Inspired by Geoff Pilling

There are 9 equally likely instances $(E,B,F)$ , (where $E,B,F$ represent the respective number of children of Ella, Bella and Frabduzella), in which $E \gt B$ , namely

$(2,0,0), (2,0,1), (2,0,2), (2,1,0), (2,1,1), (2,1,2), (1,0,0), (1,0,1), (1,0,2)$ .

In 5 of these cases $E \gt F$ as well, so the desired probability is $\boxed{\dfrac{5}{9}}$ .

In general, if each woman is equally likely to have $0,1,2,...,n$ children and $E \gt B$ , then the probability that $E \gt F$ as well is

$\displaystyle \dfrac{\sum_{k=0}^{n} k^{2}}{(n + 1) \sum_{k=0}^{n} k} = \dfrac{\dfrac{n(n + 1)(2n + 1)}{6}}{\dfrac{n(n + 1)^{2}}{2}} = \dfrac{2n + 1}{3(n + 1)}$ , which goes to $\dfrac{2}{3}$ as $n \to \infty$ , consistent with the continuous-variable inspiration problem.

Since we are given that Ella has more children than Bella there are three possibilities:

• Ella has $2$ children and Bella has $1$ child
• Ella has $2$ children and Bella has no children
• Ella has $1$ child and Bella has no children

In the first two cases, there is a $\frac{2}{3}$ probability that Frabduzella has fewer children than Ella. In the third case, the probability is $\frac{1}{3}$ . Therefore, the answer is $\frac{2}{3} \times \frac{2}{3} + \frac{1}{3} \times \frac{1}{3} = \boxed{\frac{5}{9}}$ .

[This is not a solution.]

In Geoff's problem, the answer is $\frac{2}{3}$ .
What is the answer different in this problem?
What happens if the woman are equally likely to have 0, 1, 2, and 3 children?