I have came across a question in job interview aptitude test for critical thinking. It is goes something like this:

The Zorganian Republic has some very strange customs. Couples only wish to have female children as only females can inherit the family's wealth, so if they have a male child they keep having more children until they have a girl. If they have a girl, they stop having children. What is the ratio of girls to boys in Zorgania?

I don't agree with the model answer given by the question writer, which is about 1:1. The justification was any birth will always have a 50% chance of being male or female.

Can you convince me with a more mathematical vigorous answer of $\text{E}[G]:\text{E}[B]$ if $G$ is the number of girls and B is the number of boys in the country?

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    $\begingroup$ You are correct in your disagreement with the model answer because the M:F ratio of births is different from the M:F ratio of children. In real human societies, couples who wish to only have female children will likely resort to means like infanticide or foreign adoption to get rid of male children, resulting in a M:F ratio less than 1:1. $\endgroup$
    – Gabe
    Commented Apr 16, 2014 at 6:07
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    $\begingroup$ @Gabe There is no mention of infanticide in the question, it is a mathematical excercise as opposed to a gritty analysis of a real country where murder is common place. Equally the real ratio of births of boys to girls is closer to 51:49 (ignoring social factors) $\endgroup$ Commented Apr 16, 2014 at 9:38
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    $\begingroup$ Thanks to the answers I now understand why the ratio would be 1:1, which originally sounds counter intuitive to me. One of the reason for my disbelief and confusion is that, I know villages in China have the opposite problems of too high of boys:girls ratio. I can see that realistically, couples won't be able to continue to procreate indefinitely until they get the gender of child they want. In china the law allows only 2 children maximum for people living in rural areas, so in that case the ratio will be closer to 3:2 than 1:1. $\endgroup$ Commented Apr 16, 2014 at 14:15
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    $\begingroup$ @MobiusPizza: No, the ratio is 1:1 no matter how many children you have! The reason China has a different ratio is due to social factors like infanticide, sex-selective abortion, and foreign adoption. $\endgroup$
    – Gabe
    Commented Apr 16, 2014 at 21:57
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    $\begingroup$ @newmount Simulations are good, but they mean only as much as the assumptions built into them. Displaying only the code, without any explanation, makes it difficult for people to identify those assumptions. In the absence of some such justification and explanation, no amount of simulation output will address the question here. As far as the "real world" goes, anyone making that claim will have to support it with data about human births. $\endgroup$
    – whuber
    Commented Apr 17, 2014 at 14:49

15 Answers 15


Start with no children

repeat step


Every couple who is still having children has a child. Half the couples have males and half the couples have females.

Those couples that have females stop having children


At each step you get an even number of males and females and the number of couples having children reduces by half (ie those that had females won't have any children in the next step)

So, at any given time you have an equal number of males and females and from step to step the number of couples having children is falling by half. As more couples are created the same situation reoccurs and all other things being equal, the population will contain the same number of male and females

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    $\begingroup$ I think this is an excellent way of explaining the probability distribution without relying on a rigorous mathematical proof. $\endgroup$
    – LBushkin
    Commented Apr 16, 2014 at 2:55
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    $\begingroup$ What I like is that this also explains what happened to the excess girls your intuition expects: The excess girls are desired by the parents (they are the parents who try again), but those parents (on the whole) never successfully create an excess of girls. $\endgroup$ Commented Apr 19, 2014 at 17:47
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    $\begingroup$ You could simplify even further by saying "repeat step { someone decides whether or not to have a child }". The rules by which they decide are completely irrelevant provided that everybody produces boys and girls independently with the same probability. It's not even necessary to assume a value for that probability, you could just say the frequency in the population will be the same as the frequency at birth. $\endgroup$ Commented Apr 19, 2014 at 17:52
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    $\begingroup$ @martino I don't believe this is the case, although I wouldn't be surprised if there was some very convincing math to this effect. I believe this scenario leads to a breakdown in our notion of ratios, because the expected number of children per family is infinite. You should be sceptical of your answer because of the generality with which people have answered your question in this thread. $\endgroup$ Commented Apr 21, 2015 at 19:02
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    $\begingroup$ @ martino. For fun I just ran a simulation with that stopping criterion. 10,000 families had a total of 160,693,469 boys (and that number plus 10,000 more girls) for a ratio of 0.9999377735896915. Pretty incredible stuff. $\endgroup$ Commented Apr 21, 2015 at 20:04

