A few weeks back I posted a Friday puzzle, and forgot to answer it. This was the question:
You meet a woman and ask how many children she has; she replies “two.” You ask if she has any boys, and she replies “yes”. What is the probability of the other child also being a boy?
As was figured out in the comments, the answer is 1/3. This is because there are four combinations of children: BB, BG, GB, GG. You know there is at least one boy, so you can remove the GG option, leaving BB, BG and GB. Only in BB is the other child also male, hence 1/3.
If you rephrase the question ever so slightly, the answer changes:
You meet a woman and her son, and ask how many children she has; she replies “two.” What is the probability of the other child also being a boy?
In this case, the answer is 1/2. In the first question ‘the other child’ referred to either of two children, resulting in three possibilities. This time ‘the other child’ refers to only one child – the one that isn’t the boy – which can obviously be only a boy or a girl.
I figured out the first question – it’s similar to Monty Hall – but the second confused me for ages, even after I’d seen the correct answer 🙂 This from Simon in the comments makes it clearer, I think:
The first question is similar to this. I put before you two boxes, each containing a child. I tell you one is a boy. In working out the probability the trap people will fall into is this: They will assign one of the boxes the boy and then consider the second box. They will fail to consider that the first box may have contained a girl and thus failed to acknowledge one of the three cases (if the first box contains a girl the second had to contain the boy).
The second question is more like me placing a single box containing a child in front of you and placing a boy next to it. Here there are only two cases and the boy is totally irrelevent.