Solution
The Talkative Warden
Show the problem again
Three prisoners, A, B and C, are told that exactly one of them, chosen uniformly at random, has been pardoned and the other two will be executed. A asks the warden to name one of B or C who will be executed. The warden never lies and never names the pardoned prisoner. - If B is pardoned, the warden must say "C". - If C is pardoned, the warden must say "B". - If A is pardoned, both B and C are to be executed, and the warden says "B" with probability 3/4 and "C" with probability 1/4. The warden says "B". What is the probability that A is the one pardoned? Give your answer as a reduced fraction.
Worked solution
The answer is 3/7. The probability that the warden says "B" in each case is: - A pardoned: (1/3)(3/4) = 1/4 - B pardoned: 0 - C pardoned: (1/3)(1) = 1/3 So P(A pardoned | "B") = (1/4) / (1/4 + 1/3) = 3/7. The answer is below 1/3 because the warden prefers to say "B" when he has a choice, so hearing "B" is weak evidence about A. Had he said "C", the posterior for A would be (1/12)/(1/12 + 1/3) = 1/5. Only an unbiased warden leaves A at 1/3. A's chance depends on how the information was produced, not just on what it says.
Source: Variant of the three prisoners problem, popularised by Martin Gardner in Scientific American (1959). Statement written for AxiomIQ.