Given what I know about the current situation, the probability that their other child is a girl is 66.666%

If I look out the window to that neighbor's yard and see a boy playing outside, the probability that their other child is a girl drops to 50%.

How I know the information about on data point one affected the likelihood of an unrelated one. What seems like identical knowledge on the outside, changes outcomes based on how I accrued that knowledge. If that doesn't mimic the double-slit experiment, I don't know what does.

Probability means this: "in X tries, how likely is it for Y to happen". The ambiguity here is this: what is the set of X? Is the set of all families with two children at least one boy, or the set of all families with 2 children of which you have determined one is a boy.

Those are different sets. It doesn't matter how you discovered the information. The apparent paradox is all grammar and set description.

Think of it this way - you want to test this out and actually run this experiment? Great! What's the experiment? Are you starting with all families with 2 children, or all families which 2 children at least one boy.

>> How I know the information about on data point one affected the likelihood of an unrelated one

No, it doesn't.

Somehow, you're saying that seeing the boy means you have ruled out the possibility of "a girl and a boy". But the ordering is in no way related to some arbitrary observation. It's just lexical.

For the first case here are the different combinations for two children based on sex (let's say left denotes older and right denotes younger to distinguish the difference between girl and a boy and boy and a girl.)

B B

B G

G B

G G

These outcomes are each equally likely, but since we know that one has to be a boy we can eliminate the (G G) case leaving us with two chances out of three that the other child is a girl.

The other case can be viewed as follows (S) denotes seen.

B(S) B

B B(S)

B(S) G

B G(S)

G(S) B

G B(S)

G(S) G

G G(S)

I can eliminate all cases where I see a girl out the window, leaving me with the following equally likely cases:

The neighbor has two boys and I see the older boy. The neighbor has two boys and I see the younger boy. The neighbor has an older boy whom I saw and younger girl. The neighbor has an older girl and a younger boy whom I saw. The neighbor has a girl in 50% of the remaining possible cases.

If differentiation is not possible, then they are the exact same person [1], so the parents have one child and it must be boy then ;)

[1] https://en.wikipedia.org/wiki/Identity_of_indiscernibles

One of the twins is still considered to be born first and therefore older :-)

While you may argue that one twin is always born first, this is an idealized mathematical problem, so it can't depend on particular details of physical reality in that way. There is nothing logically preventing them from being the same age, so it can't be pivotal to solving the problem.

And in fact, I experimentally got the right result without any reference to ages.

People like to write the outcome space as ordered pairs BB, BG, GB, GG (ordered by age or whatever else), because this has the advantage that all four outcomes are equiprobable, so you can calculate probabilities just by counting.

If you prefer to regard the outcome space as unordered pairs {B,B}, {B,G}, {G,G} this is absolutely fine: you just have to bear in mind that the prior probability of {B,G} is 1/2, whereas it's 1/4 for the other pairs, so a tiny bit more calculation might be required.

P({B,G}| not {G, G}) = P({B,G})/(P{B,B} + P{B,G}) = (1/2)/((1/2) + (1/4)) = 2/3 as with the other method.