3 Prisoners problem question: I think Prisoner C is correct.

Question

Wikipedia sez,

Problem: Three prisoners, A, B, and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one of the two who are going to be executed. "If B is to be pardoned, give me C's name. If C is to be pardoned, give me B's name. And if I'm to be pardoned, secretly flip a coin to decide whether to name B or C." The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who reasons that A's chance of being pardoned is unchanged at 1/3, but he is pleased because his own chance has gone up to 2/3. Which prisoner is correct?

Solution: The answer is that prisoner A did not gain any information about his own fate, since he already knew that the warden would give him the name of someone else. Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3, the same as both B and C. As the warden says B will be executed, it is either because C will be pardoned (1/3 chance), or A will be pardoned (1/3 chance) and the B/C coin the warden flipped came up B (1/2 chance; for an overall 1/2*1/3=1/6 chance B was named because A will be pardoned). Hence, after hearing that B will be executed, the estimate of A's chance of being pardoned is half that of C. This means his chances of being pardoned, now knowing B is not, again are 1/3, but C has a 2/3 chance of being pardoned...

I don't believe that "Hence after hearing that B will be executed, the estimate of A's chance of being pardoned is half that of C."

I think that regardless of before or after hearing that B will be executed, the estimate of A's chance of being pardoned is equal to that of C being pardoned.

Am I wrong? I believe in Prisoner C that A's chance of being pardoned is unchanged at $1/3$.

Answer 1

Prisoner A could expect that the probability for being pardoned was 1/3, and that the probability for one among the other two prisoners being pardoned was 2/3.

Unless the warden was lying, prisoner A was informed that prisoner C was pardoned and B named (a probability of $1/3$) or prisoner A was pardoned and B randomly named (a probability of $1/6$).

The conditional probability that A was pardoned given that B was named is therefore $1/3$. Thus the conditional probability that C was pardoned given that B was named is $2/3$.

Answer 2

Sorry, I thought the rest of the argument was obvious. So let me provide the rest of the argument to make the proof complete.

If there was no coin toss, once $A$ has been told $B$ will be executed, $A$ and $C$ have an equal chance of being pardoned, namely each has a $50\%$ chance of being pardoned. However, with the coin toss, once $A$ has been told $B$ will be executed, if $A$ is to be pardoned the coin toss came up $B$. Thus, one of the two ways $A$ can be pardoned, namely the coin toss coming up $C$, has been eliminated thereby cutting $A$'s chances of being pardoned by one-half while leaving $C$'s chances of being pardoned unchanged. Thus, $C$'s chances of being pardoned are now twice as good as $A$'s. Since $A$'s and $C$'s chances of being pardoned have to add up to one, $A$'s chances of being pardoned are now $\frac13$ and $C$'s chances of being pardoned $\frac23$.