Expected number of shared-birthday pairs

Question

Birthdays are independent and uniform over $365$ days. In a room of $n$ people, consider the number of pairs that share a birthday.

What is the expected number of matching pairs, and at what $n$ does it reach 1?

Accepted Answer

There are \binom{n}{2} pairs, and each pair shares a birthday with probability \frac{1}{365} (the second person hits the first person's day). Linearity of expectation, which needs no independence between pairs, gives E[	ext{matching pairs}] = \binom{n}{2}\cdot \frac{1}{365} = \frac{n(n-1)}{2 \cdot 365}. Set it to 1: n(n-1) = 730, so n \approx 27.5. At n = 28 the expectation is \frac{28 \cdot 27}{730} \approx 1.04 pairs. Expectation versus probability. These are different questions. The probability of at least one match crosses 	frac12 at n = 23; the expected count of matches reaches 1 at n …

Expected number of shared-birthday pairs

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