Three Prisoners Problem
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The Three Prisoners Problem appeared in Martin Gardner's Mathematical Games column in 1959. It is completely analogous to the Monty Hall problem.
There are three prisoners, A, B, and C. Two of them will be released and one will be executed. A asks the warden to tell him the name of one of the others in his cohort who will be released. As the question is not directly about A's fate, the warden obliges and says, "B will be released." Assuming the warden's truthfulness, what are A's and C's respective probabilities of dying now? To make the analogy to the Monty Hall problem more explicit -- if A could switch fates with C now, should he?
[edit] Solution
Just as in the Monty Hall problem, prisoner A, prior to hearing from the warden, estimated his chances of being executed as 1/3. Since the warden was going to tell A some name, the fact that the name that the warden told A was "B" doesn't affect A's chances of being executed (A already knew that at least 1 of his cohort would go free; having a name to associate with a free person gives A no information about himself). So after hearing that B will go free, A still estimates his own chances of being executed to be 1/3, which means that C's chances of being executed are now 2/3. If A wants to live, he's usually better off not switching fates with C.
Some people argue that this is NOT the same as the Monty Hall problem, suggesting that there is an important difference - namely that the player in Monty Hall decides for himself which door to select whereas the prisoner to be executed has already been decided and he cannot anymore affect his fate, thus forcing his chances at 50-50. However this criticism is incorrect, and misses the point of the paradox.
The Three Prisoners Problem is centered around the idea that IF a prisoner had the "choice" as to whether or not he should keep his fate the way it is, or switch it with that of the other remaining prisoner, should he? In this manner, the problem is in fact the same as the Monty Hall Problem, and thus no, he should not switch fates as there is a 2/3 chance that the other prisoner will be executed. This is sort of the opposite of the standard Monty Hall Problem in that in this case the player is trying NOT to win (ie. the two goats represent life for the prisoner, whereas winning the car represents being executed) so while the desired choices are opposite to one another, the methodology behind selecting your fate is the same.
[edit] Aids to understanding
Again, as with the Monty Hall Problem, it might be useful to see this problem from another viewpoint for better understanding. Consider the following six scenarios:
- A b c [b]
- A b c [c]
- a B c [c]
- a B c [c]
- a b C [b]
- a b C [b]
This diagram is to be read as follows: Lowercase is for the prisoners being pardoned, while a capital letter means capital punishment. Between brackets is the name of the prisoner going free as revealed by the guard when A asks. So if A himself will be the victim, the guard can either answer with B or C; let us assume that he will choose at random, so that both answers have 50% chance (case 1 and 2). When B is to be executed, of course he only can answer with C (case 3 and 4). Likewise, when C is to die, the answer is B (case 5 and 6). Note that case 3 and 4 are equal, and so are case 5 and 6; they still are mentioned because in this way all 6 scenarios have an equal chance (1/6) of occurring.
It is now clear that if the guard answers B to A, (what he will do in 3 out of the 6 cases), in one of them A will die, in two of them C will die. Is C now suddenly twice as likely to die as A? Yes, if considered from those 3 cases only. But they are only half of the truth: In the other 3 cases, when the guard answers C, it is B who is twice as likely to die than A. In other words, if the execution were changed into a torture to be repeated every day on one prisoner only, chosen at random, and A asked the guard every day, then in 50% of the cases he would find (when B is the answer) 1 chance for him to suffer, zero for B and 2 for C, and in 50% (with C the answer) 1 for him, 2 for B, and zero for C. So it remains on the average for each of them 2 out of 6.
The wording of the question throws things off considerably. Instead, consider if you had to choose to be one of the prisoners, A B or C (in a way like the Monty problem). If you were to pick at first, (A, B or C), you would have a 1/3rd chance of dying. On the other hand, if after the guard "removed" one of the choices, you would have a 50% chance of dying if you picked A or B. So it would be best to stick with A.