Let $X$ be the number of boys in a family. As soon as they have a girl, they stop, so

\begin{array}{| l |l | } \hline X=0 & \text{if the first child was a girl}\\ X=1 & \text{if the first child was a boy and the second was a girl}\\ X=2 & \text{if the first two children were boys and the third was a girl}\\ \text{and so on}\ldots &\\ \hline \end{array}

If $p$ is the probability that a child is a boy and if genders are independent between children, the probability that a family ends up having $k$ boys is $$\mbox{P}(X=k)=p^{k}\cdot (1-p),$$ i.e. the probability of having $k$ boys and then having a girl. The expected number of boys is $$ \operatorname{E}X=\sum_{k=0}^\infty kp^k \cdot (1-p)=\sum_{k=0}^\infty kp^k-\sum_{k=0}^\infty kp^{k+1}.$$ Noting that $$\sum_{k=0}^\infty kp^k=\sum_{k=0}^\infty (k+1)p^{k+1}$$ we get $$\sum_{k=0}^\infty kp^k-\sum_{k=0}^\infty kp^{k+1}=\sum_{k=0}^\infty (k+1)p^{k+1}-\sum_{k=0}^\infty kp^{k+1}=\sum_{k=0}^\infty p^{k+1}=p\sum_{k=0}^\infty p^{k}=\frac{p}{1-p}$$ where we used that $\sum_{k=0}^\infty p^k=1/(1-p)$ when $0<p<1$ (see geometric series).

If $p=1/2$, we have that $\operatorname{E}X=0.5/0.5$. That is, the average family has 1 boy. We already know that all families have 1 girl, so the ratio will over time even out to be $1/1=1$.

The random variable $X$ is known as a geometric random variable.

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    $\begingroup$ This, of course, assumes that p is the same for all families. If instead we assume that some couples are more likely to have boys than others (i.e., their p is higher) then the result changes, even if the average value of p is still 0.5. (Still, this is an excellent explanation of the basic underlying statistics.) $\endgroup$ Commented Apr 16, 2014 at 11:47
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    $\begingroup$ @Ben Your comment contains a key idea. The same thing had occurred to me, so I have edited my question to include an analysis of this more realistic situation. It shows that the limiting ratio is not necessarily 1:1. $\endgroup$
    – whuber
    Commented Apr 16, 2014 at 14:27
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    $\begingroup$ @BenHocking Indeed! And as we know from both modern statistics and Laplace's classic analysis of birth ratios, $p$ is not really equal to $1/2$ anyway. :) $\endgroup$
    – MånsT
    Commented Apr 16, 2014 at 16:21


The simple model that all births independently have a 50% chance of being girls is unrealistic and, as it turns out, exceptional. As soon as we consider the consequences of variation in outcomes among the population, the answer is that the girl:boy ratio can be any value not exceeding 1:1. (In reality it likely still would be close to 1:1, but that's a matter for data analysis to determine.)

Because these two conflicting answers are both obtained by assuming statistical independence of birth outcomes, an appeal to independence is an insufficient explanation. Thus it appears that variation (in the chances of female births) is the key idea behind the paradox.


A paradox occurs when we think we have good reasons to believe something but are confronted with a solid-looking argument to the contrary.

A satisfactory resolution to a paradox helps us understand both what was right and what may have been wrong about both arguments. As is often the case in probability and statistics, both arguments can actually be valid: the resolution will hinge on differences among assumptions that are implicitly made. Comparing these different assumptions can help us identify which aspects of the situation lead to different answers. Identifying these aspects, I maintain, is what we should value the most.


As evidenced by all the answers posted so far, it is natural to assume that female births occur independently and with constant probabilities of $1/2$. It is well known that neither assumption is actually true, but it would seem that slight deviations from these assumptions should not affect the answer much. Let us see. To this end, consider the following more general and more realistic model:

  1. In each family $i$ the probability of a female birth is a constant $p_i$, regardless of birth order.

  2. In the absence of any stopping rule, the expected number of female births in the population should be close to the expected number of male births.

  3. All birth outcomes are (statistically) independent.

This is still not a fully realistic model of human births, in which the $p_i$ may vary with the age of the parents (particularly the mother). However, it is sufficiently realistic and flexible to provide a satisfactory resolution of the paradox that will apply even to more general models.


