A393268 Decimal expansion of the probability that in a randomly selected group of 16 girls and 16 boys there is at least one girl-boy pair with a matching birthday, not considering leap years.

5, 0, 4, 4, 1, 9, 9, 2, 5, 8, 8, 8, 5, 6, 7, 4, 0, 8, 1, 5, 2, 5, 4, 8, 7, 4, 6, 0, 3, 9, 1, 1, 5, 4, 6, 4, 6, 7, 0, 9, 5, 7, 9, 9, 4, 3, 1, 2, 0, 0, 1, 4, 9, 2, 2, 1, 1, 4, 5, 9, 7, 5, 4, 1, 8, 7, 3, 8, 6, 4, 7, 4, 1, 2, 1, 5, 3, 0, 6, 6, 1, 8, 3, 7, 5, 4, 9, 1, 6, 7, 6, 8, 9, 9, 9, 2, 4, 8, 1, 1, 6, 6, 8, 3, 7

Offset

0,1

Comments

16 is the least number of pairs for which the probability exceeds 1/2.

This constant is a rational number: its numerator and denominator have 78 and 79 digits, respectively (see the Formula section).

The sequence has a period of 635003019788041508950067678118375842016205817006497843592 = 6.350...*10^56.

References

Tony Crilly, 50 Maths Ideas You Really Need to Know, Quercus, 2007, p. 134.

Links

Table of n, a(n) for n=0..104.

Tony Crilly and Shekhar Nandy, The birthday problem for boys and girls, The Mathematical Gazette, Vol. 71, No. 455 (1987), pp. 19-22.

Alan H. Rapoport, 82.23 Birthday problems - a search for elementary solutions, The Mathematical Gazette, Vol. 82, No. 493 (1998), pp. 111-114.

Eric Weisstein's World of Mathematics, Birthday Problem.

Wikipedia, Birthday problem.

Formula

Equals p(16, 365), where p(n, m) = 1 - (1/m^(2*n))* Sum_{k=1..n} (m-k)^n * m!/(m-k)! * Stirling2(n, k) is the probability for the general case of n girls and n boys, with m days, and Stirling2(n, k) are the Stirling numbers of the second kind (A008277) (Crilly and Nandy, 1987). 16 is the least value of n for which p(n, 365) exceeds 1/2.

Equals 544416179442231759781983199231458470963544567415740105745789686392230709653681 / 1079291581281624505141632177462834782600007593666071598003990948200225830078125.

Example

0.504419925888567408152548746039115464670957994312001...

Mathematica

p[n_, m_] := 1 - (1/m^(2*n)) * Sum[(m-k)^n * m!/(m-k)! * StirlingS2[n, k], {k, 1, n}];
RealDigits[p[16, 365], 10, 120][[1]]

Prog

(PARI) p(n, m) = 1 - (1/m^(2*n)) * sum(k = 1, n, (m-k)^n * m!/(m-k)! * stirling(n, k, 2));
list(len) = digits(floor(p(16, 365)*10^len));

Crossrefs

Cf. A008277, A333507, A343015, A393269, A393270, A393271.

Sequence in context: A139341 A159575 A092173 * A292902 A394726 A200417

Adjacent sequences: A393265 A393266 A393267 * A393269 A393270 A393271

Keyword

nonn,cons,easy

Author

Amiram Eldar, Feb 08 2026

Status

approved