It's actually even worse for the assertion than this, because the genders of babies from the same parents are not independent statistical events.

So, if you have 1 B, you stop.

If you have 1G, you may have another child, which is somewhat more likely to be a girl than a boy. So you'd expect this policy (in isolation) to produce a surplus of girls.

You’re totally right. I guess what I wrote above doesn’t hold water. And an intuitive way to think about it is that each child does not know the sex of the child before it—if you had a line of children and each mother simply took the next child in the line, you would expect a 50-50 sex distribution even if mothers followed the rules given above.

For gambling it would would work if you do a "double or nothing" type of bet.

Unfortunately, you'll be quickly rate limited by birth capacity, even if you somehow manage to double the amount of children per birth.

Also, asymptotically, you'll still be around a 50% split.

Assuming no selective-abortion shenanigans, every child you have is expected 50% male, 50% female (barring e.g. unusual men who only produce sperm of one gender—I think that exists but is extremely rare), so there's no possible strategy that leads to an uneven gender split, probability-wise.

But I think there were selective-abortion shenanigans, and that the policy was effectively in response to them. If we assume each family wants, as its first objective, to maximize the number of sons, and, as its second objective, to maximize the number of daughters, then each selective-aborting family would have exactly one daughter followed by one son under such a policy.

Parent was talking about having indentical multiples: 1 girl, 2 girls, 4 boys, stop.

Is that correct? The left side adds to 1 and the right to 1.5.

Rewrite it as

    .50(B) + .25(G+B) + .25(G+G) = .75(B) + .75(G)

and it balances.

50% * 1 + 25% * 2 + 25% * 2 = 1.5.

Ahh, got it.