A014088 -id:A014088 - cp's OEIS frontend

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

Eric W. Weisstein

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

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.

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).

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 *)

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

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.

A225852 The number of people required for there to be at least a 50% chance that at least 3 share a birthday in a year with n days.

Original entry on oeis.org

3, 4, 5, 6, 7, 7, 8, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 26, 27, 27, 27, 28, 28, 28, 29, 29, 29, 29, 30, 30

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 17 2013

nonn

a(365) = 88.

For n <= 1000, a(n) = 1.436 + 1.812*n^0.654 - 0.817/n^3 provides an estimate accurate to 0.6 units.

			The probability that out of 87 people 3 share a birthday in a year with 365 days is 0.4994549. The corresponding probability for 88 people is 0.5110651. Therefore a(365)=88.

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000
Christian N. K. Anderson, Table of n and exact probabilities of a(n)-1 and a(n) for n = 1..1000
Patrice Le Conte, Coincident Birthdays

Cf. A014088 (n people on 365 days), A033810 (2 people on n days), A225871 (4 people on n days).

Cf. A088141, A182008, A182009, A182010.

library(gmp);#prob of a maximum of exactly k coincident birthdays is
BigQ<-function(nday,p,k) { #nday=days in a year; p=people
    if(p1,sum(sapply(2:k-1,function(j) BigQ(nday-i,p-k*i,j))),1)
    }
    tot
}
BDaySharedByAtLeast<-function(nday,people,k) {
    if(nday<1 | people

A050255 A Diaconis-Mosteller approximation to the Birthday problem function.

Original entry on oeis.org

1, 23, 88, 187, 313, 459, 622, 797, 983, 1179, 1382, 1592, 1809, 2031, 2257, 2489, 2724, 2963, 3205, 3450, 3698, 3949, 4203, 4459, 4717, 4977, 5239, 5503, 5768, 6036, 6305, 6575, 6847, 7121, 7395, 7671, 7948, 8227, 8506, 8787, 9068, 9351, 9634, 9919, 10204

Offset: 1

Eric W. Weisstein

nonn

Comment from Stig Blücher Brink, May 18 2023: (Start)
Maximum relative approximation error for a(1) to a(10000) is 0.27%.
Maximum absolute approximation error for a(1) to a(10000) is 2126.
(End)

_Stig Blücher Brink_, Table of n, a(n) for n = 1..10000
P. Diaconis and F. Mosteller, Methods of studying coincidences, J. Amer. Statist. Assoc. 84 (1989), pp. 853-861.
Eric Weisstein's World of Mathematics, Birthday Problem

Cf. A014088, A050256.

a[n_]:=Ceiling[x /. N[Solve[x Exp[-x/(365 n)]==(365^(n-1) n! Log[2] (1-x/(365 (n+1))))^(1/n), x, Reals]]]; Flatten[Table[a[n], {n, 15}]] (* Iain Fox, Oct 26 2018 *)

a(n) is ceiling(x), where x is the real solution to x*exp(-x/(365*n)) = (log(2)*365^(n-1)*n!*(1 - x/(365*(n+1))))^(1/n). - Iain Fox, Oct 26 2018

a(43)-a(45) from Alois P. Heinz, May 17 2023

A225871 Number of people required for there to be a 50% probability that at least 4 share a birthday in a year with n days.

Original entry on oeis.org

4, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 28, 29, 30, 31, 32, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 51, 52, 53, 53, 54, 54, 55, 56

Offset: 1

Kevin L. Schwartz and Christian N. K. Anderson, May 18 2013

nonn

a(365)=187.

For n<1000, the formula a(n) = 2.79 + 2.456*n^0.732 - 1.825/n provides an estimate of a(n) accurate to 0.82.

			For a year with 365 days, a(365), the probability that out of 186 people 4 of them share a birthday is 0.495825. The corresponding probability for 187 people is 0.502685, and therefore a(365)=187.

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000
Christian N. K. Anderson, Table of n, exact probabilities of a(n)-1 and a(n) for n = 1..1000.
Patrice Le Conte, Coincident Birthdays

Cf. A014088 (n people on 365 days), A033810 (2 people on n days), A225852 (3 on n days).

Cf. A088141, A182008, A182009, A182010.

library(gmp);#prob of a maximum of exactly k coincident birthdays is
BigQ<-function(nday,p,k) { #nday=days in a year; p=people
    if(p1,sum(sapply(2:k-1,function(j) BigQ(nday-i,p-k*i,j))),1)
    }
    tot
}
BDaySharedByAtLeast<-function(nday,people,k) {
    if(nday<1 | people

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

Original entry on oeis.org

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

A091674 Numerator Q of probability P = Q(n)/365^(n-1) that two or more out of n people share the same birthday.

Original entry on oeis.org

1, 1093, 795341, 481626601, 262130079485, 132974790903865, 64157156143943045, 29808728817823292065, 13447118719710220490765, 5923562823392985950002825, 2558600264156303883127171925, 1087010123072386037371040127025

Offset: 2

Hugo Pfoertner, Feb 03 2004

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

Patrice Le Conte, Coincident Birthdays.
Mathforum at Drexel, The Birthday Problem, Ask Dr. Math: FAQ.
Eric Weisstein's World of Mathematics, Birthday Problem.

Cf. A014088, A091673 (probabilities for exactly two), A091715 (probabilities for three or more).

