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

2,2

Comments

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

Links

Table of n, a(n) for n=2..13.

Patrice Le Conte, Coincident Birthdays.

Eric Weisstein's World of Mathematics, Birthday Problem.

Formula

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

Example

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.

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 *)

Crossrefs

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

Sequence in context: A043873 A240719 A239875 * A288097 A281001 A271100

Adjacent sequences: A091670 A091671 A091672 * A091674 A091675 A091676

Keyword

frac,nonn

Author

Hugo Pfoertner, Feb 03 2004

Extensions

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

Status

approved