Solution: The Prisoner's Problem

This problem, perhaps more than any other, started the Bohr Group.

Before the guard returns, A knows that he will return, A knows that he will say either that he gave the note to B or that he gave the note to C. So A hasn't learned anything new about her chances of being executed, and that probability remains what it was before: One third.

This can be shown formally using Bayes law

That is, the probability of a given that we know b equals the probability of b given a times the probability of a, all divided by the probability of b.

Let a mean “A will be executed.” Let b mean “B was given the note.”
P(b|a)=1/2.
P(a)=1/3.
P(b)=1/2.
(1/2)×(1/3)/(1/2) = 1/3.

The probability of B getting the note if A is to be executed is 1/2 (could be B or C). The prior probability that A is to be executed is 1/3. The probability of B getting the note is 1/2. This is intuitive because of symmetry, or it can be calculated as P(A is executed, B gets the note) + P(B is executed, B gets the note) + P(C is executed, B gets the note) = (1/3 × 1/2) + (1/3 × 0) + (1/3 × 1) = 1/2.

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