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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.

%I A225852 #29 Nov 26 2021 20:35:54
%S A225852 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,
%T A225852 17,18,18,18,19,19,19,20,20,20,21,21,21,22,22,22,23,23,23,24,24,24,25,
%U A225852 25,25,26,26,26,26,27,27,27,28,28,28,29,29,29,29,30,30
%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.
%C A225852 a(365) = 88.
%C A225852 For n <= 1000, a(n) = 1.436 + 1.812*n^0.654 - 0.817/n^3 provides an estimate accurate to 0.6 units.
%H A225852 Christian N. K. Anderson, <a href="/A225852/b225852.txt">Table of n, a(n) for n = 1..1000</a>
%H A225852 Christian N. K. Anderson, <a href="/A225852/a225852.txt">Table of n and exact probabilities of a(n)-1 and a(n)</a> for n = 1..1000
%H A225852 Patrice Le Conte, <a href="/A225852/a225852.pdf">Coincident Birthdays</a>
%e A225852 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.
%o A225852 (R) library(gmp);#prob of a maximum of exactly k coincident birthdays is
%o A225852 BigQ<-function(nday,p,k) { #nday=days in a year; p=people
%o A225852     if(p<k | nday<1) return(0)
%o A225852     if(k==1) return(prod(1-(1:p-1)/nday))
%o A225852     tot=0;
%o A225852     for(i in 1:floor(p/k)) {
%o A225852         q=(1-as.bigz(i)/nday)^(p-k*i) * prod((p-as.bigz(1:(k*i))+1)/nday) * prod((nday-as.bigz(1:i)+1)/((1:i)*factorialZ(k)))
%o A225852         tot=tot+as.numeric(q)*ifelse(k*i<p & k>1,sum(sapply(2:k-1,function(j) BigQ(nday-i,p-k*i,j))),1)
%o A225852     }
%o A225852     tot
%o A225852 }
%o A225852 BDaySharedByAtLeast<-function(nday,people,k) {
%o A225852     if(nday<1 | people<k) return(0)
%o A225852     if(k==1) return(prod(1-(1:people-1)/nday))
%o A225852     prob=1; for(j in 2:k-1) prob=prob-BigQ(nday,people,j); prob
%o A225852 }
%o A225852 y=rep(0,100); for(i in 1:100) { j=ifelse(i==1,3,y[i-1]); while(BDaySharedByAtLeast(i,j,3)<.5) j=j+1; y[i]=j}; y
%Y A225852 Cf. A014088 (n people on 365 days), A033810 (2 people on n days), A225871 (4 people on n days).
%Y A225852 Cf. A088141, A182008, A182009, A182010.
%K A225852 nonn
%O A225852 1,1
%A A225852 _Kevin L. Schwartz_ and _Christian N. K. Anderson_, May 17 2013