Timeline for Google question: In a country in which people only want boys
Current License: CC BY-SA 4.0
56 events
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| S Apr 23, 2023 at 16:45 | history | suggested | Benjamin Wang | CC BY-SA 4.0 |
clarified ambiguous fraction 1/2 - 1/4k (and added \left(\right) fixes to meet minimum edit limit)
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| Apr 21, 2023 at 19:35 | review | Suggested edits | |||
| S Apr 23, 2023 at 16:45 | |||||
| Nov 2, 2021 at 5:34 | comment | added | user4722818 |
This post is all correct except for the last paragraph. This property you describe is true in general, but this problem is a simpler case. It's trivial to show that E[G]/E[G+B] = 1/2 if G and B are i.i.d variables... E[G]/E[G+B] => 1/(E[G+B]/E[G]) => 1/(E[G]/E[G]+E[B]/E[G]) => 1/(1+1) => 1/2`
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| Apr 16, 2020 at 19:55 | comment | added | Douglas Zare | @lamanon The calculation is similar to the binomial probability mass function. See the negative binomial distribution. | |
| Apr 15, 2020 at 20:09 | comment | added | anonuser01 | @DouglasZare I enumerated out a bunch of terms and it seems to simplify to what you have. For example, I enumerated the probability of family 1 having 1 girl and the probability of family 2 have 1,2,3,4, etc... girls and finding the joint probability and the values (proportion of girls) among the 2 families, and then did it for family 1 having 2 girls, and so on. Is that what you did, or did were you able to come up with that formula by inspection? If it's the latter, could you go into some of your thought process on how you got that formula quickly? | |
| Apr 15, 2020 at 19:55 | comment | added | anonuser01 | @DouglasZare Could you please explain where you got the expression from for 2 families? I understand the expression for 1 family, which is simply just the formula for expected proportion of girls for a family. | |
| Jun 21, 2019 at 15:29 | comment | added | Michael | @DouglasZare, oops, thanks, don't know what I was thinking. | |
| Jun 21, 2019 at 11:50 | comment | added | Douglas Zare | @Michael: The probability of $g^nb$ is not $2^{-n}$ but rather $2^{-(n+1)}$, and the proportion of girls in that case is $\frac{n}{n+1}$, leading to the expected proportion $\sum_{0}^{\infty} \frac{n}{n+1}2^{-(n+1)}$. | |
| Jun 19, 2019 at 18:38 | comment | added | Michael | @DouglasZare, could you explain your solution for 1 family? I got confused a bit considering the possibilities b, gb, ggb, gggb, etc, where g and b are girl and boy born in sequence. The probability of $g^nb$ would be $2^{-n}$, therefore the proportion of girls would be $\Sigma_{n=0}^{\infty}n2^{-n-1}$, which leads to an incorrect result. | |
| Sep 11, 2017 at 16:07 | comment | added | Randy Lai | With a bit algebra, one can show that the expected girl ratio can be expressed in terms of hypergeometric function: 2F1(k,1,k+1,−1). randycity.github.io/blog/girl-ratio.html | |
| Apr 26, 2015 at 2:42 | comment | added | Steven Landsburg | A tiny correction, years later: In the fourth comment, $(-1)^k$ should be $(-1)^{k+1}$. | |
| May 27, 2013 at 6:10 | comment | added | user112109 | @Jon Peterson: A winning strategy also is one which guarantees many different people to win money in short runs - in paricular if these people cannot be distinguished. | |
| May 17, 2013 at 21:51 | comment | added | Steven Landsburg | I want to make clear that the upvote for Rhett Butler's comment "You have found a roulette winning strategy..." is mine and was cast in error when I was trying to flag this comment as spam. | |
| May 17, 2013 at 11:01 | comment | added | Jon Peterson | @Rhett Butler: One more comment. The fraction of wins up until a random time isn't even the correct thing to look at in roulette. Under the strategy above, even though the expected fraction of wins would be 53% by the time of the 10-th win, the expected actual winnings would be 0 dollars (assuming the casino pays out even money, which is of course wrong). This is because $B_n-G_n$ is a martingale. | |
| May 17, 2013 at 10:35 | comment | added | Jon Peterson | @Rhett Butler: You misunderstand what it means to have a "winning strategy." A winning strategy is one which guarantees you win money in the long run. Using this as a roulette strategy only gives a strategy where the expected fraction of wins is 53% for a short period of time (up to the 10-th black). If you try this repeatedly, then you essentially have increased the number of "families" and the answer approaches 50%. Therefore, this strategy will (not surprisingly) not guarantee you make money. | |
| May 14, 2013 at 18:20 | comment | added | user112109 | @Jon Peterson: $\frac{G_n}{B_n + G_n}$ is not a martingale. But it is completely irrelvant to calculate it. $\frac{E(G)}{E(B + G)}$ is asked for. The latter is 1/2 for every country and every number of families. | |
