Monty Hall Redux

[W0X0F]

Yes, you can break it down that way, but the four possibilities don't have equal probabilities. Say the child whose gender you know is a boy, then you've eliminated 1 and 2 and you're left with 3 and 4. But 3 and 4 are not equally likely, i.e. 50-50. 4 is twice as likely as 3, because there are two ways for 4 to be true (boy oldest, then girl, or girl oldest, then boy), but only one way for 3 to be true.

[ultrafilter]

Originally Posted by Barkis is Willin'

Right, and if the daughter, whose gender you have revealed....

I haven't revealed the gender of any child. All I've told you is that they're not both boys. This is a subtle distinction, but it's at the heart of the problem.

Try this experiment on for size: Take two quarters and flip them. Write down the number of heads. After you repeat this about 100 times, look at the ratio of times you wrote "one" to times you wrote "two". It will be pretty damn close to 2 with very high probability.

If you don't want to do it by hand, here's the R code for an equivalent experiment:

M <- matrix(runif(2 * 100) < 0.5, ncol = 2)
N <- apply(M, 1, sum)
ratio <- length(N[N == 1]) / length(N[N == 2])

I ran this a thousand times, and the average ratio was 2.07. Only 10% of the ratios were less than 1.5. If the true value were one, that basically wouldn't happen.

[Barkis is Willin']

You're still not framing the question in a way that would lead one to conclude 1/3 probability in the boy/girl case. As per the wiki linked to previously, the question should be stated like this:

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.

From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.

So, to get your 1/3 probability you have to remove some of the randomization, which you have never done in your single family boy/girl problem. You did do in your quarter tossing experiment and your village of 1000 families where you narrowed it down to 750 who had at least one boy. However, even in the 1000 families problem, if you take the probability of a family having 2 boys only among the 750 families who have at least one boy, it's going to be 1/2.

[zut]

Originally Posted by Barkis is Willin'

However, even in the 1000 families problem, if you take the probability of a family having 2 boys only among the 750 families who have at least one boy, it's going to be 1/2.

Your conclusion here is that out of 1000 families with two children, 250 have two girls, 375 have two boys, and 375 have a boy and a girl. I'm not sure how you could have concluded that, particularly given the obvious asymmetry.

You don't think that, among all two-child families, 1/3 have two boys, 1/3 have two girls, and 1/3 have a boy and a girl, do you?