Sat Jan 12, 2013, 01:53 PM
cthulu2016 (10,960 posts)
Fun with Math
China has limits on how many children you can have. Hypothetically, let's say you are only allowed to have one child. (I don't know specifics of Chinese policy in 2013, but it doesn't matter for the hypothetical.)
So every couple can only have one child and, for whatever reason, people really want at least one son. This leads to people aborting female fetuses. Now assume that the state strongly objects to such selective abortion, for whatever reason. So a solution is proposed. Each couple can keep having children until they get a son. If your first child is a son, you're done. But if not you can keep going, and have 10 daughters before having a son, if that is how it works out. What would this solution do to the ratio of males and females? (Assuming, hypothetically, that males and females are a 5050 proposition) .

24 replies, 1586 views
24 replies  Author  Time  Post 
Fun with Math (Original post) 
cthulu2016  Jan 2013  OP 
Robb  Jan 2013  #1  
Publiuus  Jan 2013  #2  
cthulu2016  Jan 2013  #3  
Recursion  Jan 2013  #12  
cthulu2016  Jan 2013  #16  
Buzz Clik  Jan 2013  #4  
cthulu2016  Jan 2013  #6  
Buzz Clik  Jan 2013  #8  
Recursion  Jan 2013  #5  
cthulu2016  Jan 2013  #7  
Speck Tater  Jan 2013  #9  
DreamGypsy  Jan 2013  #21  
Speck Tater  Jan 2013  #23  
Recursion  Jan 2013  #10  
cthulu2016  Jan 2013  #14  
hfojvt  Jan 2013  #20  
Agnosticsherbet  Jan 2013  #11  
muriel_volestrangler  Jan 2013  #13  
surrealAmerican  Jan 2013  #15  
cthulu2016  Jan 2013  #17  
DreamGypsy  Jan 2013  #18  
bhikkhu  Jan 2013  #19  
cthulu2016  Jan 2013  #22  
bhikkhu  Jan 2013  #24 
Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 01:56 PM
Publiuus (31 posts)
2. Probability
Depends on the probability of having m vs f

Response to Publiuus (Reply #2)
Sat Jan 12, 2013, 01:57 PM
cthulu2016 (10,960 posts)
3. Yes, I should have speciffied a hypothetical 5050
Response to cthulu2016 (Reply #3)
Sat Jan 12, 2013, 03:37 PM
Recursion (44,973 posts)
12. Check out the effect on populations I worked out in #10 with non 50/50 distributions though
I think that's an interesting direction to take this.

Response to Recursion (Reply #12)
Sat Jan 12, 2013, 04:10 PM
cthulu2016 (10,960 posts)
16. Your post is interesting. See my reply
The effect on the adult population would also have builtin wrinkles... probably countervailing ones

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 01:58 PM
Buzz Clik (34,304 posts)
4. There would be more females.
Why are we playing this?

Response to Buzz Clik (Reply #4)
Sat Jan 12, 2013, 02:01 PM
cthulu2016 (10,960 posts)
6. Try it with a coin
Make heads and tails male and female.
It is counterintuitive, but it works out exactly 5050 males and females. Hence the headline "fun with math" It's a cool thing about probabilities. 
Response to cthulu2016 (Reply #6)
Sat Jan 12, 2013, 02:08 PM
Buzz Clik (34,304 posts)
8. Yeah, I actually thought about AFTER I posted, and I agree.
My original gut reaction was, "You will have families with 10 girls and one boy, so there will be more females." But, you will also have lots of families with only child, a boy. So, it comes out even.

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 01:59 PM
Recursion (44,973 posts)
5. It's the integral of a poisson process
Last edited Sat Jan 12, 2013, 02:56 PM  Edit history (2) IIRC that works out to 1 female per male if female and male children are equally likely. Will look it up when I get home.

