A New Twist on Prisoner Probabilities

I was reading an old Martin Gardner mathematics book and came upon an interesting problem, with a twist that wasn't explored. It is entertaining the think about, but completely meaningless.

There are three prisoners, named A,B,and C. All three are sentenced to die, but the governor decides to pardon one of them through a random selection. The warden knows who was selected, but cannot tell the prisoners.

Instead, Prisoner A asks for the name of one of the two who will be executed. If B is pardoned the warden can say C and vice versa. If A is pardoned then the warden can select either B or C, so he thinks no information is given. He says that B will be executed.

Except that does give information. Obviously B has not been pardoned, so either A or C is pardoned. Write down all the possible combinations of who is pardoned and which name the warden gives, and you will find that there is a 1/3 chance of A being pardoned and a 2/3 chance of C being pardoned. (I don't have space for a full explanation, but it is discussed in many books and websites, as a variation on the Monty Hall problem)

That is where it is usually ended, with the odd fact that C has doubled his chances of survival without doing anything himself. But here is the twist: Suppose that C did the exact same thing, and the warden also told him that B was not pardoned. Now A has a 2/3 chance of survival and C has only a 1/3 chance. What happened?

And to make it more entertaining, suppose that A and C are secretly communicating - now they each know what the other was told. By symmetry alone, each prisoner now has only a 1/2 chance of survival, so each has helped himself and hurt the other.

I will let the reader work out why this happens, but one hint is to consider what would have happened if A and C were told different names...