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

Author

Hugo Pfoertner, Feb 03 2004

Keywords

• frac
• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

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

Programs

Mathematica

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 *)

Python

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

Formula

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

Extensions

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