6 ms·

50% B + 25% GB + 25% GG = 0.75 B + 0.75 G. Can you think of a law ("stopping rule") that could even get the expected number of boys higher than the expected nu

50% B + 25% GB + 25% GG = 0.75 B + 0.75 G. Can you think of a law ("stopping rule") that could even get the expected number of boys higher than the expected number of girls?

dmichulke 5y ago
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.
- waterhouse 5y ago
  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.
  - alisonkisk 5y ago
    Parent was talking about having indentical multiples: 1 girl, 2 girls, 4 boys, stop.
dandellion 5y ago
Is that correct? The left side adds to 1 and the right to 1.5.
- areyousure 5y ago
  50% * 1 + 25% * 2 + 25% * 2 = 1.5.
  - dandellion 5y ago
    Ahh, got it.
- function_seven 5y ago
  Rewrite it as .50(B) + .25(G+B) + .25(G+G) = .75(B) + .75(G) and it balances.
dan-robertson 5y ago
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.
mlyle 5y ago
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.

50% B + 25% GB + 25% GG = 0.75 B + 0.75 G. Can you think of a law ("stopping rule") that could even get the expected number of boys higher than the expected number of girls?

dmichulke 5y ago

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.

waterhouse 5y ago
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.
- alisonkisk 5y ago
  Parent was talking about having indentical multiples: 1 girl, 2 girls, 4 boys, stop.

dandellion 5y ago

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

areyousure 5y ago
50% * 1 + 25% * 2 + 25% * 2 = 1.5.
- dandellion 5y ago
  Ahh, got it.
function_seven 5y ago
Rewrite it as .50(B) + .25(G+B) + .25(G+G) = .75(B) + .75(G) and it balances.

dan-robertson 5y ago

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.

mlyle 5y ago

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.