Although it is interesting to conduct a thorough analysis of this model, the main points become apparent even when a specific, simple (but somewhat extreme) version is considered. Suppose the population has $2N$ families. In half of these the chance of a female birth is $2/3$ and in the other half the chance of a female birth is $1/3$. This clearly satisfies condition (2): the expected numbers of female and male births are the same.

Consider those first $N$ families. Let us reason in terms of expectations, understanding that actual outcomes will be random and therefore will vary a little from the expectations. (The idea behind the following analysis was conveyed more briefly and simply in the original answer which appears at the very end of this post.)

Let $f(N,p)$ be the expected number of female births in a population of $N$ with constant female birth probability $p$. Obviously this is proportional to $N$ and so can be written $f(N,p) = f(p)N$. Similarly, let $m(p)N$ be the expected number of male births.

  • The first $pN$ families produce a girl and stop. The other $(1-p)N$ families produce a boy and continue bearing children. That's $pN$ girls and $(1-p)N$ boys so far.

  • The remaining $(1-p)N$ families are in the same position as before: the independence assumption (3) implies that what they experience in the future is not affected by the fact their firstborn was a son. Thus, these families will produce $f(p)[(1-p)N]$ more girls and $m(p)[(1-p)N]$ more boys.

Adding up the total girls and total boys and comparing to their assumed values of $f(p)N$ and $m(p)N$ gives equations

$$f(p)N = pN + f(p)(1-p)N\ \text{ and }\ m(p)N = (1-p)N + m(p)(1-p)N$$

with solutions

$$f(p) = 1\ \text{ and }\ m(p) = \frac{1}{p}-1.$$

The expected number of girls in the first $N$ families, with $p=2/3$, therefore is $f(2/3)N = N$ and the expected number of boys is $m(2/3)N = N/2$.

The expected number of girls in the second $N$ families, with $p=1/3$, therefore is $f(1/3)N = N$ and the expected number of boys is $m(1/3)N = 2N$.

The totals are $(1+1)N = 2N$ girls and $(1/2+2)N = (5/2)N$ boys. For large $N$ the expected ratio will be close to the ratio of the expectations,

$$\mathbb{E}\left(\frac{\text{# girls}}{\text{# boys}}\right) \approx \frac{2N}{(5/2)N} = \frac{4}{5}.$$

The stopping rule favors boys!

More generally, with half the families bearing girls independently with probability $p$ and the other half bearing boys independently with probability $1-p$, conditions (1) through (3) continue to apply and the expected ratio for large $N$ approaches

$$\frac{2p(1-p)}{1 - 2p(1-p)}.$$

Depending on $p$, which of course lies between $0$ and $1$, this value can be anywhere between $0$ and $1$ (but never any larger than $1$). It attains its maximum of $1$ only when $p=1/2$. In other words, an expected girl:boy ratio of 1:1 is a special exception to the more general and realistic rule that stopping with the first girl favors more boys in the population.


If your intuition is that stopping with the first girl ought to produce more boys in the population, then you are correct, as this example shows. In order to be correct all you need is that the probability of giving birth to a girl varies (even by just a little) among the families.

The "official" answer, that the ratio should be close to 1:1, requires several unrealistic assumptions and is sensitive to them: it supposes there can be no variation among families and all births must be independent.


The key idea highlighted by this analysis is that variation within the population has important consequences. Independence of births--although it is a simplifying assumption used for every analysis in this thread--does not resolve the paradox, because (depending on the other assumptions) it is consistent both with the official answer and its opposite.

Note, however, that for the expected ratio to depart substantially from 1:1, we need a lot of variation among the $p_i$ in the population. If all the $p_i$ are, say, between 0.45 and 0.55, then the effects of this variation will not be very noticeable. Addressing this question of what the $p_i$ really are in a human population requires a fairly large and accurate dataset. One might use a generalized linear mixed model and test for overdispersion.

If we replace gender by some other genetic expression, then we obtain a simple statistical explanation of natural selection: a rule that differentially limits the number of offspring based on their genetic makeup can systematically alter the proportions of those genes in the next generation. When the gene is not sex-linked, even a small effect will be multiplicatively propagated through successive generations and can rapidly become greatly magnified.

Original answer

Each child has a birth order: firstborn, second born, and so on.

