# How many sibling-pairs are likely in a semi-random group of 100,000 people?

In a group of, say, 100,000 Americans at an event (yes, a pretty big event!) in the US, how many sibling-pairs are likely to be attending? Consider a variation: Vegas has over 100,000 people visiting every day. How many of those visitors would be sibling-pairs (meaning, at least two siblings in Vegas on that day)? I realize that the Vegas variation may have more sibling-pairs, since families tend to holiday together more often than they would attend a conference. I'm looking for a method more than a specific number, but a number with some background would be terrific!

• Interesting question. You need to have an estimate on what the probability of two people being siblings is. Sep 6, 2016 at 8:49
• One input may be that in the US, about 60% of families have more than 1 child. Does that help? Sep 6, 2016 at 9:59

In expectation : Let $N$ be the number of americans. Let $M$ be the number of sibling pairs among the number of pairs of americans. Let $V(x)$ be a random variable which is $1$ if $x$ goes to the event, $0$ otherwise. Then the expectation of the number of sibling pairs is $$\mathbb E[\sum_{x,y \text{ siblings}} V(x)V(y)] = M \mathbb E[V(x)V(y)]$$
Suppose $V$ is a uniform selection of $n$ people among the $N$ americans. Then $\mathbb E[V(x)V(y)]=\frac{n(n-1)}{N(N-1)}$ (exercise : why ?)
So the number we want is $\frac{M n(n-1)}{N(N-1)}$. But we need to know $M$. According to your statistic, about 60% of families have more than one child. Suppose it has always been the case, and suppose everyone still has all his siblings alive. So at least $\frac{60+60}{40+60+60}N$ americans belong in families of at least $2$ siblings. So the number of sibling pairs is at least $M\geq \frac{60+60}{2(40+60+60)}N = \frac 3 8 N$.
So the expected number of sibling pairs is larger than $\frac{3 n(n-1)}{8(N-1)} \sim \frac{3n^2}{8N} \sim 11.75$ when $N=318.9\times 10^6$, $n=100000$.