Three Share a Birthday

Worked solution

The answer is 88. Count assignments in which no day is used more than twice. With j days used twice and s = n − 2j days used once, the number of assignments is n! / 2^j · 365! / (j! s! (365 − j − s)!), and summing over j and dividing by 365^n gives P(no triple). Exact computation gives P(some triple) ≈ 0.4995 at n = 87 and ≈ 0.5111 at n = 88. For intuition: with n people there are C(n,3) triples, each matching with probability 1/365^2, so the expected number of triple coincidences is C(n,3)/365^2. Setting that equal to ln 2 gives n ≈ 84, which is the right scale, though the Poisson approximation is crude here. The jump from 23 to 88 reflects the extra factor of 365 per additional coincident person.