[seano]

The difference is that you have specified that the child picked up from the police station is a girl, thus only the sex of the other child is unknown. This other child is either a boy or a girl, presumably with a 50/50 chance either way, resulting in a 50% chance of one being a boy and one being a girl. Concretely, using a capital letter to denote the sex of the child picked up from the station, there are only two possible permutations Gg and Gb, each of equal probability, and 50% of those are girl and boy (Gb). On the other hand if we only know that at least one is a girl we have the permuations, gg, gb, bg - resulting in the 66% chance.

[DougBTX]

How much do you need to know about the person to know which child is which?

I don't quite understand it, but apparently anything which can be used to distinguish the children will do.

Possibilities with two children:

    Gg, Bg, Gb, Bb

If one of them has a distinguishing mark, they have an apostrophe: (in jail, has red hair, or born first)

    G'g, B'g, G'b, B'b, Gg', Bg', Gb', Bb'

Then note that the marked one is a girl:

    G'g, G'b, Gg', Bg'

So, there is a 50% change that the children are a boy and a girl. Only if there is no way to distinguish them, do you get the 66% behaviour, where the set is:

    Gg, Bg, Gb

[seano]

No, that is not it. With G'g and Gg' you are repeating the same permutation in your set above!

It makes no difference if they have distinguishing marks or not. It matters if you are told that a particular child is a girl (50%) or if you are only told that at least one child is a girl (66%).

[DougBTX]

How can you be told information about a particular child if you have no way to distinguish them?

I partially understand your point about G'g vs Gg' now: if having a prime is the only way to distinguish the children, then G and g must be indistinguishable, so G = g.

[seano]

Think about it this way: I take two coins out of my pocket and hold them inside my hand so that neither of us has seen them. I show you the coin in my left hand and you see that it is tails, what are the odds of the coin in the other hand being heads? 50%. This is the chance of a head/tail combo in this case.

I now put the coins back into my pocket, shuffle them about, and again take them out inside my hands. This time I look inside both my hands, not letting you see, and tell you (truthfully) that at least one is tails. Given that information, you can deduce three mutually exclusive possibilities each of equal probability - both are tails, only the coin in my right hand is tails or only the coin in my left hand is tails. Hence we have the odds in this situation of 2/3 for a head/tail combo.

It is easy to see that the first situation is akin to knowing that a particular child is female, whilst the second is akin to knowing that at least one of the children is female. Also, in either case it does not matter if the coins are distinguishable - one could be a euro and the other a pound.

[dinosaur]

No, I don't think the label "picked up from a police station" affects the problem the way you are thinking. What I believe you are not accounting for is that there are twice as many mixed families in the population as there are pure girl families.

Take 1000 two-child families (so as to intuitively ignore fluctuations). Families 1-250 are girl-girl, 251-500 had a girl then a boy, and 501-750 had a boy then a girl. Any of those 750 families could have made the quote in my post, yet there are 500 families with boy-girl, and 250 with girl-girl.

Or another example that I think speaks more directly to your post: you see a woman with a T-shirt reading "Proud Mother of Two" next to a girl who is obviously her daughter. What is the probability of her other child being a boy? Again, since there are twice as many mixed families as pure girl families, the odds are 2/3.

[seano]

"Or another example that I think speaks more directly to your post: you see a woman with a T-shirt reading "Proud Mother of Two" next to a girl who is obviously her daughter. What is the probability of her other child being a boy? Again, since there are twice as many mixed families as pure girl families, the odds are 2/3."

The odds are 50%, do you seriously think different?

Your mistake is that in families 1-250 the girl next to the mother could be either daughter.

In families 1-750 there are 1000 daughters, the daughter standing next to the mother is equally likely to be any of those and half have brothers, half have sisters.

[dinosaur]

Sorry; you are definitely right about that. I was so focused on the mom with the T-shirt that I did not even think about being able to count the states with the girl, but yes, the answer is obviously 50%.