I passed by this question today, in which I'd like to have some help or hints regarding part b)

So the story is as follows:

The residents of a certain country value boys over girls, and every couple makes sure a boy is born in the family. So, if the first child is a boy, they stop there. If the first child is a girl, they have another child, and keep on having children until the first boy. So, the progression of children for each family ends with a boy. Some examples would be B, GB, GGB, GGGGB. At birth, the probability of a child being a girl or a boy is equal (½). (a) What is the expected number of children of a family? (b) What is the expected value of the proportion of males to the total population in this country?

Part a) is solved easily, as it's equal to the $\sum_{i\ge1}{ip^i}$ which leads to $2.$

In part b) I solved according to the ratio between the average of boys and the average number of family children. I'm not sure if it shall be solved as follows : $$\frac{E(\#\text{boys} )}{E(\#\text{boys})+E(\#\text{girls})}$$

I'm not sure if the parents shall be ignored in part b) as well

Can someone guide me if that's true ?


  • 1
    $\begingroup$ I don't think so, since larger families have more girls. $\endgroup$ Jun 5, 2020 at 20:59
  • $\begingroup$ It seemed the same for me as well, still I'm not sure how to address it $\endgroup$
    – mandez
    Jun 5, 2020 at 21:03
  • $\begingroup$ @herbsteinberg : But there would be very few of those. $\endgroup$ Jun 5, 2020 at 21:10
  • $\begingroup$ It would be easy to do this by brute-force computation, but I'm wondering if there's a slick way to do it by properties of expected values. $\endgroup$ Jun 5, 2020 at 22:09
  • $\begingroup$ Actually I was setting up my matlab to start evaluating it, haha but it's already late so I might leave it till tomorrow $\endgroup$
    – mandez
    Jun 5, 2020 at 22:19

2 Answers 2


$\newcommand{\e}{\operatorname E}$ Let $B$ be the total number of births, a random variable.

Let $I$ be a random variable whose conditional distribution given $B$ is uniform in the set $\{ 1, \ldots, B\},$ so that each of those numbers has probability $1/B$ of being the value of $I.$

The conditional expected value $$ \e \left( \dfrac{\#\text{boys in the $I$th birth}}{\#\text{boys in the $I$th birth} + \#\text{girls in the $I$th birth}} \mid B, I \right)$$ is clearly $1/2,$ i.e. for each birth separately it's $1/2.$

So the marginal (or "unconditional") expected value of that fraction is the expected value of $1/2,$ i.e. it's $1/2.$

The fact that large families have more girls than boys is countered by the fact that there would be few of those, plus the fact that half the families have only one child and that one is a boy, and in the other half of the families there are also boys.

  • $\begingroup$ Hi, if may I ask what does I represent in this illustration $\endgroup$
    – mandez
    Jun 6, 2020 at 10:39
  • $\begingroup$ @MohamadMisto : Sorry --- I omitted something. It's there now. $\endgroup$ Jun 7, 2020 at 18:55
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    $\begingroup$ I have now written $\#\text{boys in the $I$th birth}$, meaning the $I$th baby is either $1$ boy or $0$ boys, and similarly for girls. $\endgroup$ Jun 7, 2020 at 18:57

Expected number of girls${}=\sum_{i=0}^\infty ip^{i+1}=1$ for $p=\frac{1}{2}$ so the ratio is $1:2$.

  • $\begingroup$ Thank you Herb. So, you think that in other words, the $\frac{E(boys)}{E(boys) + E(girls)}$ works ? $\endgroup$
    – mandez
    Jun 5, 2020 at 21:25
  • $\begingroup$ Could you translate that bit after the word "is" into English? $\endgroup$ Jun 5, 2020 at 21:34
  • $\begingroup$ I think he meant 1, if you're referring to ! ? $\endgroup$
    – mandez
    Jun 5, 2020 at 21:35
  • 1
    $\begingroup$ ! was typo. It is the "capital" 1. I had hit the shift key by mistake. It has been corrected, $\endgroup$ Jun 6, 2020 at 2:48

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