Schlagwort-Archive: Probability

Don’t talk about „Chances“!

I came across a YouTube video featuring another well-known paradoxical problem that is complete bogus. The video is done in a very stressful style with a sort of speaking that I can’t stand very long. I am still not sure, but I think the author just wanted to confuse.

The problem is rather simple. Assume you pick two strangers on the streets of New York and ask them to play a game: They should open their purses and count the money they are carrying. The one with the smaller sum wins the money of the one with the larger sum.

The confusion is now spread by claiming that each of the two strangers must be interested in the deal because the possible loss is always less than the possible win by the very nature of the game.

That’s nonsense. The argument falls apart if you think of a person with an empty purse, who would be overjoyed to play this game, while someone who just picked up a large amount of money from the bank should be very hesitant to make the deal.

The reason for the confusion is that it does not make sense to talk about „expected“ outcome without the assumption of an a priori distribution. I have written about this before in the two-envelopes-problem. This is just a variation.

The best way to avoid this error is to think of a simulation of the experiment on a computer. For this, you would have to fix a distribution which determines the amount of money that you put into the purses of the two strangers. With that knowledge, you can compute the expected outcome of the experiment. Without that knowledge, it does not make sense to talk about „expectation“.

Understand Bayesian arguments!

I just stumbled over another problem of probability theory on the channel MindYourDecisions. The video is rather old already, but mathematical problems never age.

„A nursery has 3 boys and a number of girls at one day, plus an additional birth at night. The next day a child from the day before is picked at random, and it is a boy. What is the probability that the child born at night is a boy?“

I hate the form that this question for a probability is posed. For the casual audience this does not make much sense. They will just answer 50%, and I would not be angry about this answer. It is sort of correct. You need a mathematical education to understand the problem to start with. Let me explain.

Everyone who has not met this kind of problems before will argue that the probability of the night child being a boy is 50%. That is true. The fact that the survey picks a boy the next day does not change this at all. I mean, the randomly picked boy was a boy. So what?

The problem is that the question is asking something more involved. It asks you to consider all possible random experiments. I.e., you assume a random child at night and a random pick. Then you discard those experiments that do not fit to the story.

An extreme case would be that there are no boys at daytime. Then a girl at night would not fit to the story, and all experiments that fit feature a boy at night. The Bayesian mathematics calls this a 100% probability for a boy.

For the actual case of three boys at daytime, let us assume g girls. I now follow the explanation of the video, but with different wordings and a completely different concept of thinking.

There are two cases now.

Case 1: We have a boy at night. We expect this case to happen in half of the experiments. The chance to pick a boy the next day is then 4/(4+g). We throw away all experiments where a girl is picked, i.e., g/(4+g) of them on average.

Case 2: We have a girl at night. Again this happens in half of the experiments on average. The chance to pick a boy next day is 3/(4+g). We throw away all experiments where a girl is picked.

Now imagine both cases together. In the experiments that we did not throw away we have an overhead of 4 to 3 for case 1. Thus, the Bayesian mathematics says that the chance for case one is 4/7.

There are other ways of writing this up. But they are only shortcuts of the above thought experiment. And for me, defining a precise experiment is the clearest way to think about such problems.

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.