# What is the probability that at least 2 people in class have the same birthday?

The title states the full question I was provided with. Am I able to just assume the class contains more than 12 students? If so, would we just do:

(n choose 2?); where n = number of students in class?

If not, how else could I go about this? Do I need to break it into more than 1 case? I know the solution is probably very simple, I just don't know if I am thinking about this in the right manner.

• This is called the birthday problem. The wiki page contains the information you need.
– user347499
Jun 14, 2016 at 18:39
• "Am I able to just assume the class contains more than 12 students?" - Now why would you ever assume that??? If anything, you may want to assume "more than $1$ student" (otherwise, obviously, the probability is $0$). In order to find the general formula for $n$ students, calculate $1$ minus the probability of the complementary event: $1-\frac{365!}{(365-n)!\cdot365^{n}}$. BTW, with $n=23$, the probability is just a little over $50\%$, which is why this is generally referred to as the Birthday Paradox (considered a paradox because people normally expect a lot more than $23$ here). Jun 14, 2016 at 19:50
• You assume the class has n students an's express the answer in terms of n. If you assume n=12 you will get a precise number. If you assume n=20 you will get another. Not sure where you got 12 from. Jun 14, 2016 at 20:48
• n choose 2 isn't the answer by the way. It doesn't make sense. Jun 14, 2016 at 20:50