Assuming equal probabilities of male and female births and no correlations among the genders, the Weak Law of Large Numbers asserts there will be close to a 1:1 ratio of firstborn females to males. For the same reason there will be close to a 1:1 ratio of second born females to males, and so on. Because these ratios are constantly 1:1, the overall ratio must be 1:1 as well, regardless of what the relative frequencies of birth orders turn out to be in the population.

  • $\begingroup$ Interesting; this seems to be because although no rule can change the ratio from the natural ratio it can change the number of resulting children and that number of children is dependent on the natural ratio. So in your example you have two populations of parents and they are affected differently. (That said this this feels like a situation outside the scope of the implied fictional country which is more of a mathematical exercise) $\endgroup$ Commented Apr 17, 2014 at 11:14
  • $\begingroup$ @Richard It might feel like that only because, for the sake of exposition, I have oversimplified. In reality one would model the population with a distribution of $p_i$ having a mean of $1/2$. Unless the variance of that distribution is zero, the same analysis implies the same conclusions, including that the expected girl:boy ratio will be strictly less than $1$. This shows that the popular conclusion (that the ratio must be 1:1) depends crucially on the no-variation assumption. I won't apologize for using mathematics to reason about this, which does not diminish the interest of the result. $\endgroup$
    – whuber
    Commented Apr 17, 2014 at 14:45
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    $\begingroup$ nor should you apologise, this is a very interesting result (I did actually think wow when I read it). I would just prefer it in the form "Original result", "More realistic situation". The way its written it feels like cheating (which is unfair because as i say it's very interesting) because I could just as easily say "Well obviously it's not 1:1 because male births are more common" (I believe due to our historical tenancies to die in armed conflict) $\endgroup$ Commented Apr 17, 2014 at 14:54
  • $\begingroup$ @Richard That's a good point. I refrained from discussing more realistic versions of the question, such as changing the mean of the $p_i$ to about $0.51$ (which is unrelated to armed combat, by the way: it has a biological explanation), because the post was over-long as it is and it should be clear how to generalize its methods to that case. I would prefer to keep the focus on resolving the paradox, which is finding a natural (but perhaps overlooked) mechanism that clarifies and explains the apparent conflict among multiple seemingly-valid answers. $\endgroup$
    – whuber
    Commented Apr 17, 2014 at 15:00
  • $\begingroup$ @whuber Thanks for the informative answer. I do not understand why in your calculation you split the population into 2 families with different probability of giving birth to girls though. According to point 1 of your model assumption, the p_i should be the same for all families. So, why did you split the population into 2 kind of families? $\endgroup$ Commented Mar 27, 2015 at 12:20

The birth of each child is an independent event with P=0.5 for a boy and P=0.5 for a girl. The other details (such as the family decisions) only distract you from this fact. The answer, then, is that the ratio is 1:1.

To expound on this: imagine that instead of having children, you're flipping a fair coin (P(heads)=0.5) until you get a "heads". Let's say Family A flips the coin and gets the sequence of [tails, tails, heads]. Then Family B flips the coin and gets a tails. Now, what's the probability that the next will be heads? Still 0.5, because that's what independent means. If you were to do this with 1000 families (which means 1000 heads came up), the expected total number of tails is 1000, because each flip (event) was completely independent.

Some things are not independent, such as the sequence within a family: the probability of the sequence [heads, heads] is 0, not equal to [tails, tails] (0.25). But since the question isn't asking about this, it's irrelevant.

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    $\begingroup$ As stated, this is incorrect. If the genders were unconditionally independent, in the long run there would be as many girl-girl sequences in births among the families as there are boy-boy-sequences. There are many of the latter and never any of the former. There is a form of independence, but it is conditional on birth order. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:39
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    $\begingroup$ @whuber We're not asked how many girl-girl sequences there are. Only the ratio of girls to boys. I did not state that the sequence of births by an individual mother is a series of independent events, like coin flips. Only that each birth, individually, is an independent event. $\endgroup$
    – Tim S.
    Commented Apr 15, 2014 at 16:42
  • $\begingroup$ You will need to be much clearer about that. I mentioned the sequences to demonstrate the lack of independence, so the burden is on you to state exactly in what rigorous sense "independence" applies here. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:47
  • $\begingroup$ @whuber The events are independent in the same way coin flips are. I've expounded on this in my answer. $\endgroup$
    – Tim S.
    Commented Apr 15, 2014 at 19:29
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    $\begingroup$ @whuber the girl-girl sequences turn up if you put all births in a line; after one couple finishs the next go in etc etc $\endgroup$ Commented Apr 15, 2014 at 20:06

Imagine tossing a fair coin until you observe a head. How many tails do you toss?