Response to Recursion (Reply #5)
Sat Jan 12, 2013, 02:07 PM
cthulu2016 (10,960 posts)
7. Your recollection is correct
The flaw in intuition is forgetting that each sequence ends with a male.
Every couple gets a male. Half of couples get no females. The unlikely FFFFFFM type sequences are, in aggregate, balanced by the very commonplace M sequence. 
Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 02:30 PM
Speck Tater (10,618 posts)
9. Outcomes enumerated are...
1/2 of all families = S = 0.5( 0 + 1 )
1/4 of all families = DS = 0.25( 1 + 1 ) 1/8 of all families = DDS 0.125( 2 + 1 ) 1/16 of all families = DDDS 0.0625( 3 + 1 ) 1/32 of all families = DDDDS 0.03125( 4 + 1 ) ... The number of sons per family is always 1. The number of daughters per family is zero for half the families, 1 for 1/4 of the families, 2 for 1/8 of the families, 3 for 1/16 of the families... 0 + 1/4 + 2/8 + 3/16 + 4/32 + 5/64 + ... = 1 (Work it out on your calculator if you don't believe it.) Of course this assumes that an infinitesimal number of families will have infinite daughters! But for a finite population, the ratio to any reasonable number of decimal places will be 50:50. 
Response to Speck Tater (Reply #9)
Sat Jan 12, 2013, 04:38 PM
DreamGypsy (2,211 posts)
21. Of course, there is another constraint in the problem as posed...
...since the scenario one couple, the series is terminated at the maximum number of children (girls, in particular) that a woman can deliver before she stops ovulating.
Some guidance from Wikipedia: Vassilyev and his first wife, whose name is unknown, holds the record for most children a couple has parented. She gave birth to a total of 69 children. She gave birth to 16 pairs of twins, 7 sets of triplets and 4 sets of quadruplets between 1725 and 1765, in a total of 27 births. 67 of the 69 children born are said to have survived infancy.
No ratio of gender for the children is given and we also don't learn the distribution of identical and fraternal siblings in the multiple births, but sixty nine girls is definitely a safe upper bound to to number of terms in the sequence. Thomas Greenhill was the last child of 39 by his mother Elizabeth and William Greenhill. The family consisted of 7 sons and 32 daughters. Not only is this a large number of live newborns but unusual in that apart from one pair of twins they were single births.
Assuming that the distribution of male children among female children is random (which is, of course, an underlying assumption in this whole thread), this example suggests that a male birth after every say 4 to 7 female births could be valid in bounding the sequence to estimating the population outcome. 
Response to DreamGypsy (Reply #21)
Sat Jan 12, 2013, 05:10 PM
Speck Tater (10,618 posts)
23. I also didn't account for twins and triplets... etc.
If a couple had triplet sons, for example. But those could be considered rare enough that they wouldn't tilt the statistics significantly, especially since sons would probably been born with about the same frequency of triplet daughters.

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 03:12 PM
Recursion (44,973 posts)
10. So, the generalized answer, given the unequal birth rates we observe, plus the effect on population
EDITED: I screwed up the population calculation at first.
Assuming each childbirth is male with probability p and female with probability q = (1  p), this is a series of identically distributed independent Bernoulli trials, you have a male on average every 1 / p births. So in the simple case where p = 0.5, you have 1 male every 2 births, and therefore as many males as females. We observe in most populations that p = 0.5025 or so (presumably we evolved to produce slightly more males to account for higher male mortality), so you would have on average 1.05 males per every female (which is roughly what we have in the US). The point is, your plan keeps the ratio of males to females at whatever it is "naturally". Now, what is the expected population? Given that every couple stops after a male, you will have Half of your couples having 1 child A quarter having 2 children An eighth having 3 children A sixteenth having 4 children, i.e., the probability that a couple has n children is 2 ^ (n). So, with X couples, the number of children in the next generation is 0.5 X + (0.25 * 2) X + (0.125 * 3) X + ... ie, sigma (n = 0 > infinity) of n * 0.5 ^ n, which in closed form is 0.5 / (1  0.5)^2 = 0.5 / 0.25 = 2. So you will have twice as many children as you have couples, which is a bit under the population replacement rate (since not everyone has children, and not all children survive to adulthood). With the actual uneven birthrates, instead of 0.5 ^ n you have 0.5025 ^ n, which yields sigma (n = 0 > infinity) of n * 0.5025 ^ n, which in closed form is 0.5025 / (1  0.5025) ^ 2 = 0.5025 / 0.2475= 2.03 children per couple Imagine we switch the sexes, and couples had boys until they got a girl? Then you have p = 0.4975, 1 / p = 2.01, and the series is n * 0.4975 ^ n which in closed form is 0.4975 / (1  0.4975) ^ 2 = 0.4975 / 0.2525 = 1.97 children per couple. So, as you can see, even small perturbations in the gender probabilities can have fairly large impacts on the population size with this scheme. 
Response to Recursion (Reply #10)
Sat Jan 12, 2013, 04:08 PM
cthulu2016 (10,960 posts)
14. Interestingly... Fisher's law is usually expressed wrong
Last edited Sat Jan 12, 2013, 04:42 PM  Edit history (1) Fisher's law is usually expressed is that a sexually reproducing species will tend toward 5050 births.
But the evolutionary pressure involved doesn't apply to births. It applies to births that go on to result in grandchildren. The driver of Fisher's law is what being of one sex or another is "worth" in terms of subsequent genetic survival through offspring. So the real 5050 balance should be found in the breeding population, not the infant population. If male and female children died (or were otherwise removed from the future viable breeding population) at the same rate the two populations would be the same. But if boys died at a slightly higher rate before sexual maturity we would expect to see more male births. And they do, and we do. In terms of the number and quality of sperm that can make a male vs. a female, the femalemaking human sperm are actually handicapped so that the number of protomale zygotes is MUCH higher. Males have a high failure rate (probably because they require the additional developmental step of turning male from the female default). Most spontaneous abortions are male or protomale. Handicapping some sperm with slower speed to achieve a high male zygote rate since males fail more often, to result in a birthrate that will, when male children die at a slightly higher rate, result in a 5050 population at sexual maturity is the sort of halfassed jerryrigging that only evolution would do. It is so clearly not designed, but arising through a series of steps that don't know what the next step will be. (The fact that the tube from the testes to the penis is tied in a knot is another sort of fuck up that can only sensibly arise from an imperfect stepbystep process.) 
Response to Recursion (Reply #10)
Sat Jan 12, 2013, 04:28 PM
hfojvt (36,564 posts)
20. then for 100 couples
50 boys + (25 boys + 25 girls) + (12 boys + 24 girls) + (6 boys + 18 girls) + (3 boys + 12 girls) + (2 boys + 10 girls) + (1 boy + 6 girls) + (1 boy + 7 girls) or 100 boys + 102 girls or really the ratio is still 5050.

