A033810 Number of people needed so that probability of at least two sharing a birthday out of n possible days is at least 50%.

2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy
• nice

Comments

a(365) = 23 is the solution to the Birthday Problem.

Links

• T. D. Noe and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (366 terms from T. D. Noe)
• S. E. Ahmed and R. J. McIntosh, An Asymptotic Approximation for the Birthday Problem, Crux Mathematicorum 26(3) 151-5 2000 CMS.
• D. Brink, A (probably) exact solution to the Birthday Problem, Ramanujan Journal, June 2012, Volume 28, Issue 2, pp. 223-238. - From _N. J. A. Sloane_, Oct 08 2012
• Mathforum, The Birthday Problem.
• Eric Weisstein's World of Mathematics, Birthday Problem.

Crossrefs

Essentially the same as A088141. See also A182008, A182009, A182010.

Cf. A014088 (n people on 365 days), A225852 (3 on n days), A225871 (4 people on n days).

Cf. A380129 (Strong Birthday Problem).

Programs

Mathematica

lst = {}; s = 1; Do[Do[If[Product[(n - i + 1)/n, {i, j}] <= 1/2, If[j > s, s = j]; AppendTo[lst, j]; Break[]], {j, s, s + 1}], {n, 86}]; lst (* Arkadiusz Wesolowski, Apr 29 2012 *)
A033810[n_] := Catch@Do[If[1/2 >= n!/(n - m)!/n^m, Throw[m]], {m, 2, Infinity}]; Array[A033810, 86] (* JungHwan Min, Mar 27 2017 *)

Python

from math import comb, factorial
def A033810(n):
    def p(m): return comb(n,m)*factorial(m)<<1
    kmin, kmax = 0, 1
    while p(kmax) > n**kmax: kmax<<=1
    while kmax-kmin > 1:
        kmid = kmax+kmin>>1
        if p(kmid) <= n**kmid:
            kmax = kmid
        else:
            kmin = kmid
    return kmax # Chai Wah Wu, Jan 21 2025

Formula

a(n) = ceiling(sqrt(2*n*log(2)) + (3 - 2*log(2))/6 + (9 - 4*log(2)^2) / (72*sqrt(2*n*log(2))) - 2*log(2)^2/(135*n)) for all n up to 10^18. It is conjectured that this formula holds for all n.