A091674 -id:A091674 - cp's OEIS frontend

A091673 Numerator Q of probability P = Q(n)/365^(n-1) that exactly two out of n people share the same birthday.

1, 1092, 793884, 480299820, 261163522620, 132358677731280, 63798093049771080, 29612552769907347240, 13345042642324219106280, 5872442544965392834838400, 2533775368098060137659608000, 1075256447734638237381213700800

Offset: 2

Hugo Pfoertner, Feb 03 2004

A 365-day year and a uniform distribution of birthdays throughout the year are assumed.

			a(3)=1092 because the probability that in a group of 3 people exactly two of them share the same birthday is (1/365^3)*3!*binomial(365,1)*binomial(364,1)/2 = (1/365^2)*3*364 = (1/365^2)*1092.

Patrice Le Conte, Coincident Birthdays.
Eric Weisstein's World of Mathematics, Birthday Problem.

Cf. A014088, A091674 gives probabilities for two or more coincidences, A091715 gives probabilities for three or more coincidences.

P[n_] := (n! Sum[ Binomial[365, i]*Binomial[365 - i, n - 2i] /2^i, {i, 1, Floor[n/2]}]/365); Table[ P[n], {n, 2, 13}] (* Robert G. Wilson v, Feb 09 2004 *)

from math import factorial, comb
from fractions import Fraction
def A091673(n): return int(factorial(n)*sum(Fraction(comb(365,i)*comb(365-i,n-(i<<1)),365<>1)+1))) # Chai Wah Wu, Jan 22 2025

P(n) = n!*Sum_{i=1..floor(n/2)} binomial(365, i)*binomial(365-i, n-2*i)/2^i.

More terms from Robert G. Wilson v, Feb 09 2004

A091715 Numerator Q of probability P = Q(n)/365^(n-1) that three or more out of n people share the same birthday.

Original entry on oeis.org

1, 1457, 1326781, 966556865, 616113172585, 359063094171965, 196176047915944825, 102076077386001384485, 51120278427593115164425, 24824896058243745467563925, 11753675337747799989826426225

Offset: 3

Hugo Pfoertner, Feb 04 2004

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.

			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.
The probability exceeds 50% for n > A014088(3) = 88.

Patrice Le Conte, Coincident Birthdays.
The Math Forum at Drexel, Three Share a Birthday, Ask Dr. Math.
Eric Weisstein's World of Mathematics, Birthday Problem.

Cf. A014088, A091673 (probabilities for exactly two), A091674 (probabilities for two or more).

a(n) = A091674(n) - A091673(n).

cp's OEIS Frontend

A091673 Numerator Q of probability P = Q(n)/365^(n-1) that exactly two out of n people share the same birthday.

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