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 03:24 PM
Agnosticsherbet (7,460 posts)
11. China's preference of male babies and one child policy has skewed the ratio...
http://www.guardian.co.uk/world/2011/nov/02/chinasgreatgendercrisis
Now, there are between 120 and 130 men born for every 100 women. (One assumes selective abortion, though historically girls in Chinz were exposed at birth and allowed to die.) China today is short about 35 million women. Your hypothetical suggestion might mean more women, but only if families are happy to have bunches of girls for every boy. But a willingness to use selective abortion in China would probably still keep most educated families small as it has across the rest of the world. http://www.prb.org/Educators/TeachersGuides/HumanPopulation/Women.aspx 
Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 03:52 PM
muriel_volestrangler (78,799 posts)
13. Without selective abortion, the chances of each birth are still 50:50
and it doesn't matter if the child has elder siblings or not, that will still mean a 50:50 population.

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 04:10 PM
surrealAmerican (9,722 posts)
15. Even without genderselective abortion ...
... in a realworld situation, this would yield more males than females. Families have a actual limits on the number of children they will produce. They may only be able to afford to feed two children; they may have started reproducing at an age where their fertility won't exceed three; complications in the first pregnancy could make it their only child; etc.
Just because you are legally "allowed" to have 10 daughters before having a son, doesn't mean anybody will actually do that. 
Response to surrealAmerican (Reply #15)
Sat Jan 12, 2013, 04:12 PM
cthulu2016 (10,960 posts)
17. That is an excellent point
Yes, in the reproductive real world it would favor males, despite the intuitive sense of the opposite

Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 04:15 PM
DreamGypsy (2,211 posts)
18. Rosendcrantz and Guildenstern are dead.
Last edited Sun Jan 20, 2013, 11:12 AM  Edit history (1) A great play by Tom Stoppard, starring the two eponymic minor characters in Hamlet, is a fine and funny journey into probabilities in a world slightly awry.
(Script and some commentary available here.) Each of them has a large leather money bag.
Guildenstern's bag is nearly empty. Rosencrantz's bag is nearly full. The reason being: they are betting on the toss of a coin, in the following manner: Guildenstern (hereafter 'GUIL') takes a coin out of his bag, spins it, letting it fall. Rosencrantz (hereafter 'ROS') studies it, announces it as "heads" (as it happens) and puts it into his own bag. Then they repeat the process. They have apparently been doing it for some time. As they game continues, Guildenstern begins to muse about the situation ROS: Heads. (He picks it up and puts it in his money bag. The process is repeated.) Heads. (Again.)
ROS: Heads. (Again.) Heads. (Again.) Heads. GUIL (flipping a coin): There is an art to the building up of suspense. ROS: Heads. GUIL (flipping another): Though it can be done by luck alone. ROS: Heads. GUIL: If that's the word I'm after. ROS (raises his head at GUIL): Seventysix love. (GUIL gets up but has nowhere to go. He spins another coin over his shoulder without looking at it, his attention being directed at his environment or lack of it.) Heads. GUIL: A weaker man might be moved to reexamine his faith, if in nothing else at least in the law of probability. <snip> GUIL (understanding): Games. (Flips a coin.) The law of averages, if I have got this right, means that if six monkeys were thrown up in the air for long enough they would land on their tails about as often as they would land on their  ROS: Heads. (He picks up the coin.) <snip> GUIL: It must be the law of diminishing returns... I feel the spell about to be broken. (Energizing himself somewhat.) (He takes out a coin, spins it high, catches it, turns it over on to the back of his other hand, studies the coin  and tosses it to ROS. His energy deflates and he sits.) Well, it was a even chance... if my calculations are correct. ROS: Eightyfive in a row  beaten the record! <big snip> ROS: Eightynine. GUIL: It must be indicative of something, besides the redistribution of wealth. (He muses.) List of possible explanations. One: I'm willing it. Inside where nothing shows, I'm the essence of a man spinning doubleheaded coins, and betting against himself in private atonement for an unremembered past. (He spins a coin at ROS.) ROS: Heads. GUIL: Two: time has stopped dead, and a single experience of one coin being spun once has been repeated ninety times... (He flips a coin, looks at it, tosses it to ROS.) On the whole, doubtful. Three: divine intervention, that is to say, a good turn from above concerning him, cf. children of Israel, or retribution from above concerning me, cf. Lot's wife. Four: a spectacular vindication of the principle that each individual coin spun individually (he spins one) is as likely to come down heads as tails and therefore should cause no surprise that each individual time it does. (It does. He tosses it to ROS.) ROS: I've never known anything like it! GUIL: And syllogism: One, he has never known anything like it. Two: he has never known anything to write home about. Three, it's nothing to write home about... Home... What's the first thing you remember? ..... and the fun continues from there. I won't reveal the punchline. But as you might guess, the play ends with 'thanks'. 
Response to cthulu2016 (Original post)
Sat Jan 12, 2013, 04:22 PM
bhikkhu (9,750 posts)
19. Done that way: 100% chance of having a son, 50% chance of having a daughter
So I would imagine (without thinking too hard on it), the odds of any family having one child  a boy  would be 50%. That would be half of families with one male child.
The other half would have a girl, then a 50% chance of having a second child who was a boy  so you'd have 25% of the total with one boy and one girl. Then that 25% of the total (with two girls each now) would have a 50% chance of a girl or a boy, winding up with 12.5% of the total having a boy and two girls...and so forth. Without working it all out, I wonder if it would wind up with 2.718 times more girls than boys? 
Response to bhikkhu (Reply #19)
Sat Jan 12, 2013, 04:39 PM
cthulu2016 (10,960 posts)
22. You started out very well, but
it works out to dead even (if this sterile hypothetical where the birth odds are exactly 5050 and woman can give birth to any number of children)
M = 50% of cases / population = MMMMMMMMMMMMMMMM FM = 25% / population = MMMMMMMMMMMMMMMMMMMMMMMMFFFFFFFF FFM = 12.5% / population = MMMMMMMMMMMMMMMMMMMMMMMMMMMMFFFFFFFFFFFFFFFF FFFM = 7.25% / population = MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMFFFFFFFFFFFFFFFFFFFFFF FFFFM = 3.625% / population = MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMFFFFFFFFFFFFFFFFFFFFFFFFFF etc. As you extend it out the females keep stepping closer and closer to 50%, with the two ending up at parity. In the real world, of course, people have to stop somewhere. But as a stylized math problem it is a surprising 5050. 
Response to cthulu2016 (Reply #22)
Sat Jan 12, 2013, 08:03 PM
bhikkhu (9,750 posts)
24. Of course...
I should have known it would be one of those "common sense doesn't work" problems. I had to sit down and make a table like yours to see that girls proliferate enough to compensate for the shrinking percentages of the population left, and it does work out even.
