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Comment Re:0.5 (Score 1) 981

You're implying in a random sample of 1000 pairs of children, 1/3 will be boy/boy, 1/3 girl/boy and 1/3 girl/girl...

That's incorrect - 1/2 of the pairs will be one girl and one boy. It's pretty easy to write some code to generate random pairs to convince yourself of this.

Comment Re:0.5 (Score 1) 981

But if I say that the youngest is a boy, then the probability that the oldest is also a boy is 1/2 (not 1/3). If I just say one of the children is a boy, with no other information, then I could be refering to either one in the boy/boy case, and the probability is 1/3 of both being boys.

Being younger or older is independent of the gender, but does affect the statistical outcome.

It's not easy to understand (and goes against instinct, especially as the pieces of information appear unrelated), but I wouldn't call it a trick.

Comment Re:The other problem posed in TFA (Score 3, Insightful) 981

That's incorrect - you've just skewed the population!

In 1000 pairs of children you'll have 250 girl/girl, 250 boy/boy, 500 girl/boy.

Of those, the ones that have at least one boy are the 250 boy/boy and 500 girl/boy pairs. So there's a 33% chance it's boy/boy if you know one is a boy.

The whole point is you could be talking about either of the boys in the 250 boy/boy pairs - it doesn't increase the probability that it's boy/boy instead of girl/boy (you're still twice as likely to have a girl/boy pair relative to a boy/boy pair). If you specify more about the boy you're talking about - for example (ironically) saying his name is Peter - then the boys are no longer interchangable and the probability tends towards 1/2.

It is tricky ;-)

Comment Re:He is wrong (Score 1) 981

The {known/unknown} bit is applied after the gender is fixed. By adding a fourth boy/boy option you're skewing the relative probabilities (rather than just removing the cases that no longer apply).

Maybe think of it this way: suppose there are 1000 pairs of children taken randomly from the population. On average, 250 of the pairs will be two boys, 250 two girls, and 500 a boy and a girl.

So if one pair is taken randomly from that group, and you are told at least one of the pair is a boy, then you know it's one of the 750 pairs other than girl/girl. 250 of those pairs are boy/boy, the remaining 500 are girl/boy - leaving you with a probability of 33% that it's boy/boy.

Comment Re:Ordering and Convergence (Score 1) 981

Nicely put!

So at one extreme you have BB, BG, GB (33%). At the other you have B*B, BB*, B*G, GB* (50%) where B* is the boy who fits the extra criteria and it is impossible for both boys to fit it. By making it improbable that both boys fit the criteria you're separating the BB into B*B and BB* to some degree, so the probabilities will tend towards 1/2.

Comment Re:Well? (Score 2, Informative) 981

If you ignore the Tuesday bit, then eliminating the option that you know is not possible (FF won't work since one child is a boy) you have three options remaining, each with equal probability, giving 33% chance.

Or to put it another way: If it was 25% chance like you say then so is FM and MF. There are no other possibilities so it should add up to 100%, but you only get 75%...

Comment Re:Ordering and Convergence (Score 1) 981

The number of days in the week does have 7 possibilities though... so saying it's a Tuesday narrows it down to one in seven children (on average) that fits that criteria.

That means it's less likely that both boys fit the critera (in the cases where there are two boys), which in turn alters the probabilities.

Comment Re:Ordering and Convergence (Score 1) 981

Just because it's an arbitrary division of possibilities doesn't mean it isn't useful.

Saying one of the children is a boy who's favourite number between 1 and 7 (inclusive) is 3 would give the same probability of 13/27.

You're just narrowing down which boy you're talking about - the more specific you get the closer the probability tends towards one.

"The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1)."
The wisdom of EldavoJohn

Comment another way of looking at it? (Score 1) 981

I'm trying to understand this in laymans terms. Is this a valid way to interpret it? If you know that a specific child is a boy (e.g. the youngest), then the probability the other one is a boy is 1/2. If you only know that one of the two children is a boy, and you don't know which, then the probability they are both boys is 1/3. So the more you pin down details towards identifying a specific child is a boy the closer the probability tends towards 1/2 for both boys. In this case you're specifying a day, and as that reduces the ambiguity a lot it gets you quite close to 1/2.

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