## three birthdays and a numeral

The riddle of the week on The Riddler was to find the size n of an audience for at least a 50% chance of observing at least one triplet of people sharing a birthday, as is the case in the present U.S. Senate. The question is much harder to solve than for a pair of people but the formula exists!, as detailed on this blog entry, this X validated entry, or my friend Anirban Das Gupta’s review of birthday problems. If W is the number of triplets among n people,

$\mathbb P(W =0) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{365! n!}{i! (n-2i)! (365-n+i)! 2^i 365^n}$

which returns n=88 as the smallest population size for which

P(W=0|n=88)=0.4889349, while P(W=0|n=87)=0.5005451.

A simulation based on 10⁶ draws confirms this boundary value,

P(W=0|n=88)≈0.4890849 and P(W=0|n=87)≈0.5006471.

### 4 Responses to “three birthdays and a numeral”

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