# 3 Birthdays

This code computes the probability that three (and possibly more) people in a group have the same birthday. The simple case of having two (or more) with the same birthday is widely discussed; it is one minus the probability that no one shares a birthday. Having three birthdays in a day has to take into account one minus no one sharing a birthday and no two people on any days sharing a birthday. For all of the math, there is a website: http://mathforum.org/library/drmath/view/56650.html Make sure you read all of the corrections to the original post. But here is the R code to go with it.

`n=50 #group size, must be even`

K=n/2

## one birthday per day

one=1

for(i in 1:n){ one=one*(365-i+1)/(365) }

## two share a birthday

1-one

### three share

pnk=0

for(k in 1:K)

{ mult2=1; mult3=1

for(i in 1:(n-2*k)){ mult3=mult3* (365-k-i+1)/365 }

mult3=mult3/(365^(2*k))

for(i in 1:(2*k)){ mult2=mult2* (n-i+1)/2}

mult2=mult2*k^(n-2*k)

mult22=1

ind=seq((n-2*k+2), n, length=(n-(n-2*k+2))/2+1)

for(i in ind){ mult22=mult22*choose(i,2) }

pnk=pnk+choose(365,k)*mult22*mult3

}

## three share

1-one-pnk