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Two of three prisoner A, B and C will be executed, A asks the name of one other than A himself who will be executed. Jailer says that it is B. Merely by asking the question, A reduced the probability that he will be executed from 2/3 to 1/2, regardless of which answer.

What is wrong with the reasoning? I think the key is jailer answer the question by excluding the situation of A. So the answer just reduced the probability that C will die. A had better ask a question like if B will be executed but I want to ask the accurate reasoning by some formula.

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Ops, nevermind. –  M.B. Jun 15 '12 at 12:43
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"merely by asking the question,A reduced the probability that he will be executed from 2/3 to 1/2,regardless of which answer.what is wrong with the reasoning?" What is wrong with which reasoning? There is no reasoning in your question up to that point, just a bald (and wrong) assertion that the probability suddenly changes. –  Henning Makholm Jun 15 '12 at 12:59
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up vote 4 down vote accepted

Just to have an answer, and to elaborate on Henning Makholm's comment above:

There is no reasoning given to suggest the probability changes to 1/2 that A will be executed, and evidently, it won't be.

After the jailer names B, there is one other name left on the execution list: there is a 2/3 chance that it is A, and a 1/3 chance that it is C.

The general lesson to be drawn is that, just because there are two possibilities for an event (in this case, A or C being the second name on the execution list), they won't necessarily occur with equal probability.

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thank you ,all.I think it can be explained by bayes' formular just like link given by Levon Haykazyan. –  perry zhu Jun 15 '12 at 13:36
    
Notice that the Prisoner's Dilemma problem which is older in origin than the Monty Hall problem is really the same problem with prisoners replacing doors and not being executed replacing the big prize! –  Michael Chernick Jun 15 '12 at 15:04
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P(AB)=1/3 P(AC)=1/3 P(BC)=1/3 AB:A and B will be executed. b:jailer names B. P(b/AB)=1 P(b/AC)=0 P(b/BC)=1/2. P(AB/b)+P(AC/b)=(1*1/3)/(1*1/3+0*1/3+1/2*1/3) +(0*1/3)/(1*1/3+0*1/3+1/3*1/2)=2/3=P(A) enter image description here

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