Three Prisoners

Question

Prisoners A, B, and C await execution; exactly one, chosen uniformly at random, has been secretly pardoned. Prisoner A begs the warden, who knows the outcome, to name one of the other two who will be executed, arguing this reveals nothing about A himself. The warden truthfully says "B will be executed" (if both B and C are doomed, he names one of them with equal probability). What is the probabili

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Condition properly on the warden's statement, not on the event "B is executed." The warden says "B" with probability: - A pardoned: B and C both doomed, he picks B half the time → frac12. - B pardoned: he can never name B → 0. - C pardoned: B is the only nameable prisoner → 1. By Bayes, with the uniform frac13 prior: P(A mid textsays B) = fracfrac13 cdot frac12frac13 cdot frac12 + frac13 cdot 0 + frac13 cdot 1 = frac1/61/2 = frac13. A's chances are unchanged, but C's have jumped to frac23. A learned nothing about himself precisely because the warden's answer was guaranteed to name someone else; all the information flowed to the third prisoner. The isomorphism This is Monty Hall in prison clothes: A's pardon is "your door," the warden is the host, and C is "the other door" that you should switch to. The structural lesson is identical, a constrained revelation carries information only about the parties it could have implicated. The trap "Two prisoners remain, so each is frac12" ignores the asymmetry of the warden's protocol: hearing "B" was twice as likely under C's pardon as under A's. Whenever an informant couldn't have said otherwise about you, his statement doesn't move your own

Three Prisoners

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