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Comment by bad_user | Hacker News Reader

No there aren't ... notions of order like older/younger don't enter the equation (as the problem was stated).

Oh well, I guess this is what the article is talking about :)

Older/younger doesn't exactly enter into it, but some ordering is necessary to properly describe the set of probabilities. This can be arbitrary, but birth order makes the most intuitive sense. The set of probabilities of sex distribution in two-child families is [(B, B), (B, G), (G, B), (G, G)], for any arbitrary ordering, this cannot be reduced to [(B,B), (B,G), (G,G)] as you have done without adding a variable to double the probability of (B,G).

Assuming a coin-flip probability for boy/girl distribution, you get the 1/3 answer if we select for two-child families where at least one child is a boy: [(B,B), (B,G), (G,B)]. If we don't pre-select for having at least one boy, (i.e. if we select the family because we just met the father socially), the probability rises to 1/2, because we have two cases to consider, each with a 1/2 probability: [(B,B), (B,G)], and [(B,B), (G,B)].

Yes, but this is wrong ... the sex of the second child is in no way connected to the sex of the first child.

He didn't ask ... what's the sex of the second child? No ... he asked ...

  p(boys = 2 | boy >= 1) == p(child = boy)

Choosing a distribution that involves both children in this case is wrong, hence my answer that ordering doesn't matter. 33% is just wrong.

What I wrote is not wrong unless, when you wrote: "So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl" you are asserting that in the absence of any information about the sex of either child, 2 boys is a 1/3 probability, instead of (1/2)*(1/2)= 1/4.

Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.

Suppose I roll two dice until at least one of them shows a 6. What's the odds of both being 6's? I've said nothing about the red die versus the blue die, but the underlying truth requires that the situations are kept separate. It's only - as far as we know - in quantum mechanics where you deliberately lose the distinction.

I've done these as real world experiments as I explore them with kids, and I have a lot of direct experience. If you disagree then I'd be delighted to gamble with you.

> Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.

You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head?

Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin.

That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.

Here's the problem as Gary originally stated it, and as the article quotes it:

    > I have two children, one of whom is a son born
    > on a Tuesday.  What is the probability that I
    > have two boys?

You say:

the problem as stated in the article is that you know one coin is going to be a head,

No. The point of the article is that you don't know how or why you are given this information.

Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3.

I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous.

It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.

> The probability that they are both heads is 1/3.

The probability they are both heads is 1/2.

You're trying to formulate this as ....

  P(head AND head | one is a head) =
  P(head AND head) / P(one is a head) =
  1/4 * 4/3 = 4/12 = 1/3

But this is wrong ... you know that you have a head ... which makes ...

  P(one is a head) = 1 && P(head_a AND head_b) = P(head_a) = P(head_b)

1/3 would be the probability only on the first try (instead of stopping when you've got a head).

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