Answer: The Prisoner's Dilemma

Key Idea / Intuition

The mistake is subtle: $A$ forgets that the warder's choice of words is itself random data. When $A$ and $B$ are both to be released, the warder must say "$B$" (he can't name $A$). But when $B$ and $C$ are both to be released, the warder only says "$B$" with probability $\tfrac{1}{2}$. So hearing "$B$" is more likely under the scenario $\{A,B\}$ than under $\{B,C\}$, which tilts the posterior back in $A$'s favor. The answer is: $A$'s probability of release is still $\tfrac{2}{3}$ — asking changes nothing!

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Formal Proof / Solution

Step 1 — Prior sample space.

The three equally likely release pairs are:

$\{A,B\},\quad \{A,C\},\quad \{B,C\}, \qquad \text{each with probability } \tfrac{1}{3}.$

Step 2 — Warder's behavior (the key modeling step).

The warder must name someone other than $A$ who will be released.

| Event | Probability of event | Prob. warder says "$B$" given event | |---|---|---| | $\{A,B\}$ released | $\tfrac{1}{3}$ | $1$ (only $B$ is available) | | $\{A,C\}$ released | $\tfrac{1}{3}$ | $0$ (must say "$C$") | | $\{B,C\}$ released | $\tfrac{1}{3}$ | $\tfrac{1}{2}$ (names $B$ or $C$ equally) |

Step 3 — Joint probabilities.

$P(\{A,B\} \text{ and warder says } B) = \tfrac{1}{3}\cdot 1 = \tfrac{1}{3}$

$P(\{A,C\} \text{ and warder says } B) = \tfrac{1}{3}\cdot 0 = 0$

$P(\{B,C\} \text{ and warder says } B) = \tfrac{1}{3}\cdot \tfrac{1}{2} = \tfrac{1}{6}$

Step 4 — Bayes' theorem.

$P(\text{warder says } B) = \tfrac{1}{3} + 0 + \tfrac{1}{6} = \tfrac{1}{2}.$

$P(A \text{ released} \mid \text{warder says } B) = \frac{P(\{A,B\} \text{ and warder says }B)}{P(\text{warder says }B)} = \frac{\tfrac{1}{3}}{\tfrac{1}{2}} = \boxed{\tfrac{2}{3}}.$

Conclusion. $A$'s probability of release is still $\tfrac{2}{3}$, exactly as before asking. The information "$B$ will be released" tells $A$ nothing new about his own fate — but it does update the relative probability between $\{B,C\}$ and $\{A,B\}$ in exactly the right way to cancel $A$'s naive worry.

This is a cousin of the Monty Hall problem: the host/warder's protocol for choosing what to reveal is essential to the calculation.