Q[n_] := (1 - Product[(1 - i/365), {i, 1, n - 1}])365^(n - 1); Table[ Q[n], {n, 2, 13}] (* Robert G. Wilson v, Feb 05 2004 *)

from math import prod
def A091674(n): return 365**(n-1)-prod(365-i for i in range(1,n)) # Chai Wah Wu, Jan 22 2025

Q(n) = (1 - Product_{i=1..n-1} (1-i/365))*365^(n-1).

More terms from Robert G. Wilson v, Feb 05 2004

A050256 a(n) = floor(47*(n-3/2)^(3/2)).

Original entry on oeis.org

16, 86, 185, 307, 448, 606, 778, 965, 1164, 1376, 1599, 1832, 2077, 2331, 2595, 2868, 3150, 3440, 3739, 4047, 4362, 4685, 5016, 5354, 5699, 6052, 6411, 6777, 7150, 7530, 7916, 8309, 8708, 9113, 9524, 9941, 10364, 10793, 11227, 11667, 12113, 12565, 13022, 13484

Offset: 2

Eric W. Weisstein

nonn

Mentioned in the Diaconis-Mosteller article.

Iain Fox, Table of n, a(n) for n = 2..10000
P. Diaconis and F. Mosteller, Methods of studying coincidences, J. Amer. Statist. Assoc. 84 (1989), pp. 853-861.
Eric Weisstein's World of Mathematics, Birthday Problem

Cf. A014088, A050255.

a:= n-> floor(47*(n-3/2)^(3/2)):
seq(a(n), n=2..55);  # Alois P. Heinz, May 17 2023

vector(50,n,n++;floor(47*(n-1.5)^(3/2))) \\ Derek Orr, Sep 05 2015

a(n) = floor(47*(n-1.5)^1.5) \\ Charles R Greathouse IV, Sep 05 2015

First term removed by Derek Orr, Sep 05 2015
Offset corrected by Iain Fox, Nov 16 2017
Entry revised by N. J. A. Sloane, Jun 21 2023

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).

A380129 Strong Birthday Problem: Number of people needed so that probability of everyone sharing a birthday out of n possible days is at least 1/2.

Original entry on oeis.org

2, 4, 8, 12, 16, 21, 26, 31, 36, 41, 47, 52, 58, 64, 69, 75, 81, 87, 93, 100, 106, 112, 119, 125, 131, 138, 144, 151, 158, 164, 171, 178, 184, 191, 198, 205, 212, 219, 226, 233, 240, 247, 254, 261, 268, 275, 283, 290, 297, 304, 312, 319, 326, 334, 341, 348, 356

Offset: 1

Mike Sheppard, Jan 13 2025

nonn

The answer to the strong birthday problem is 3064, that is, a(365) = 3064. This is the number of people that need to be gathered together before there is a 50% chance that everyone in the gathering shares their birthday with at least one other person.

Chai Wah Wu, Table of n, a(n) for n = 1..1000
(Note: OEIS uses definition of m and n which are switched from original references)
Mario Cortina Borja, The strong birthday problem, Significance 10.6 (2013): 18-20.
Anirban DasGupta, The matching, birthday and the strong birthday problem: a contemporary review, Journal of Statistical Planning and Inference 130.1-2 (2005): 377-389.

Cf. A014088, A033810.

(* a(n)=m, n number of days, m number of people *)
p[n_, m_] :=Sum[((-1)^i*n^(-m)*((-i + n)^((-i) + m))*n!*m!)/(i!*((-i + n)!)*((-i + m)!)), {i, 0, m}];
a[n_] := Module[{m = n + 1, prob = 0}, While[prob < 0.5, prob = p[n, m]; m++;]; m - 1];
Table[a[n], {n, 1, 20}]

p(n,m) = sum(i=0, n, (-1)^i*(n-i)^(m-i)*binomial(n,i)*m!/(m-i)!)/n^m;
a(n) = my(ma, mb=n+1, md); while(2*p(n,mb)<1, mb<<=1); ma=mb\2; while(mb>ma, md=(ma+mb)\2; if(2*p(n,md)<1, ma=md+1, mb=md)); ma; \\ Jinyuan Wang, Jan 24 2025

from math import comb, factorial
def A380129(n):
    def p(m): return sum((-1 if i&1 else 1)*comb(m,i)*comb(n,i)*factorial(i)*(n-i)**(m-i) for i in range(m+1))<<1
    kmin, kmax = n, n+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

a(n) ~ -n LambertW(-1, -Log(2)/n).

A333507 Decimal expansion of 1 - ((365!) / ((365 - 23)! * 365^23)).

Original entry on oeis.org

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

Offset: 0

Kritsada Moomuang, Mar 25 2020

This is the probability that at least two people in a room of 23 randomly selected people share a birthday, ignoring leap days and assuming days are equiprobable.

This is a rational number; numerator and denominator both have 53 digits. - Charles R Greathouse IV, Jul 11 2020

			0.5072972343239854072254172283370325...

Eric Weisstein's World of Mathematics, Birthday Problem

Cf. A014088, A343015.

RealDigits[1- ((365!) / ((365 - 23)! * 365^23)), 10, 120]

1. - 364!/(342! * 365^22) \\ Charles R Greathouse IV, Mar 25 2020

cp's OEIS Frontend

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

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A225852 The number of people required for there to be at least a 50% chance that at least 3 share a birthday in a year with n days.

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A050255 A Diaconis-Mosteller approximation to the Birthday problem function.

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A225871 Number of people required for there to be a 50% probability that at least 4 share a birthday in a year with n days.

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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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A091674 Numerator Q of probability P = Q(n)/365^(n-1) that two or more out of n people share the same birthday.

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A050256 a(n) = floor(47*(n-3/2)^(3/2)).

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