You meet a random stranger who tells you that he has two children. You ask, “Is at least one of them a boy?” and he says yes. What is the probability that the other one is also a boy?
It seems the answer is that the second is equally likely to be a girl or a boy.
Well, this is actually a closeted variation on the Monty Hall Problem.
Possible Set-Ups: |Boy Boy| |Boy Girl| |Girl Girl|
Prior Probabilites: P = ¼ P = ½ P = ¼
New Evidence: The situation of two girls is not in our probability space anymore because we know that there is at least one boy
Therefore: the probability of |Boy Boy| (¼) is now being measured out of a new probability space ¾ of the size of the first. P(Boy&Boy|new knowledge) = ⅓
Similarly: the probability of |Boy Girl| is now being measured out of a new probability space ¾ of the size of the first. P(Boy&Girl|New Knowledge) = ⅔
But: This probability (of boy and girl) can be subdivided into half for what the second of the two will be because we know that one of the two is a boy. Therefore, the second is equally likely within the original boy and girl case to be a girl and a boy → P(Boy now for second|Original Boy&Girl) = ⅓, P(Girl now for second|Original Boy&Girl) = ⅓
Summing up all these split things for whatever will give us the second child as being a boy in the new probability space we have
- P(Boy&Boy|new knowledge) = ⅓
- P(Boy now for second|Original Boy&Girl) = ⅓
Therefore the total probability now that the second child is also a boy is now ⅔
Once again, reality surprises.