$P(0 \text{ tails}) = \frac{1}{2}, P(1 \text{ tail}) = (\frac{1}{2})^2, P(2 \text{ tails}) = (\frac{1}{2})^3, ...$

The expected number of tails is easily calculated* to be 1.

The number of heads is always 1.

* if this is not clear to you, see 'outline of proof' here


Couples with exactly one girl and no boys are the most common

The reason this all works out is because the probability of the one scenario in which there are more girls is much larger than the scenarios where there are more boys. And the scenarios where there are lots more boys have very low probabilities. The specific way it works itself out is illustrated below

NumberOfChilden Probability  Girls   Boys  
1               0.5           1       0  
2               0.25          1       1  
3               0.125         1       2  
4               0.0625        1       3  
...             ...           ...     ...  

NumberOfChilden Probability   Girls*probabilty   Boys*probabilty 
1               0.5           0.5                0
2               0.25          0.25               0.25
3               0.125         0.125              0.25
4               0.0625        0.0625             0.1875
5               0.03125       0.03125            0.125
...             ...           ...                ...  
n               1/2^n         1/(2^n)            (n-1)/(2^n)

You can pretty much see where this is going at this point, the total of the girls and boys are both going to add up to one.

Expected girls from one couple${}=\sum_{n=1}^\infty(\frac{1}{2^n})=1 $
Expected boys from one couple${}=\sum_{n=1}^\infty(\frac{n-1}{n^2})=1 $

Limit solutions from wolfram

Any birth, whatever family is it in has a 50:50 chance of being a boy or a girl

This all makes intrinsic sense because (try as couples might) you can't control the probability of a specific birth being a boy or a girl. It doesn't matter whether a child is born to a couple with no children or a family of a hundred boys; the chance is 50:50 so if each individual birth has a 50:50 chance then you should always get half boys and half girls. And it doesn't matter how you shuffle the births between families; you're not going to affect that.

This works for any1 rule

For due to the 50:50 chance for any birth the ratio will end up as 1:1 for any (reasonable1) rule you can come up with. For example the similar rule below also works out even

Couples stop having children when they have a girl, or have two children

NumberOfChilden Probability   Girls   Boys
1               0.5           1       0
2               0.25          1       1
2               0.25          0       2

In this case the total expected children is more easily calculated

Expected girls from one couple${}=0.5\cdot1 + 0.25\cdot1 =0.75$
Expected boys from one couple${}=0.25\cdot1 + 0.25\cdot2 =0.75$

1As I said this works for any reasonable rule that could exist in the real world. An unreasonable rule would be one in which the expected children per couple was infinite. For example "Parents only stop having children when they have twice as many boys as girls", we can use the same techniques as above to show that this rule gives infinite children:

NumberOfChilden Probability   Girls   Boys
3               0.125         1       2
6               1/64          2       4
9               1/512         3       6
3*m             1/((3m)^2     m       2m

We can then find the number of parents with a finite number of children

Expected number of parents with finite children${}=\sum_{m=1}^\infty(\frac{1}{1/(3m)^2})=\frac{\pi^2}{54}=0.18277\ldots. $

Limit solutions from wolfram

So from that we can establish that 82% of parents would have an infinite number of children; from a town planning point of view this would probably cause difficulties and shows that this condition couldn't exist in the real world.

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    $\begingroup$ That the births are not independent is evident by examining sequences of births: the sequence girl-girl never appears while boy-boy sequences occur often. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:45
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    $\begingroup$ @whuber I see your point (although arguably it is the decision to have a child at all that is dependant, rather than the outcome of the event itself) possibly it would be better to say "a future birth's probability of being a boy is independant from all past births" $\endgroup$ Commented Apr 15, 2014 at 16:48
  • $\begingroup$ Yes, I think there is a way to rescue the use of independence here. But this gets--I think--to the heart of the matter, so it seems that to honor the OP's request for a "vigorous" (rigorous?) demonstration some careful reasoning about this issue is needed. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:50
  • $\begingroup$ @whuber To be honest that first paragraph is the handwavey bit, the further paragraphs (and specifically the limits) are supposed to be the rigourous bit $\endgroup$ Commented Apr 15, 2014 at 16:51
  • $\begingroup$ No argument there--but the latter material has already been covered in the same way in answers at stats.stackexchange.com/a/93833, stats.stackexchange.com/a/93835, and stats.stackexchange.com/a/93841. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:53

