This writeup basically just works out the probabilities involved in the situation. To simplify let us assume that each year has 365 days uniformly .
Now what is the chance that two people selected at random will have the same birthday. The total possibilities for the birthday of the first person are 365 and this is also true for the birthday of the second person. Therefore the total number of possibilities for both their birthdays are 365 2 . Of these, 365 possibilities involve the two people having the same birthday. So the probability of two people having the same birthday is 1/365.
Now comes the second problem. Instead of calculating the probability of two or more people having the same birthday(which is difficult) let us instead calculate the probability that no two people have the same birthday. When subtracted from 1 this probability will give us the answer we want. If you have n people the total no. of possibilities for their birthdays is 365 n . Let us choose one person and label him person 1. He can have a birthday on any of the 365 days. Now let us take another person(person 2). In order that the birthdays dont clash person 2 has only 364 possibilities for his(her) birthday. Similarly there are only 363 days of the year left for person 3. We can continue this way but note that if there are 366 people in the room two of them have to share the same birthday
Anyway with the above calculation we find that the probability is
1- (365 P n / 365 n )
The
distribution is not a
binomial distribution.
I must make a correction here. In my previous WU, I accused The Cow of being wrong. However, thanks to hobyrne, I now realise that this was unfair. The Cow is addressing a different problem, namely the probability of one person in a room having a birthday on a particular day. My WU deals with the probability of any two people sharing a birthday.