= Solution
Use the equally likely release prior, and write $s$ for the <probability> that the guard names B when A is released. If B is released the guard cannot name B; if C is released the guard must name B. The <law of total probability> therefore gives
$$
P(B\text{ named})=\frac{s}{3}+\frac13=\frac{1+s}{3}.
$$
A remains jailed and B is named precisely when C is released. Its <probability> is $1/3$, so the <conditional probability> is
$$
\boxed{P(A\text{ remains}\mid B\text{ named})=\frac1{1+s}.}
$$
In \b[(a)] unbiased naming has $s=1/2$, giving \b[$2/3$]: the guard's claim is false and A's chances have not improved. In \b[(b)] preferentially naming B has $s=1$, giving \b[$1/2$]: that naming rule makes the claim correct. In \b[(c)] preferentially naming C has $s=0$, giving \b[$1$]: naming B then reveals that C is the released person, so A certainly remains.
This is the <three prisoners problem>. The observed name includes information about how the guard selects a name; it is not merely the event that B remains. With general release priors $\pi_A,\pi_B,\pi_C$, the same calculation gives $\pi_C/(s\pi_A+\pi_C)$, making explicit where the equal-prior convention enters.
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