| May 14, 2013 at 18:11 | comment | added | user112109 | @Jon Peterson: If for 10 rounds/families always stopping at b (for black or boy) will supply 47 % red/girl and 53 % black/boy, then you have a winning strategy. Always bet the same amount of money on black. You will win more frequently than you will lose. | |
| May 14, 2013 at 16:35 | comment | added | Douglas Zare | @Rhett Butler: Do you really think I have no idea that the sex of a child is modeled as $1/2$ independently of what came before, even though this is used in my calculations? You think that triviality is what everyone else is missing, too? How stupid. The mathematically interesting thing to me is that when the families can choose how many children to have based on the previous sexes, the proportion $B/(B+G)$ is a biased estimator of that $1/2$, as I stated in my answer, which means the expected value is not $1/2$. And $E[B/(B+G)]$ is what the OP's summation tried to calculate. | |
| May 14, 2013 at 10:54 | comment | added | Jon Peterson | @Rhett Butler: You never addressed the main point of my remark - that $\frac{G_n}{B_n + G_n}$ is not a martingale. My point is that agreeing that Douglas's answer is correct does not imply that one has a "winning roullete strategy." Secondly, I agree that when $n$ is large this fraction is very unlikely to deviate from $1/2$. However, you go too far when you say "the population will never deviate by more than the statistical fluctuations from the 50:50 equipartition." This is plainly false. In fact the law of the iterated logarithm shows that there will always be some such deviations. | |
| May 14, 2013 at 6:56 | comment | added | user112109 | @Douglas Zare: Whatever you did, you did not correctly answer the question whether family planning can influence the equilibrium beteen boys and girls. The answer is no. But you claim that this answer is correct only for large populations and that your answer is different and the only correct one for smaller populations. Once you will recognize that the sex of a child is in no way dependent on the intentions or history of the mother, you should see your error. Your calculation of the weighted average over the ratios is not what you claim it was, namely an answer to the original question. | |
| May 13, 2013 at 18:10 | comment | added | Douglas Zare | @Rhett Butler: It sounds like you are confused about basic probability. I assure you that I am not. You ask what the definition is of expected value. You say expected value is only defined for a large number of trials. You incorrectly try to apply the optional stopping theorem. EV is a basic idea I have taught many times and which you can find explained in many introductions or my book. It is not restricted to large repeated samples. I explained multiple times that $B/(B+G)$ is not a martingale, unlike $B-G$, so the OST for martingales does not apply to $B/(B+G)$, and the conclusion fails. | |
| May 13, 2013 at 14:50 | comment | added | user112109 | @Jon: The probability of a girl to be born is a martingale, completely independent of the number of girls and of the history of their mothers. The expectation value of additional girls within the next 200 births is 100 with an error marge of 10. Same holds for boys. Therefore the population will never deviate by more than the statistical fluctuations from the 50:50 equipartition. | |
| May 13, 2013 at 14:39 | comment | added | user112109 | @Jon: The answer calculated the average of the percentage of girls. But the question asks for the percentage of all girls. Further what is an expectation value? It has no meaning for a single case but only for a big number of cases. If 31 % girl expectation for a single family would be correct, then the ensemble of all families of the country would get close to it. If you don't believe in my explanations, then play roulette. Always bet 5 bucks on black and stop a sequence after black has appeared. Within 3000 sequences you should have earned more than 1000 bucks. Good luck! | |
| May 13, 2013 at 12:36 | comment | added | Jon Peterson | @Rhett: I'm not sure I understand your comment about the "winning roulette strategy," but maybe this will explain things. You seem to want to apply some martingale theory where it isn't appropriate. The fraction of girls in the population is not a martingale. That is, let $G_n$ and $B_n$ be the number of girls and boys respectively in the first $n$ births, and let $X_n = \frac{G_n}{B_n+G_n} = \frac{G_n}{n}$. It's easy to see that $E[X_n] = \frac{1}{2}$. However, $X_n$ is not a martingale, since $E[X_n | X_{n-1} ] = \frac{n}{n+1}X_n + \frac{1}{2(n+1)}$. | |
| May 13, 2013 at 10:40 | comment | added | user112109 | @Douglas: You have found a roulette winning strategy without raising the stakes. No doubts? | |
