Your neighbours has two children, one of which is a girl.
What is the probability of the other child to be a girl?
This is oft-mentioned but with this wording it's ambiguous. Here are three possible interpretations:1. Your neighbor has two children. We observed both and found that there was exactly one girl. What is the probability that both are girls? (Answer: 0)
2. Your neighbor has two children. We observed one and found that she was a girl. What is the probability that the other (unobserved) child is a girl? (Answer: 1/2)
3. Your neighbor has two children. We observed both and found that there was at least one girl. What is the probability that both are girls? (Answer: 1/3)
Interpretation #1 is kind of silly but it serves to illustrate the ambiguity of the plain English wording. Interpretation #2 is how most people interpret the problem, and it's bolstered by the use of the word "other" in your prompt. For there to be an "other" child, there must have been a specific girl observed originally, rather than a general observation that there was at least one girl. Interpretation #3 is how you intended the problem.
Edit: however now that I look into it, I should rather have used "conditional probabilities" name rather than Bayes'. I always mix both of them, my bad.
NOTE I've since edited the Wiki page, and deleted the sentence - since it isn't relevant and there was no citation for the claim!
The way you have phrased the question, the probability is 1/2.
So I guess what makes this counterintuitive is that we think of the example as independent variables actually affecting each other, but what happens is really that we're drawing from a collection without replacement, and obviously the mix of the collection is going to change as we do so.
This should apply also on a humankind, world-wide scale: if you have met three women in a row, the fourth person you meet is more likely to be a man. Gambler's fallacy! Or just drawing without replacement. A roulette table is drawing with replacement. :)
If you flip two coins, you get HH, HT, TH, and TT with equal probability. Try it.
HT is equal to TH, because in this situation we only care whether the two coins are equal to each other or different from each other.
I think this problem is actually equivalent to the Monty Hall problem. But it can get much weirder.
A. Similarly, counting all possible cases, one does not get 1/2 or 1/3, but rather 13/27.