100% was the right answer.
It's easy to get people to argue when you give them an almost-ambiguous word problem; they're not arguing about the math, they're arguing about the mapping from ambiguous words to math.
I think this would be a better quote of what the person might have said:"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" Pulling out the information corresponding to gender and family size would give only the information given in Jeff's post.
When applying math to the real world, you have to pull out the important information and deal with just that information. But here you are doing the opposite--trying to find a real world situation that applies to the math problem. In my opinion, your example does not quite apply.
(I don't know how the probabilities change when you account for hermaphrodites, but if it changes significantly enough so that approximately 66% is a bad answer, I would find that very interesting!)
If I am correct about that, then it matches the conditions discussed in the article and the answer would be 2/3 for a boy and a girl.
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
G'g, B'g, G'b, B'b, Gg', Bg', Gb', Bb'
G'g, G'b, Gg', Bg'
Gg, Bg, GbIt 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%).
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.
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.
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.
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.