You can also use simulation:

for (i in 1:10000){
  while(a != 1){   #Stops when having a girl
    a<-as.numeric(rbinom(1, 1, 0.5))   #Simulation of a new birth with probability 0.5
    p=p+1   #Number of births
(p-10000)/10000   #Ratio
  • 1
    $\begingroup$ Simulation results are good in that they can give us some comfort we haven't made a serious mistake in a mathematical derivation, but they are far from the rigorous demonstration requested. In particular, when rare events that contribute a lot to an expectation can occur (such as a family with 20 boys before a girl appears--which is highly unlikely to emerge in a simulation of just 10,000 families), then simulations can be unstable or even just wrong, no matter how long they are iterated. $\endgroup$
    – whuber
    Commented Apr 15, 2014 at 16:44
  • $\begingroup$ Recognizing the geometric distribution of # of boys in the family is the key step to this problem. Try: mean(rgeom(10000, 0.5)) $\endgroup$
    – AdamO
    Commented Apr 15, 2014 at 19:53

Mapping this out helped me better see how the ratio of the birth population (assumed to be 1:1) and the ratio of the population of children would both be 1:1. While some families would have multiple boys but only one girl, which initially led me to think there would be more boys than girls, the number of those families would not be greater than 50% and would diminish by half with each additional child, while the number of one-girl-only families would be 50%. The number of boys and girls would thus balance each other out. See the totals of 175 at the bottom. Children ratio


What you got was the simplest, and a correct answer. If the probability of a newborn child being a boy is p, and children of the wrong gender are not met by unfortunate accidents, then it doesn't matter if the parents make decisions about having more children based on the gender of the child. If the number of children is N and N is large, you can expect about p * N boys. There is no need for a more complicated calculation.

There are certainly other questions, like "what is the probability that the youngest child of a family with children is a boy", or "what is the probability that the oldest child of a family with children is a boy". (One of these has a simple correct answer, the other has a simple wrong answer and getting a correct answer is tricky).




be the sample space and let

$\text{X: $\Omega\longrightarrow\mathbb{R}$; $\omega\mapsto\vert\omega\vert$-1}$

be the random variable that maps each outcome, $\omega$, onto the number of boys it involves. The expected value of boys, $\text{E(X)}$, comes then down to


Trivially, the expected value of girls is 1. So the ratio is 1, too.


It's a trick question. The ratio stays the same (1:1). The right answer is that it does not affect birth ratio, but it does affect the number of children per family with a limiting factor of an average of 2 births per family.

This is the kind of question you might find on a logic test. The answer is not about birth ratio. That's a distraction.

This is not a probability question, but a cognitive reasoning question. Even if you answered 1:1 ratio, you still failed the test.

  • $\begingroup$ I have recently edited my answer to show that the solution is not necessarily 1:1, which explicitly controverts your assertions. $\endgroup$
    – whuber
    Commented Apr 16, 2014 at 14:24
  • $\begingroup$ I read your answer. You have introduced a predicate that is not stated in the problem (variance in birth rate of females). There is nothing in the problem that asserts Zorganian Republic is representative of the human population or even humans. $\endgroup$ Commented Apr 16, 2014 at 17:30
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    $\begingroup$ That is correct--but there equally well is nothing that justifies the oversimplified assumption that all birth probabilities are the same. Assumptions have to be made in order to provide an objective, defensible answer so at a minimum a good answer will be explicit about the assumptions it makes and provide support for those assumptions. Claiming "this is not a probability question" does not address the issues, but overlooks them entirely. $\endgroup$
    – whuber
    Commented Apr 16, 2014 at 17:45
  • $\begingroup$ @whuber - The birth ratio in this problem is an invariant. The variant in the problem is the number of births per family. The question is a distraction, it is not part of the problem. <br/> Lateral thinking, is the ability to think creatively, or “outside the box” as it is sometimes referred to in business, to use your inspiration and imagination to solve problems by looking at them from unexpected perspectives. Lateral thinking involves discarding the obvious, leaving behind traditional modes of thought, and throwing away preconceptions. [fyi> I am a principal scientist in Lab] $\endgroup$ Commented Apr 16, 2014 at 19:11
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    $\begingroup$ You may, then, have overlooked a key point in my answer: its assumptions also keep the population-averaged chance of a female birth invariant at 1:1 (in a specific way that I hope was clearly described). I would maintain there is substantial "lateral thinking" involved in any resolution of a paradox in which assumptions are critically examined: it requires imagination and good analytical skills to see that one is making assumptions in the first place. Dismissing any question outright as a mere "trick," as you do here, would seem antithetical to promoting or celebrating such thinking. $\endgroup$
    – whuber
    Commented Apr 16, 2014 at 19:24

I am showing the code I wrote for a Monte Carlo simulation (500x1000 families) using `MATLAB' software. Please scrutinise the code so that I did not make a mistake.

