# Solve birthday problem with probability theory and combinatorics

I got the following problem:

How many persons have to be in a room so the probability that at least two persons have birthday in the same month is over 50%.

Through trying out I figured there have to be at least 5 persons in the room for a probability over 50% (for 5 people it would be exactly 62%):

$$1-\frac {12*11*10*9*8} {12*12*12*12*12} = 0,62$$

I heard there is a direct way to compute this through probability theory (where every month has the same probability to be a birth month) and there is another way through combinatorics. Is this correct?

• The way you computed was direct, wasn't it? What do you mean by direct? Oct 28, 2015 at 18:43
• Done many times here, full results on wiki. Oct 28, 2015 at 18:44
• that is how you solve it direction. Prob2withsamemonth = 1 - Probwithalldifferentmonth. Probwithalldifferentmonth(n people) = $(12\*...\*(12 -n))/12^n = \frac{12!}{(12-n)!*12^n}$. Set to .5 and solve for n. Oct 28, 2015 at 18:52
• Well I just thought there would be also a way to compute this though combinatorics. Oct 29, 2015 at 19:56