On the other hand, once you make that assumption, the ** proportion** of two-boy families, among families with at least one, is 1/3. The

On the gripping hand, probability is not the same thing as proportion. In the equally paradoxical game show problem, the proportion of games where (1) you choose Door X, (2) there is a goat behind Door Y, and (3) switching will win, is 1/2. But the probability you will win by switching after the host opens Door Y to show a goat is 2/3. The difference is because, if the prize is behind Door X, the host will not always open Door Y. Sometimes, he will open Door Z. The actual answer is 1/(1+Q), where Q is the probability he will open Door Y in that case. It varies from 1/2 when Q=1, to 1 when Q=0. The accepted answer is 2/3, when Q=1/2, because you can only assume the host is unbiased.

Similarly, the answer to the Jones/Brown questions depend on how you learned the facts. They are 1/(1+2Q) and (1+12Q)/(1+26Q), respectively, where Q is the probability you will learn the stated fact when it applies to only one child. Both are 1/2 is Q=1/2. And the reason the 1/3 and 13/27 answers seem unintuitive, is because Q=1/2 is the better assumption if you aren’t told how to determine it. Just like in the Game Show Problem: you can only assume the method is unbiased. And frequently it is not even an assumption, based on wordings like “If I tell you that…”

]]>But I find it unfortunate that some visible mathematicians have publicly stated that the problem isn’t well-defined, since how the information was obtained hasn’t been stated.

What an utter red herring! The problem is interesting and tricky enough without muddying the waters with this kind of nonsense. Of course, any statement in English is bound to be ambiguous in some way if you look hard enough. But there is a long tradition of interpreting probability problems of the form “Given a random member M of a population P satisfying conditions C_1,…,C_n, what is the probability that M also satisfies additional condition C ?” in a completely standard way: What fraction of the members of P that satisfy C_1,…,C_n also satisfy C ?” Or in other words, exactly what Stephen Morris wrote above.

I find it irksome that some people find it necessary to display their cleverness by pointing out an unintended interpretation of this problem. ]]>

Going back to Mike’s ‘irksome’ comment, isn’t this the precise point of the problem? The solution derives from the information you have. ‘I have a sibling’ is different information from ‘I have a sibling and my mother has a son’, etc. ‘There are two people in a room, at least one of them is a male’ is different from ‘There are two people in a room, one of them is the Pope’.

]]>

‘Mrs X has two children. One is a boy who is an identical twin. What is the chance that they are both boys?’ ]]>

We can generalise to:

‘Mrs X has two children. One is a boy who satisfies condition A. What is the chance that they are both boys?’

where condition A has a probability of p for each child.

Then the answer is (2-p)/(4-p)

This varies from 1/2 to 1/3 as p varies from 0 to 1.

We would like to understand all three cases using this formula. Cases 2 and 3 are straightforward, we have p=1 and p=1/7 respectively.

For case 1, Mrs Smith, we need a condition with p=0. At this point it breaks down.

I think we need to introduce a new concept which is related to ‘dependency’. Specifically put q = probability that Condition A is true for one child given it is true for the other child.

The answer is (2-q)/(4-q). Note that p does not appear.

Now we can solve case 1 for Mrs Smith. Condition A is that the child is the eldest. As only one child can be the eldest we have q=0 and so the answer is 1/2. In cases 2 and 3 we have q=p and so we get the previous results.

So the answer does not depend on the actual probability of the condition being true, it only depends on the probability of it being true for one child given it is true for the other.

]]>Mrs Brown has two children. One is a boy born in a non-leapyear. What is the chance that both are boys?

15/39, if I have it right, or ~.38, compared to 3/7 (~.43) or 13/27 (~.48), or Chaim’s example approaching .50. The lower limit is 1/3, yes?

The key is the number of combinations in which both siblings meet the stated criterion.

]]>To see why this is, analyze the above knowing that one child is a boy born at 12:01:00 pm on a Tuesday.

]]>Mrs Brown has two children. One is a boy born in an odd year. What is the chance that both are boys?

There are 16 combinations of odd/even boy/girl. 7 of them have an odd boy. Of those 7, three are BB and four are BG or GB, so the answer is 3/7.

Try writing down all the combinations and counting them.

There’s something about the nature of a proof in all this (Monty Hall comes to mind, of course): a proof does not compel its own acceptance. Or something like that…. ]]>