The result is generated and plotted below. It shows the simulated girl birth probability has very good agreement with the underlying natural birth probability regardless of the stopping rule for a range of natural birth probability.

enter image description here

Playing around with the code it is easier to understand one point I didn't quite do before---as other's point out, the stopping rule is a distraction. The stopping rule only affects the number of families given a fixed population, or from another point of view the number of child births given a fixed number of families. The gender is solely determined by dice roll and hence the ratio or probability (which is independent of number of children) will solely depend on the natural boy:girl birth rato.

N=uint32(100000); %Used to approximate probability distribution
M=1000; %Number of families
L=500; %Monte Carlo repetitions
for l = 1:L
    j=1; %Index variable for the different bgratio
    for bgratio=testRange
    k=1; %Index variable for family in each run (temp family id)
    vec(1:N*bgratio,1)=1; %Approximate boy:girl population for dice roll, 
    %1 = boy

    vec=vec(randperm(s,N)); %Random permutation, technically not necessary 
    %due to randi used later, just be safe
    bog = vec(randi(N)); %boy or girl? (God's dice roll)

    while k<M %For M families...
        if bog == 1 %if boy:
            boys(j) = boys(j)+1; %total global boys tally
            girls(j)=girls(j)+1; %total global girls tally
            %Family stops bearing children
            Nfamily(j) = Nfamily(j)+1; %total global family tally
            k=k+1; %temp family id
            %Next family...
        bog=vec(randi(N)); %Sample next gender (God's dice roll)

    j=j+1; %Index variable for the different bgratio
hold on
line([0 1],[0 1],'LineStyle','--','Color','k')
axis([0.44 0.56 0.44 0.56])

Let the random variable denoting the $i^{th}$ child in the country be $X_i$ taking on values 1 and 0 if the child is a boy or girl respectively. Assume that the marginal probability that each birth is a boy or girl is $0.5$.

The expected number of boys in the country = $E[\sum_i X_i] = \sum_i E[X_i] = 0.5 n$ (where $n$ is the number of children in the country.)

Similarly the expected number of girls = $E[\sum_i (1- X_i)] = \sum_i E[1-X_i] = 0.5 n$.

The independence of the births is irrelevant for the calculation of expected values.

Apropos @whuber's answer, if there is a variation of the marginal probability across families, the ratio becomes skewed towards boys, due to there being more children in families with higher probability of boys than families with a lower probability, thereby having an augmentative effect of the expected value sum for the boys.


I independently also programmed a simulation in matlab, prior to seeing what others have done. Strictly speaking it is not a MC because I only run the experiment once. But once is sufficient to obtain results. Here is what my simulation yields. I don't take a stand on the probability of births being p=0.5 as a primitive. I let the birth probability vary over a range of Pr(Boys=1)=0.25:0.05:0.75.

My results show that as the probability deviates from p=0.5, the sex ratio is different from 1: in expectation the sex ratio is simply the ratio of the probability of a boy's birth to the probability of a girl's birth. That is, this is a geometric random variable as identified previously by @månst. This is what I believe the original poster was intuiting.

My results closely mimic what the above poster with the matlab code has done, matching the sex ratios at the 0.45, 0.50, and 0.55 probabilities that a boy is born. I present mine as I take a slightly different approach to get at the results with faster code. To accomplish the comparison I omitted the code section vec=vec(randperm(s,N)) since s is not defined in their code and I don't know the original intention of this variable (this code section also seems superfluous - as originally stated).