| Jan 5, 2011 at 4:26 | comment | added | Tom Church | @Steve Landsburg: There were some very interesting and thoughtful comments on your post. However, since each argument continues until the participants finally agree, I expected that the fraction of thoughtful comments would be just over half. Sadly this seemed not to be the case... | |
| Jan 3, 2011 at 14:17 | comment | added | T.. | @DZ: the statement "[allowing unfinished families] ... there is a higher proportion of girls when the population is larger" is in general false. It is true only in your a priori asymmetrical model conditioned on the number of families. The asymmetry arises not from the stopping rule, but because the stopping rule allows phrasing of boy/girl asymmetric conditions ("the number of boys is at most $k$") in equivalent terms without direct reference to $B$ or $G$ (i.e., "the number of families is $k$", as in your model allowing unfinished families). This asymmetry is foreign to the Google puzzle. | |
| Dec 28, 2010 at 0:21 | comment | added | Steven Landsburg | I'm not entirely sure this comment is appropriate here, and I'll happily delete it (or let someone else delete it) if more experienced users tell me to, but my blogpost citing this response has stirred up a considerable firestorm of comment, some fraction of which is thoughtful: thebigquestions.com/2010/12/27/win-landsburgs-money | |
| Dec 21, 2010 at 20:11 | comment | added | Douglas Zare | @Steven Landsburg: Thanks for pointing that out. That paragraph has been bugging me since it isn't even the independence of G with G+B which is needed. Expectation just doesn't commute with most operations. | |
| Dec 21, 2010 at 16:04 | comment | added | Steven Landsburg | Your last sentence says that G and G+B are not independent, even though G and B are. This strikes me as (ever so slightly) misleading, because the independence of G and B is a red herring, in the sense that all the same phenomena would occur whether or not we had this independence. (E.g. in a model where everybody stops after two children, this independence goes away but all the really interesting phenomena remain.) That teeny quibble aside, thanks for this extremely enlightening explanation. | |
| Dec 21, 2010 at 10:22 | comment | added | Douglas Zare | @Lucas K. No, that simplification is not a necessary assumption. The expected value of $G/(G+B)$ is not $1/2$ even if you allow unfinished families. If the population size is not constant, and there is a higher proportion of girls when the population is larger, then girls tend to make up a smaller portion of the population, so $E[G/(G+B)] \lt 1/2$. I invite you to make explicit computations as I did instead of just stating your intuition. Many people find this counterintuitive. | |
| Dec 20, 2010 at 23:27 | comment | added | Lucas K. | I just wanted to say, that you assume that all children of one family are born instantaneous (with the last child a boy). If you take into account "unfinished" families, than the proportion is directly 50/50 (I think, because how matter what, the change of boy for a child is 50/50). | |
| Jul 16, 2010 at 18:13 | comment | added | T.. | Jensen's inequality does clarify the circularity of the argument, though. $E[G/(G+B)] \leq 1/2$ is equivalent to bivariate convexity of $f(G,B)= -G/(G+B)$. But $f$ is convex for fixed $B$ (fixing the number of boys biases $E$ downward), concave for fixed $G$ (setting the number of girls first biases $E$ upward, $\geq 1/2$), and linear for constant $x+y$ (no bias in models with population size determined prior to $B$ and $G$). To get $E \leq 1/2$ in any model you have to somehow build in an asymmetry that concentrates the number of boys and disperses the number of girls. | |
| Jul 16, 2010 at 8:09 | comment | added | T.. | Stipulating the number of families carries (artificially introduced, exogenous) information on the number of boys and this fully accounts for the "bias" in the models displayed thus far. The number of boys and girls is not constant in the gender-symmetric models that first determine the population size and then fill the families, so clearly some additional assumptions are needed to get $E < 1/2$, and the question is whether they are just another form of artificial conditioning. i.e., is the bias from the model selection or internal to the model itself. | |
| Jul 16, 2010 at 7:12 | comment | added | Douglas Zare | By the way, in the formalizations where there are a fixed number of boys, Jensen's inequality applied to $G/(G+B)$ implies that $E[G/(G+B)] \le 1/2$ with equality only if the number of girls is constant. Again, this applies even to more complicated models where the number of boys is not constant, such as asking about the expected value of $E[G/(G+B)]$ after 10 years of following the rule. By contrast, I do not think asking about the first 100 children is a reasonable interpretation of the problem. | |
