A variation on the three prisoners problem Three prisoners hear that one of them will be executed (the exact person who will be executed is determined upfront, and cannot be changed), while the other two will be released. Prisoner A asks the guard (who knows who will be executed) to tell one of the names of the two prisoners that will be released.
The guard refuses to do this, because he claims that this will influence the probability that prisoner $A$ will be executed.
Is the guard right? You can assume that the guard will never tell prisoner $A$ that he will be executed, and that if $A$ will be executed, then he tells the name of $B$ or $C$ with probability $1/2$.
Attempt: Call the prisoners $A,B,C$. Let $E_A$ be the event that $A$ will be executed and let $R_A$ be the event that the guard says that $A$ will be released. Similarly, define $R_B, R_C.$
Assume, wlog, that the guard tells us that $B$ will be released. Then we calculate using Bayes formula
$$P(E_A\mid R_B) = \frac{P(R_B\mid E_A)P(E_A)}{P(R_B\mid E_A)P(E_A) + P(R_B\mid E_B)P(E_B) + P(R_B\mid E_C)P(E_C)}$$
$$= \frac{1/2*1/3}{1/2*1/3 + 0*1/3 + 1*1/3}= \frac{1}{3}$$
so the probability that prisoner $A$ is executed when the guard tells him that prisoner $B$ is released is $1/3$, and thus the probability seems unaltered.
However, did I actually calculate the correct probability?
Maybe what I really want to calculate is
$$P(E_A\mid \textrm{Guard gives us the desired information})?$$
Under the assumption that the guard gives us the desired information, we have (here the event $E_A$ is calculated under this extra information)
$$P(E_A)= P(E_A\mid V_B) P(V_B) + P(E_A\mid V_C) P(V_C)$$
which I believe gives another probability. I think I'm mixing all kinds of things up. Any help will be highly appreciated!
 A: The problem is slightly ill-specified at the following clause:

because he claims that this will influence the probability that prisoner $A$ will be executed.

Influence the probability for whom?

*

*If the probability for the guard, then the probability is $1$ or $0$ either way, and is not influenced. Carrying out the Monty-hall procedure (revealing the name of a non-A, non-condemned party) will not change this probability.


*If the probability for A, then once again, this does not change the probability. There are three equally likely scenarios: A is condemned, B is condemned, and C is condemned. In each case, the guard reveals the name of someone who is not condemned (technically this now results in 6 equally likely outcomes, since the case of A being condemned has 2 equally probable outcomes, B being revealed and C being revealed). A remains in danger in exactly $\frac13$ of outcomes.
Here are the six equally likely outcomes I am referring to, written out more helpfully:
A executed -> B revealed
A executed -> C revealed
B executed -> C revealed
B executed -> C revealed
C executed -> B revealed
C executed -> B revealed

If the problem would have asked about the probability for B or C, that is where the situation would have changed. By revealing the name of a non-condemned non-A prisoner, say B, C becomes unfortunately more likely to be executed. C sees that B is not executed (this occurs in 3 out of 6 equally likely outcomes above) and we can see that C is the condemned prisoner in 2 out of these remaining 3 cases.
