[tijsvd]

With 2 children, there are 4 configurations of equal probability. The one with 1 boy 1 girl occurs twice. Take away the 2 girl case, then 2 boys is 1 in 3.

[vlovich123]

Yeah, the way the problem is formulated though there’s absolutely no indication that order matters so how are there two configurations within which there’s 1 boy and 1 girl?

[da39a3ee]

Order doesn't matter in the sense that the observed data set is unordered (just counts of girls and boys). What matters is how many ways there are that the universe can give rise to those unordered data sets. And in fact, there are more ways that the universe can give rise to the unordered state 1 boy 1 girl, than to the unordered state 2 boys. For similar reasons , there are more ways in which your papers can be in a mess across your desk than ways in which your papers can be neatly piled up.

And to count how many ways the universe can give rise to the unordered data sets, the usual technique is to expand the unordered data sets into all the equivalent ordered data sets, and count the latter.

[maxov]

Another way to think about it is counting the probability of getting k boys out of 2 children.

  0 boys - 1/4
  1 boy - 1/2
  2 boys - 1/4

There's a half chance of getting exactly one boy, and one way to calculate this is by noticing there are two different ways to get one boy if we take order in account. You are right that the orderings don't matter in this case, so we could also e.g. model this with a binomial distribution. Once you know there are >= 1 boys, the chance you have two is 0.25/(0.25+0.5) = 1/3.

[kgwgk]

Because the order exists even if it doesn't matter (at least for two children, maybe not for two quantum particles).

With the risk of being accused of binarism, there are four distinct possibilities with (close to) equal a priory probability of 25%: older boy/younger boy, older boy/younger girl, older girl/younger boy, and older girl/younger girl.

Discarding the girl/girl case leaves three equally probable cases.