| Jul 16, 2010 at 6:38 | comment | added | Douglas Zare | @Daniel Asimov: Again, I maintain that larger populations have more girls which means girls are weighted less in E[G/(G+B)]. The SLLN is an asymptotic result which does not tell you that the proportion is 50%, and you can check that it is not 50% by explicit calculations. Suppose there are 50 families, and they have time to have 1-5 children each. If there are 150 children, then there are more girls than if there are 70 children. The same phenomenon occurs if you take the union of several populations spread out in time. Once again, when the population size can vary, then there is a bias. | |
| Jul 15, 2010 at 20:18 | comment | added | T.. | Note also that if the answer is (exactly) 1/2 conditional on any fixed finite value of $n$, it would then be the same without the conditioning. The question appears to be whether there is any a dynamical model of how a finite population evolves, not fixing $n$ or $k$ or stipulating the completion of any number of families, that leads to an expected proportion lower than 1/2, or a gender-asymmetrical finite population. | |
| Jul 15, 2010 at 18:04 | comment | added | T.. | re: "the bias depends on the formalization of the problem, but not its existence". Consider formalizations where the population size n is chosen first (such as fixing it at one million, or selecting n from a particular probability distribution), and then n births are a stream of random coin tosses served in some fashion to an unending queue of "families" eligible under the stopping rule. [Families not fixed in advance, but constructed dynamically to fit the births.] Boy/girl distribution in such populations will be symmetrical by construction, and in particular E[G/(B+G)] = 1/2. | |
| Jul 15, 2010 at 9:26 | comment | added | Daniel Asimov | "... larger populations have more girls ..." No, since the population size is meaningful only if something is known about the number of families. In order to use the Strong Law of Large Numbers we assume infinitely many families #1,#2,#3,... with each one's sequence of births (G^n)B being concatenated in order of family #. Then the stochastic process generating this infinite sequence of G's and B's is isomorphic (up to measure 0) with repeated flips of a fair coin. Hence the SLLN implies the asymptotic fraction of B's or G's is 1/2, each with probability = 1. This is, to me, conclusive. | |
| Jul 13, 2010 at 3:53 | comment | added | T.. | Also, the claim that "larger populations tend to have more girls" is not only false (populations of any size have E[G/(B+G)]=1/2, as the stopping rule can't break boy/girl symmetry in the population), but circular as a justification of the purported bias. Positive correlation between fraction of girls G/(B+G) and population size B+G is, by definition and in any model of the problem, equivalent to E[G/(B+G)] < E[G]/E[B+G]. The latter equals 1/2, also in any model. So the correlation statement is not evidence that the bias actually exists, but only a rephrasing of the alleged existence. | |
| Jul 11, 2010 at 16:34 | comment | added | T.. | @DZ: Your comment "the number of boys equals the number of families" is clearly false. At any given time a positive proportion of families in the population will have only girls (and thus will eventually have more children under the stopping rule). As has been pointed out a number of times, your statements are true only for to the union-of-($k$)-completed-families model, which pre-biases the calculation by excluding families with higher (100 percent) proportions of girls. Do you know of a precisely specified model that restores those families but gives calculable answers less than 1/2? | |
| Jul 11, 2010 at 11:43 | comment | added | Douglas Zare | @T: You have made a lot of false statements. Perhaps you should think about this problem more carefully. My statement that larger populations have more girls is true, and almost a tautology when you look at the whole generation since the number of boys equals the number of families. Larger populations mean more girls. It is also true if you assume that only enough time has passed for each family to have up to 3 children, or if you assume that the population is not broken into generations. I'm afraid it does not appear worthwhile to continue this exchange. | |
| Jul 11, 2010 at 8:55 | comment | added | T.. | That should read "...dropping to 1 -(1/familysize)". Typo or no, the point should be clear that intuitions from a model computing expectation on a biased subset (i.e., one that excludes the incomplete families, which have 100% girls) do not reflect what happens in computing E[G/(B+G)] on an unbiased subset, such as the whole population. | |
