There’s a culture where couples must continue to procreate until they have one boy.
What is the boy/girl ratio?
My teacher says the answer is 1:1 because it’s still the same as flipping coins, after each birth the chance of the next baby being born is the same regardless of whether of what sex the previous baby was.
I say there must be more boys because once a couple has a boy, they stop procreating and there are no chances of having a girl. But if the couple has a girl, they continue until they have a boy. So because some couples will have a boy and no girl, there won’t be any couples with a girl and no boy.
She is correct in a large enough population because each birth event is independent of the outcomes of any previous births. If you flip a coin until you get ‘heads’ then repeat this task a couple thousand times, the ratio of heads to tails will still be ~50:50 because each flip has no relation to any of the previous flips.
You’re neglecting the fact that some couples will have multiple girls and no boy (yet no couple will have multiple boys). You could imaging flipping a coin a billion times to build up a list of babies:
GGBGBGBGGGGGBGBGBBGGB…
This has a 1:1 ratio. Now hand them out to couples in order (where “|” separates families)…
If my math is right your teacher is correct. Assuming that 100 couples give birth to babies and the chances of having one of either gender are equal, presumedly 50 would be boys and 50 would be girls. The fifty couples then have a second child, of which 25 are boys and 25 are girls. If you stopped right there that’s a 1:1 ratio, 75 each. When the 25 couples with two girls try again, it’ll split down the middle again (although obviously one side will have one extra as you can’t split a odd number evenly. And so on and so forth.
Every time whatever subset of couples tries again and successfully has a boy, an equal number will have a girl, so it’ll always remain a 1:1 ratio.
The couple must stop having children after their first boy?
Well, the expected number of boys per family is obviously 1. What about the girls? There’s a 50% chance of no girls(B), a 25% chance of 1 girl(GB), a 12.5% chance of 2 girls(GGB), etc.
So E(G) = sum(i=0 to infinity) of 0.5^(i + 1) * i
A quick application of the ratio test shows that the series converges. I’m not sure what value it converges to, but I can tell you that it will be greater than 1.
So there will actually be more girls than boys.
ETA: Actually, I can’t tell you that it will be greater than 1, because the first term in the series will be 0. Hm, let me try and estimate it. Give me a minute.
Aha! The trick is realizing that there are constantly new couples having their first baby. If you just had 1000 couples, with no babies, and said “go”, it would skew toward girls. I think.
Actually, I’m pretty sure that it converges to exactly 1.
The only way you get out of having an even boy/girl split in this situation is if you acknowledge the possibility that some parents are naturally more likely to have boys than girls, even if it evens out to 50-50 for the population as a whole. In this case, by encouraging those more likely to have girls to keep cranking them out, you can let this bias show.
Think about how each couple effects the b/g ratio.
Half the couples (of 1000, say) have just one boy. 500 boys.
Half of the remaining couples have a girl and a boy. 625 boys, 125 girls.
So, 75% counted, and the boys have a five to one lead.
Makes it easy to see how it balances out.
Nope. Lets say we have 1000 couples and have them get busy, popping out one baby a year until they have a boy, then stopping. Lets also say that we get the exactly 50% male female ratio each year (like we should with a large enough group). Lets look at what the families look like after each year.
Year 1:
500 B
500 G
total: 500 B, 500 G
Year 2:
500 B
250 GB
250 GG
total: 750 B, 750 G
Year 3:
500 B
250 GB
125 GGB
125 GGG
total: 825 B, 825 G
My perhaps unsophisticated logic says that there will be more girls in this system.
Let’s say you want as few children as possible (let’s assume, for the sake of argument we’re in a fascist country with overpopulation concerns), but cultural norms require that you bear at least one boy (let’s make our country heavily patriarchal/misogynist). Here are your possibilities:
B
GB
GGB
GGGB
GGGGB
GGGGGB
GGGGGGB
GGGGGGG(Mao sends you to prison)…
As you can see, there are more girls in this population.
Actually, reading over, this is exactly what Rysto is saying.
ETA: And as it turns out, my logic was unsophisticated.
What you’re not considering is that the GGGGGGB case is significantly less likely than the B case. Once you bring the probabilities into it, we see that the few families with a lot of girls balance out the many families with few girls exactly.
Oh man, this is a good one. At first, I had an immediate and very strong intuitive response that you’re selecting for boys and of course there will be more boys. and then I read up to Rysto’s and I was thinking, okay, at the end of this process, the median number of boys in each family will be one and the median number of girls will be higher. But I think that flight’s analysis is correct, and I evaluated Rysto’s expression and got one, and if nothing else helps, remember that every single birth is still an independent event. The strategies for the families again makes every family have a single boy, but the distribution of girls (from zero to infiniti) still averages out to one as well.
So props to the OP and this thread; I spent at least 30 seconds bewitched with the vehement belief that more boys, then more girls, and then that an equal number of either sex would be born.
But there are no “G” families. The boys get a huge head start. The two most likely of your families (“B” and “GB”) give an advandage to the boys side. Each other family get less and less likely, and they only just manage to catch up with the boys.
I did a quick simulation and it obviously converges to 1, meaning that, theoretically, there would be the same number of boys and girls. Of course, this runs on several assumptions:
The chance of a boy or girl are exactly even for every couple.
The probability of multiple male births (eg, male twins) is zero.
Every couple will ALWAYS terminate their birthing string with a boy. That is, they will not die or become unfertile prior to having a male child.
Obviously, the first one could throw the ratio in any direction, the second toward boys, and the last toward girls, but since they weren’t mentioned in the problem, they’re probably over complications.
As someone mentioned upthread, this is fundamentally the same as a completely random string of Gs and Bs, the increasing lengths of G substrings will have a probability of appearing equal to 1/2^n. So, just to reiterate, that point, this problem ultimately reduces to that intuitive problem.
Based on your logic, I think there would be more girls. Since a girl means have another baby, and a boy means stop. In an individual family (According to your logic) no matter how many children they have only one will be a boy.
This seems intuitively true, but let’s crunch some numbers, you’ll see it quickly converges. Using the assumptions I stated above. The running sum is the sum of the ratios multiplied by their probability.
Family Probability G/B Sum
B 0.5 0/1 0
GB 0.25 1/1 0.25
GGB 0.125 2/1 0.5
GGGB 0.0625 3/1 0.6875
...
G^20 B 4.77 x 10^7 20/1 0.9999895
...
