answers:
if the contestant switches: 1/3 x 0 + 2/3 x 1 = 2/3
if the contestant doesn't switch: 1/3 x 1 + 2/3 x 0 = 1/3
The conclusion in this case is that there is a very significant advantage in switching. The reason for this is that in Case 2, the higher-probability case, there is a certainty of winning if the contestant decides to switch. Of course switching was counterproductive in Case 1, but that was the less probable case anyway.
This article by
Peter Moylan
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