The question I am working on is:

Three brothers and their wives decide to have children until each family has two female children. What is the pmf $X=$ the total number of of male children born to the brothers? What is E(X), and how does it compare to the expected number of male children born to each brother?

So, the probability that we have x amount of failures preceding the final birth, which will be the birth of the 6th girl, is $P(X=x) = nb(x;6,p(S)) = {{x+6-1}\choose{6-1}} \cdot[p(S)]^6 \cdot [p(F)]^x$

And the expected value would be $E(X)= \frac{6 \cdot p(F)}{p(S)}$

I know that this make be an incredulous claim, but I can't figure out why the answer to the very last question is 6. I know 6 represents the total number of girls among the three families, two per family, but what does the ratio of $\frac{p(F)}{p(S)}$ portray? Intuition tells me the number of boys in each family will be the same...

  • $\begingroup$ The problem statement is ambiguous to me. Does each couple continue breeding until all have two daughters, even after they themselves do have two daughters? And what is your $p(F)$ and $p(S)$? Myabe simply the probability of birth of girl or boy, respectively? These can typically be approximated by $p(F)=p(S)=\frac12$. $\endgroup$ Feb 21, 2013 at 16:19
  • $\begingroup$ @HagenvonEitzen $p(F)$ is the probability of failure, and $P(S)$ is the probability of success. I don't think the question wants me to assume the probability of success and failure; I think it asks for a general expression. Yes, the couples continue to breed, until each family has two daughters of their own. The key words are "to have children until." $\endgroup$
    – Mack
    Feb 21, 2013 at 16:23
  • 2
    $\begingroup$ I hope they don't name all the boy children "Failure".... $\endgroup$ Feb 21, 2013 at 16:58
  • $\begingroup$ Is this perhaps too difficult of a problem? $\endgroup$
    – Mack
    Feb 21, 2013 at 18:02

1 Answer 1


This post consists of two parts: a long introduction, and a short solution.

The problem is not optimally worded. We make an interpretation, partly based on the answer provided.

Each family keeps breeding until it has achieved its goal of $2$ girls and then stops, whether or not the other families have managed to meet their quota. And we are invited to assume that the probability that a birth results in a girl is $\dfrac{1}{2}$, and to make the usual assumption of independence.

Introduction: There are a couple of unfortunately different descriptions of the negative binomial distribution. We either count the total number of trials until the $r$-th time that a certain event, often called a "success," occurs. Or else we count the number of what are usually called "successes" until the $r$-th failure. Note that the notion of success and failure are reversed, and even after we do the reversal, the answers differ.

We use the second interpretation of "negative binomial." I checked, this is the interpretation described in Wikipedia. The probability of success is often called $p$. In our case, we end up calling the birth of a girl a failure. Sorry about that!

Let $X_1,X_2,\dots,X_m$ be independent negative binomials, where $X_i$ measures the number of successes until the $r_i$-th failure. Suppose that for each of these $X_i$, the probability of success is $p$. Let $Y=X_1+X_2+\cdots +X_m$. Then $Y$ has negative binomial distribution, with the same "$p$," and $r=r_1+r_2+\cdots+r_n$.

One can prove this by a calculation (the case $m=2$ is enough). But it is also intuitively clear. Because it is important to think the right way about these things, we give a brief explanation.

Take a general negative binomial $W$, the number of successes until the $r$-th failure. Then $W$ is a sum $H_1+H_2+\cdots +H_r$ of $r$ independent random variables, where $H_j$ is the number of successes until the first failure.

Thus $X_1+X_2+\cdots+X_m$ is a sum of $r_1+r_2+\cdots+r_m$ independent random variables of type "$H$," and is therefore negative binomial.

Solution of the problems: So the total number of successes is the number of boys until the $6$-th girl. All of the analysis in the introduction was probably not necessary. The situation is the same as if a single family kept breeding until it had $6$ girls. But the problem is a good excuse for discussing the general situation.

The probability that the total number of boys is $k$ is $$\binom{k+5}{k}p^k(1-p)^6,$$ where $p$ is the probability of a boy, in this case $\dfrac{1}{2}$.

The mean total number of boys is $\dfrac{6p}{1-p}$, in this case $6$.

  • $\begingroup$ I understand what you are doing for the most part, I just don't understand why it's necessary to use so many different types of random variables. Aren't $G_i$ and $X_i$ the same? $\endgroup$
    – Mack
    Feb 25, 2013 at 13:11
  • $\begingroup$ Also, just to be clear, the way you are solving this is by considering a girl's birth a failure, and a boys birth a success? And our random variable will count the number of successes, until the $r-th$ failure? $\endgroup$
    – Mack
    Feb 25, 2013 at 13:38
  • $\begingroup$ Wait, are you defining a random variable for each family? Also, I just noticed that you have $k-th$ failure, and $r-th$ failure, what's the difference? $\endgroup$
    – Mack
    Feb 25, 2013 at 13:45
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    $\begingroup$ Yes, $Y$ is the sum of the negative binomial random variables $X_i$. I was giving a general proof that a sum of any number of independent negative binomials with the same "$p$" is negative binomial. That shows in particular that $Y$ is negative binomial, which is what we needed to solve the problem. $\endgroup$ Feb 25, 2013 at 17:33
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    $\begingroup$ No, that is not correct. The mean number of boys in each family is the same for all the families. The probability distribution of the number of boys is the same for each family. But the actual number of boys will not necessarily be the same. It is quite unlikely that the numbers will be the same. $\endgroup$ Feb 25, 2013 at 18:19

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