The Three Prisoners Problem

Three prisoners, A, B and C, are told one of them will be pardoned and the other two executed, each equally likely. Prisoner A asks the guard, who knows the outcome, to name one of B or C who will be executed — reasoning that since at least one of them must be, this reveals nothing about A's own chances. The guard says "B will be executed." Has A's chance of survival risen from 1/3 to 1/2, since only A and C remain in question?

It has not — A's chance is still exactly 1/3, and the answer depends entirely on how the guard chooses which name to give when both B and C are due to be executed. If A is the one pardoned, the guard must pick between naming B or C, say with a coin flip; if C is the one pardoned, the guard is forced to name B. Naming "B" therefore comes up with probability 1/2 in the "A pardoned" world but with probability 1 in the "C pardoned" world — so hearing "B" is twice as likely under "C pardoned" as under "A pardoned," and by Bayes' rule the posterior odds become 1/3 for A against 2/3 for C. C's odds have doubled; A's have not moved at all.

This is exactly the logic behind the Monty Hall problem, just relabelled: naming "not the pardoned prisoner" plays the same role as the host opening a goat door.