| Jul 11, 2010 at 8:41 | comment | added | T.. | @DZ: The statement "larger populations tend to have more girls" is false; the proportion of girls in each family decreases as more children are born (either it stays constant at 0, or stays at 1 until dropping to 1/familysize). It is true only in your model that excludes families-in-progress, a model equivalent to computing the proportion of Heads in a series of coin tosses that ends with Heads. The latter condition does bias the proportion of Tails (girls) downward, but the problem about the whole population corresponds to computing the proportion with no such symmetry-breaking condition. | |
| Jul 11, 2010 at 4:31 | comment | added | Douglas Zare | @T: There isn't a boy-girl symmetry. The timing of the births depends on the sexes of the past children. The number of children after the second year depends on the sexes of the children born in the first year. The populations which are larger will be those which have more girls born in the first year. The result is that E[G/(G+B)] is less than 1/2, however counterintuitive that may be. Whether you use "all births in a generation" or "all births in the first 5 years" or "all births by 2010 from several generations" the expected proportion of girls is lower than 50%. | |
| Jul 10, 2010 at 7:28 | comment | added | T.. | To interpret the problem means to specify over what finite set of births (that were subject to the stopping rule) the expected value of G/(G+B) is to be calculated. The canonical choices are "all births", "all births of living persons", or "all births in some given (long) time interval". These lead to boy-girl symmetry and consequently to E[G/(G+B)]=1/2. If you know of any other specification of a set of births that is a reasonable interpretation of the problem but (under this or any other stopping rule) leads to an expected value different from 1/2, what is it? | |
| Jul 10, 2010 at 3:53 | comment | added | Douglas Zare | @T, you are right that the bias depends on the formalization of the problem, but not that its existence depends on this particular one. If you assume that some families have not stopped reproducing, there is still a bias. If the size of the population is not constant, and larger populations tend to have more girls, then girls tend to be underweighted when you compute G/(G+B), and the expected value of G/(G+B) is under 1/2. I do not see a reasonable way to interpret the problem so that the population size is fixed or does not depend on the sexes of the children, but feel free to point one out. | |
| Jul 9, 2010 at 0:28 | comment | added | T.. | You've changed the question by assuming that the population (i.e., the set of births whose G/(G+B) we are interested in) is a union of families that have stopped reproducing. All we know is that in $k$ families, some number $n$ of births have occurred through time $t$, each birth equivalent to a fair coin toss. This of course implies a symmetrical distribution of $(G,B)$ and consequently an expected value of exactly 1/2 for $G/(G+B)$ --- independent of $n, k$ and $t$. The difference in answers between the original problem and the completed-families problem is the "bias" you calculated. | |
| Apr 5, 2010 at 16:43 | comment | added | Douglas Zare | A simpler expression for the average for a sample of $k$ families is $k*|\log 2 - (1 - 1/2 + 1/3 - ... + (-1)^k/k)|$. | |
| Mar 13, 2010 at 18:49 | comment | added | Douglas Zare | Actually, I was planning to write my next GammonVillage column on that case, which resembles the win/loss record for someone who plays $k$ single-elimination tournaments with $N$ rounds. I haven't come up with an explicit formula yet, but the win/loss record after any finite number of tournaments is still a biased estimator of the player's chance to win each game. | |
| Mar 13, 2010 at 18:38 | comment | added | Joel David Hamkins | Douglas, in your calculation, you assume (as instructed by the question) that every family continues having children until they have a boy, even if the mother has already had, say, hundreds of girls. But this is not realistic. After all, parents may die before ever having their first boy, or perhaps the child policy is recent. (This seems related to the stopping-time issue in the gamble you mention in comments below.) Could I kindly ask that you tell us the statistics of the situation if parents have only had N chances to have babies? I would be interested in your remarks. | |
| Mar 12, 2010 at 16:43 | history | edited | Douglas Zare | CC BY-SA 2.5 |
Fixed typo in index. Added 1/2-1/4k.
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| Mar 12, 2010 at 14:10 | comment | added | David E Speyer | This is really nice! I voted to close because this is an old chestnut, but you have found new life in it. | |
| Mar 12, 2010 at 14:08 | vote | accept | nkrkv | ||
| Mar 12, 2010 at 10:49 | history | edited | Douglas Zare | CC BY-SA 2.5 |
Added digamma.
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| Mar 12, 2010 at 9:37 | history | answered | Douglas Zare | CC BY-SA 2.5 |