Something like the Monty Hall Dilemma

[Somefantastik]

[problem]
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailor to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailor's reasoning?

[solution]

Let Ci be the ith prisoner will be executed. Ji be that jailor tells that the ith will be free. Suppose the 1st prisoner is asking. Then

P(Ci) = 1/3
P(J3|C2) = 1
P(J2|C3) = 1
P(J2|C1) = P(J3|C1) = 1/2

Compute P(C1|J2) = \frac{P(C1 & J2)}{P(J2)} = \frac{P(J2|C1)P(C1)}{(P(J2|C1)P(C1) + P(J2|C3)P(C3))} = 1/3

By the same token P(C1|J3) = 1/3 => Jailor's reasoning is wrong.

I don't understand how the probabilities in red were computed.

 

[dhris]

Take P(J3|C2). This means that C2 will be executed. Therefore, the jailor has to free number 3 because there's no other choice - he's not going to free prisoner A in this scheme. So P(J3|C2)=1. The same applies to the next one. As for the last one, it's prisoner A to be executed, so each of the others will be released with probability 1/2.

 

[HallsofIvy]

[problem]
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailor to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailor's reasoning?

[solution]

Let Ci be the ith prisoner will be executed. Ji be that jailor tells that the ith will be free. Suppose the 1st prisoner is asking. Then

P(Ci) = 1/3
P(J3|C2) = 1
P(J2|C3) = 1
P(J2|C1) = P(J3|C1) = 1/2

The notation is a bit confusing. You referred to the prisoners as "A", "B", and "C" but here they are being referred to as 1, 2, and 3. I'm going to call them "A1", "B2", and "C3"!Also remember that the jailor is certainly NOT going to tell A1 that he will be set free.
P(J3|C2) is the probability that the jailor tells him that prisoner C3 will be set free given that prisoner B2 will be executed. If prisoner B2 is to be executed, and the jailer will NOT tell A1 that he is going to be released, he MUST answer that C3 is going to be released: P(J3|C2)= 1.
If, instead, it is C3 who will be executed, and the jailer will NOT tell A1 himself that he is going to be released, then he MUST anwer that B2 is going to be release: P(J2|C3)= 1.

However, if it is, in fact, A1 himself who is going to be executed, then the jailer can answer either B2 or C3- and they are equally likely: P(J3|C1)= P(J3|C1)= 1/2.

Compute P(C1|J2) = \frac{P(C1 & J2)}{P(J2)} = \frac{P(J2|C1)P(C1)}{(P(J2|C1)P(C1) + P(J2|C3)P(C3))} = 1/3

By the same token P(C1|J3) = 1/3 => Jailor's reasoning is wrong.

I don't understand how the probabilities in red were computed.

 

[HallsofIvy]

Take P(J3|C2). This means that C2 will be executed. Therefore, the jailor has to free number 3 because there's no other choice - he's not going to free prisoner A in this scheme.

That's not quite correctly stated. He is in fact going to free both prisoner C and prisoner A- he just isn't going to tell prisoner A that he is going to be freed!

So P(J3|C2)=1. The same applies to the next one. As for the last one, it's prisoner A to be executed, so each of the others will be released with probability 1/2.

 

[dhris]

Yes, poor wording on my part. It may not make any difference to the calculation, but it's a huge difference if you happen to be Prisoner A!