I post my code

clear all; rng('default')

prob_of_boy = 0.25:0.05:0.75;
prob_of_girls = 1 - prob_of_boy;

iterations = 200;

sex_ratio = zeros(length(prob_of_boy),1);
prob_of_girl_est = zeros(length(prob_of_boy),1);
rounds_of_reproduction = zeros(length(prob_of_boy),1);

for p=1:length(prob_of_boy)

    pop = 1000000;

    boys = zeros(iterations,1);
    girls = zeros(iterations,1);
    prob_of_girl = zeros(iterations,1);

    for i=1:iterations

        x = rand(pop,1);

        %count the number of boys and girls
        num_boys = sum(x(x==1));

        boys(i) = num_boys;
        girls(i) = pop - num_boys;

        prob_of_girl(i) = girls(i)/(pop);

        %Only families that had a boy continue to reproduce
        x = x(x==1);

        %new population of reproducing parents
        pop = length(x);

        %check that there are no more boys 
        if num_boys==0



    prob_of_girl_est(p) = mean(prob_of_girl(prob_of_girl~=0));
    sex_ratio(p) = sum(boys)/sum(girls);
    rounds_of_reproduction(p) = length(boys);

hold on
title('Est. vs. True Probability of a Girl Birth')
ylabel('Est. Probability of Girl Birth')
xlabel('True Probability of Girl Birth')
line([0 1],[0 1],'LineStyle','--','Color','k')
axis([0.2 0.8 0.2 0.8])

hold on
title('Sex Ratio as a function of Girls')
xlabel('Probability of Girls Birth')
ylabel('Sex Ratio: $\frac{E(Boys)}{E(Girls)}$','interpreter','latex')

hold on
title('Rounds of Reproduction a function of Girls')
xlabel('Probability of Girls Birth')
ylabel('Rounds of Reproduction')

The following graph is expected given the strong law of large number. I reproduce it, but the graph that matters is the second graph.

enter image description here

Here, a population probability other than 0.5 for the birth of either sex of a child will alter the sex ratio in the overall population. Assuming that births are independent (but not the choice to keep reproducing), in each round of conditional reproduction the population probability governs the overall make up of the outcomes of boy and girl births. So as others have mentioned, the stopping rule in the problem is inconsequential to the population outcome, as answered by the poster who identified this as the geometric distribution.

enter image description here

For completeness, what the stopping rule does affect is the number of rounds of reproduction in the population. Since I only run the experiment once, the graph is a bit jagged. But the intuition is there: for a given population size, as the probability of a girl's birth increases we see that families need less rounds of reproduction to obtain their desired girl before the entire population stops reproducing (obviously the number of rounds will depend on the population size, since it mechanistically increases the likelihood that a family will have, for example, 49 boys before they get their first girl)

enter image description here

The comparison between my calculated sex ratios:

[sex_ratio' prob_of_boy']

0.3327    0.2500
0.4289    0.3000
0.5385    0.3500
0.6673    0.4000
0.8186    0.4500
1.0008    0.5000
1.2224    0.5500
1.5016    0.6000
1.8574    0.6500
2.3319    0.7000
2.9995    0.7500

and those from the previous poster with the matlab code:

[boys./girls testRange']

0.8199    0.4500
0.8494    0.4600
0.8871    0.4700
0.9257    0.4800
0.9590    0.4900
1.0016    0.5000
1.0374    0.5100
1.0836    0.5200
1.1273    0.5300
1.1750    0.5400
1.2215    0.5500

They are equivalent results.


It depends on the number of families.

Let $X$ be the number of children in a family, it is geometric random variable with $p=0.5$, i.e., $$ P(X = x) = 0.5^x, x=1,2,3... $$ which implies $E(X) = 2$

Suppose there are $N$ families in the country, the girl ratio is $$ \frac{N}{ \sum X_i} $$

Since $\sum X_i /N \rightarrow E(X) = 2$ (law of large number), the ratio coverages to 1/2 if $N \rightarrow \infty$.

If there are only finite families, let $T$ be the total number of children of the country: $T = \sum X_i$, then $T$ has a negative binomial distribution with pmf $$ P(T=t) = C^{t-1}_{N-1} 0.5^t, t = N, N+1... $$

It implies $$ E\left[ \frac{N}{\sum X_i} \right] = E\left[ \frac{N}{T} \right] = \sum_{t=N}^{\infty} \frac{N}{t} C^{t-1}_{N-1} 0.5^t = {_2F_1} (N, 1, N+1, -1) $$ where $_2F_1$ is the hypergeometric function.

Therefore the expected girl ratio is ${_2F_1} (N, 1, N+1, -1) $.


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