[spadros]

Yes, I found this one easy. Was surprised my data management intuition came back after all these years since school. There’s really only three options:

- boy - boy

- boy - girl

- girl - girl

So it must be 1/3 chance. If you’re looking at permutations in order, that’s a different question.

[AIPedant]

This intuition is wrong even if turned out to get the right answer. The three unordered options do not have equal probabilities, boy+girl is twice as likely to occur as boy+boy and girl+girl.

To get the right answer you must be careful about conditional probabilities (or draw out the sample space explicitly). The crux of the issue is that you are told extra information, which changes your estimate of the probability.

(This question as written is very easy to misinterpret. The Monty Hall problem, which illustrates the same thing, is better since the sample selection is much more carefully explained.)

[taeric]

Oddly, this is a part I'm sticking with on this problem.

Specifically, if you know that one is a girl, then the unordered options seem like they are back on equal footing? That is, it isn't twice as likely if you know that one ordering can't happen? (Or, stated differently, you don't know which version of two girls you are looking at.)

So, for this one, you know that either the youngest is a girl (so, girl-boy is not possible) or that the oldest is a girl (so boy-girl is out). That puts you back to the rest of the possibilities. Boy-boy is out, sure, as you have a girl. But every other path remains? So, you have one of (boy-girl(known), girl-girl(known), girl(known)-boy, girl(known)-girl). Which drops you back to 50/50?