Πέμπτη, 30 Δεκεμβρίου 2010

Ο Monty Hall και το μαθηματικό παράδοξο


     Με αφορμή μια συζήτηση της χθεσινής ημέρας, παραθέτουμε τη μαθηματική ανάλυση του γνωστού παράδοξου των "Τριών Επιλογών" ή αλλιώς the "Monty Hall problem". Πρόκειται για το γνωστό πρόβλημα επιλογής μιας εκ των τριών πορτών, όταν μια από αυτές κρύβει ένα βραβείο. Αν κάποιος επιλέξει μία πόρτα και μετά αφαιρεθεί η λανθασμένη επιλογή από τις άλλες δύο πόρτες, τον συμφέρει να επιμείνει στην αρχική του επιλογή ή να αλλάξει επιλογή και να διαλέξει την δεύτερη πόρτα; Ακολουθεί μια εύπεπτη ανάλυση του προβλήματος από τον μαθηματικό Keith Devlin, Mathematical Assosiation of America:

"Monty Hall, would present the contestant with three doors. Behind one door was a substantial prize; behind the others there was nothing. Monty asked the contestant to pick a door. Clearly, the chance of the contestant choosing the door with the prize was 1 in 3. So far so good.
Now comes the twist. Instead of simply opening the chosen door to reveal what lay behind, Monty would open one of the two doors the contestant had not chosen, revealing that it did not hide the prize. (Since Monty knew where the prize was, he could always do this.) He then offered the contestant the opportunity of either sticking with their original choice of door, or else switching it for the other unopened door.
The question now is, does it make any difference to the contestant's chances of winning to switch, or might they just as well stick with the door they have already chosen?
When they first meet this problem, most people think that it makes no difference if they switch. They reason like this: "There are two unopened doors. The prize is behind one of them. The probability that it is behind the one I picked is 1/2, the probability that it is behind the one I didn't is also 1/2, so it makes no difference if I switch."
Surprising though it seems at first, this reasoning is wrong. Switching actually DOUBLES the contestant's chance of winning. The odds go up from the original 1/3 for the chosen door, to 2/3 that the OTHER unopened door hides the prize.
There are several ways to explain what is going on here. Here is what I think is the simplest account.
Suppose the doors are labeled A, B, and C. Let's assume the contestant initially picks door A. The probability that the prize is behind door A is 1/3. That means that the probability it is behind one of the other two doors (B or C) is 2/3. Monty now opens one of the doors B and C to reveal that there is no prize there. Let's suppose he opens door C. Notice that he can always do this because he knows where the prize is located. (This piece of information is crucial, and is the key to the entire puzzle.) The contestant now has two relevant pieces of information:
1. The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3.
2. The prize is not behind door C.
Combining these two pieces of information yields the conclusion that the probability that the prize is behind door B is 2/3.
Hence the contestant would be wise to switch from the original choice of door A (probability of winning 1/3) to door B (probability 2/3).
Now, experience tells me that if you haven't come across this problem before, there is a probability of at most 1 in 3 that the above explanation convinces you. So let me say a bit more for the benefit of the remaining 2/3 who believe I am just one sandwich short of a picnic (as one NPR listener delightfully put it).
The instinct that compels people to reject the above explanation is, I think, a deep rooted sense that probabilities are fixed. Since each door began with a 1/3 chance of hiding the prize, that does not change when Monty opens one door. But it is simply not true that events do not change probabilities. It is because the acquisition of information changes the probabilities associated with different choices that we often seek information prior to making an important decision. Acquiring more information about our options can reduce the number of possibilities and narrow the odds.
(Oddly enough, people who are convinced that Monty's action cannot change odds seem happy to go on to say that when it comes to making the switch or stick choice, the odds in favor of their previously chosen door are now 1/2, not the 1/3 they were at first. They usually justify this by saying that after Monty has opened his door, the contestant faces a new and quite different decision, independent of the initial choice of door. This reasoning is fallacious, but I'll pass on pursuing this inconsistency here.)
If Monty opened his door randomly, then indeed his action does not help the contestant, for whom it makes no difference to switch or to stick. But Monty's action is not random. He knows where the prize is, and acts on that knowledge. That injects a crucial piece of information into the situation. Information that the wise contestant can take advantage of to improve his or her odds of winning the grand prize. By opening his door, Monty is saying to the contestant "There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3."