# A Problem on Probability

I just found an old problem in Spiegel Online (see here). The problem is absolutely mind-boggling. It goes as follows.

Three prisoners A, B and C are told that two of them will come free on the next day. You are prisoner A. You cannot stand the waiting time and ask a guard to tell you the name of one of the prisoners who will not come free, but not your name, of course. The guard says: B will come free. Now, what is the chance that you will come free too?

The impulsive answer is 1/2. After some thought, it is 2/3 again. Or maybe it is still 1/2? There are good arguments for both. And does „probability“ make any sense at all?

The decision about the two lucky prisoners has been made before A asked, hasn’t it? Asking does not change the chance. So it is still 2/3, assuming that the two prisoners are chosen at random with equal probabilities among the three. Right?

If you have decided in favor of 2/3, I change the question a bit to confuse you. Assume, it was 10 prisoners and 9 become free. After you ask the guard, he sets 8 of them free. You are alone with the last one and claim that the chance for both of you to get free is 9/10. Would you think this is a sensible claim?

To confuse a bit more assume that the guard is free to say any name. He says A. Now what is the meaning of you saying that your probability to become free is 2/3?

What makes this problem hard to treat is the notion of „probability“. For me, probabilities make sense only if there is an experiment going on. I am what they call a frequentist. Now, what could be the experiment in this case? Clearly, it is the selection of the two prisoners by an external force. Without further knowledge, A is among the selection with probability 2/3, i.e., in 2/3 of the cases on average.

The important question is if the knowledge that B is among the selected prisoners changes our experiment. We can argue that it does. The options that A/C are selected is no longer possible. We have only two options left, A/B and C/B. So in 1/2 of the possible outcomes of our experiment, A will come free. What this tells us is that the probability for A to come free is 1/2 provided we know that B will come free.

But does this change of our experiment (discarding A/C) really reflect what is happening in the problem? This can only be clarified by studying the exact question in the problem.

In the online journal, the question was formulated carefully: Does it make sense for A to ask? Will he know more after he gets an answer?

And this is not a question we can answer without assumptions. The reason is that it depends on the preference of the guard in the case B/C. If he does not select to say B or C with equal probability A does indeed know more after asking. But if he does A cannot gain any further knowledge. Our chance is still 2/3.

Let me elaborate that. We start our experiment by selecting 2 of the 3 prisoners. A is among the selected with probability 2/3. Now we ask the guard. In the B/C case, we assume the guard says B with probability p. Now a bit of thinking shows that he will say B in 1/3+p/3 of all outcomes of our experiment, and C in 1/3+(1-p)/3.

• Assuming he says B, A will be set free in (1/3)/(1/3+p/3) = 1/(1+p) of these cases. For p=1/2 this is 2/3 as expected. For p=1 it is 1/2.
• Assuming he says C, A will be set free in (1/3)/(1/3+(1-p)/3)=1/(2-p) of these cases. For p=1/2 this is 2/3 again. For p=1 it is 1. Indeed, if the guard always says B in the case B/C, we know for sure that we get free if he says C.

As always, the problem turns out to be more complicated than it appeared at first sight.

For those of you who still think this is rubbish and the probability must be 2/3 to get free because the selection has been made beforehand, I have good news. You are also right!

Let us compute. A comes free in 1/(1+p) of the cases where the guard says A, and he says that in (1+p)/3 of the cases. A does also come free in 1/(2-p) of the cases where the guard says B, and he does so in (2-p)/3 of all cases. If you add all the cases where A gets free you end up with 2/3.

We have just fine tuned our knowledge about the chances if the guard says B or C. In the case p=1/2, the result is quite easy. The guard will say B or C with probability 1/2, and A gets free in 2/3 of the cases, no matter if the guard says B or C.

## 2 Gedanken zu „A Problem on Probability“

1. rodrigob

„The guard says: B will come free“ -> „The guard says: B will not come free“ ?

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