Answer: The Lazy Prisoner: A Probability Paradox

Key Idea / Intuition

The warden's statement is not symmetric information about A and C. The warden was forced to name B if A is the one pardoned (since both B and C would be executed), but also sometimes names B when C is pardoned. Carefully tracking the warden's coin-flip behavior via Bayes' theorem reveals that A's probability stays at $\frac{1}{3}$, while C's probability jumps to $\frac{2}{3}$. The key insight: the information "B will not be pardoned" was almost guaranteed if C is the one pardoned (warden had no choice), but only had probability $\frac{1}{2}$ of being said if A is pardoned. So hearing "B" is stronger evidence for C than for A.

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Formal Proof / Solution

Setup. Let $P$ = event A is pardoned, $Q$ = event C is pardoned, $R$ = event B is pardoned. Each has prior probability $\frac{1}{3}$.

Let $W_B$ = event that warden names B (says "B will not be pardoned").

Compute $P(W_B \mid \text{each case})$:

• If A is pardoned: both B and C will be executed, so warden flips a fair coin.
  $P(W_B \mid P) = \frac{1}{2}$

• If B is pardoned: warden cannot name B (B is being pardoned), so must name C.
  $P(W_B \mid R) = 0$

• If C is pardoned: warden must name B (only non-pardoned candidate available, since A can't be named).
  $P(W_B \mid Q) = 1$

Apply Bayes' theorem:

$P(W_B) = P(W_B \mid P)\cdot\frac{1}{3} + P(W_B \mid R)\cdot\frac{1}{3} + P(W_B \mid Q)\cdot\frac{1}{3}$

$= \frac{1}{2}\cdot\frac{1}{3} + 0\cdot\frac{1}{3} + 1\cdot\frac{1}{3} = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

Posterior for A:

$P(P \mid W_B) = \frac{P(W_B \mid P)\cdot P(P)}{P(W_B)} = \frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}} = \boxed{\frac{1}{3}}$

Posterior for C:

$P(Q \mid W_B) = \frac{P(W_B \mid Q)\cdot P(Q)}{P(W_B)} = \frac{1\cdot\frac{1}{3}}{\frac{1}{2}} = \boxed{\frac{2}{3}}$

Conclusion:

| Prisoner | Prior | Posterior (after warden names B) | |----------|-------|----------------------------------| | A | $1/3$ | $1/3$ (unchanged!) | | B | $1/3$ | $0$ | | C | $1/3$ | $2/3$ |

Why does A not gain? A already knew that either B or C would be executed — hearing which one adds no information about A's own fate. But it transfers all of B's probability mass to C, since C's execution was guaranteed to produce the answer "B." This is a classic instance of the Monty Hall phenomenon: the agent who had no choice is the one whose posterior surges.

The intuition that "now it's 50-50 between A and C" is the famous fallacy: it ignores the asymmetry in how the warden was constrained.