Let's say I flipped two coins. I tell you I did this. You ask me, "Did heads come up at least once?" I answer yes.

What are the odds that the other coin was tails? 2/3, because there are three scenarios (HH, HT, TH) that satisfy my first answer to you.

If you flip a pair of coins a few thousand times, and search the flip pairs for all the ones that had a heads among them, you'll find a tails in those pairs 2/3 of the time, and a pair of heads 1/3 of the time.

Same with the girl-boy problem.

Now if you ask me if the left coin came up heads—and I say yes—that doesn't give you any information whatsoever on the right coin. And if you search through the thousands of pair flips looking for an "H" in the left column, the right column will 50/50 have an H or a T.

That's a good explanation.

> Now if you ask me if the left coin came up heads—and I say yes—that doesn't give you any information whatsoever on the right coin.

Exactly. "At least one" gives information about the whole set, implicitly giving information about the second one whereas "the first one is X" gives no information about the second one.

Because if you select a random family from the set of familes with two children, the 4 permutations (BB,BG,GB,GG) are equally likely, at 25% each, so it's the simplest way to model it. If you model it in your way with 3 possible scenarios you'll find that the probabilities are BB: 25%, one of each: 50%, GG: 25%, and dealing with those differing probabilities is harder than just splitung the one of each scenario into two.

Try it with 2 coins flipped at the same time and you'll see it's the case. You'll get 2 unmatching coins about twice as often as you get either HH or TT.

I read the wikipedia article and then it clicked for me. Here is the intuition I got:

If you know one of the kids is a boy then there is a 2/3 chance of there being a girl kid. I will assume you are happy with the reasoning behind this (3 options BG GB BB, two of which yield a girl).

Now why does spotting one of the boys change the odds? You already knew they had a boy right?

Well the boy being spotted conveys some information!

In particular you DIDN'T see a girl. You could have seen one, it would be possible right... but it didn't happen.

That fact means it is a bit more likely that they have two boys. How much is that bit, well 1/6th to be precise. So 1/3+1/6 = 1/2 chance they have two boys. And 2/3-1/6 = 1/2 chance they have two girls.

That gives me the intuition and makes this no longer a paradox intuitively. Although it was never a paradox logically.

CAREFUL:

The conditions under which the boy was 'revealed' to you matter.

If it was random 50-50 chance of seeing the girl or the boy then the above applies.

If the boy was picked, i.e. you said "ah you have at least 1 boy, show him to me". Then no information has been conveyed.

If something else situational (e.g. girls birthday party and all the girls are inside in the room, and boys are playing soccer outside [sorry for stereotyping but my imagination isn't so good] and so you see him) there may be a non 50-50 and non 100-0 probability then there are some more calculations to do :-).

>The way I see it, it should be modeled as {both-boys, boy-and-girl, both-girls}; by telling me that one of the kids is a boy, you eliminated one possibility out of three, leaving us with 1/2 as an answer.

This would be like saying "When I buy a lottery ticket, the only outcomes are {lose lottery, win lottery}; therefore, I have a 50% chance of winning the lottery."

The difference is that boy-girl is twice as likely to occur of those three choices.

It's not the ordering that matters, it's simply that the odds of boy-and-girl are twice the odds of boy-and-boy.