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!

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

1 Answer 1


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 $N=318.9\times 10^6$, $n=100000$.

The number seems small, but it is based on the bold assumption that the subset of people chosen is uniform. Imagine some kind of US-wide random Army draft among everybody including infants and elders. 100000 over the whole population is not that big, and it is not probable at all for one family to see multiple children being drafted.

  • $\begingroup$ So, about 12 sibling-pairs out of the 100,000, assuming: the subset is uniform, 60% of families have always (at least for the last two generations) had two or more children (the percentage was higher till recently, so we're OK there), all attending individuals still have their siblings alive. Great! One question: can you elaborate on "subset of people chosen is uniform"? In what manner? Demographically? What factors may throw that assumption out? $\endgroup$ Sep 6, 2016 at 14:01
  • $\begingroup$ It means, drawn without replacement from the urn containing all americans. Choose 1 american uniformly at random from the 320000000, then a second one uniformly at random from the 319999999 others, etc. $\endgroup$
    – justt
    Sep 7, 2016 at 13:06

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