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%I #30 Mar 01 2023 04:44:47
%S 1,1457,1326781,966556865,616113172585,359063094171965,
%T 196176047915944825,102076077386001384485,51120278427593115164425,
%U 24824896058243745467563925,11753675337747799989826426225
%N Numerator Q of probability P = Q(n)/365^(n-1) that three or more out of n people share the same birthday.
%C A 365-day year and a uniform distribution of birthdays throughout the year are assumed. The probability that 3 or more out of n people share a birthday equals the probability A091674(n)/365^(n-1) that 2 or more share a birthday minus the probability A091673(n)/365^(n-1) that exactly 2 share a birthday.
%H Patrice Le Conte, <a href="/A225852/a225852.pdf">Coincident Birthdays.</a>
%H The Math Forum at Drexel, <a href="https://web.archive.org/web/20180805132402/http://mathforum.org:80/library/drmath/view/56650.html">Three Share a Birthday</a>, Ask Dr. Math.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BirthdayProblem.html">Birthday Problem</a>.
%F a(n) = A091674(n) - A091673(n).
%e The probability that 3 or more people in a group of 10 share the same birthday is a(10)/365^9 = 102076077386001384485/114983567789585767578125 ~= 8.87744913*10^-4.
%e The probability exceeds 50% for n > A014088(3) = 88.
%Y Cf. A014088, A091673 (probabilities for exactly two), A091674 (probabilities for two or more).
%K frac,nonn
%O 3,2
%A _Hugo Pfoertner_, Feb 04 2004
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