Nothing novel or unique about this problem.  This just extends the problem to measure the probability to three or more people sharing the same birthday using simulation approaches.

For two people it’s fairly straight forward and with a group of about 22 people the probability that two people share the same birthday is about 0.5.  For groups approaching 50 there is an extremely high probability that two people share the same birthday

When determining that three (or more) people have the same birthday the probability decreases fairly quickly compared to measuring only two people.  A fairly large group would be needed to find three people with the same birthday.

Here is some R code to determine these probabilities.

n.rep = 5000
theta.val = 75
doy = seq(from=1, to=365, by=1)
sim.mat = matrix(NA, nrow=theta.val, ncol=4)

getProb = function(n){
q = 1 - seq(0,n-1)/365
p = 1 - prod(q)
}

theta.list = seq(from=2, to=75, by=1)
p.graph = sapply(theta.list, getProb)
fifty.fifty = which(p.graph >.5)[1]
plot(p.graph, main="Probability Two People Have the Same Birthday", ylab='Probability', xlab="Number of People in Group")
lines(p.graph)
abline(h=.5, v=fifty.fifty)

 

## For matching multiple people

## Runs a little slow.  If I had more time I would find a more efficient way to write this.
for(i in 2:theta.val){
bday = replicate(n.rep, sample(doy, size=i, replace=T) )
bday[1,]
bday.table = apply(bday, 2, table)

sim.2 = ifelse( unlist( lapply(bday.table, max) ) >=2, 1, 0)
sim.3 = ifelse( unlist( lapply(bday.table, max) ) >=3, 1, 0)
sim.4 = ifelse( unlist( lapply(bday.table, max) ) >=4, 1, 0)

sim.mat[i,1] = i
sim.mat[i,2] = sum(sim.2)/length(sim.2)
sim.mat[i,3] = sum(sim.3)/length(sim.3)
sim.mat[i,4] = sum(sim.4)/length(sim.4)

}

graph.sim = t( sim.mat[,2:4] )
colnames(graph.sim) = sim.mat[,1]

barplot(graph.sim[1,], ylim=c(0,1), col="red",
main="Probability of Having Multiple People with the Same Birthday",
xlab="People with Birthday",
ylab="Probability")
barplot(graph.sim[2,], ylim=c(0,1), col="blue", add=T)
barplot(graph.sim[3,], ylim=c(0,1), col="black", add=T)
abline(h=.5)
legend("topleft", c("2","3","4"), col=c("red","blue","black"), lwd=3)