For this problem, it is easier to calculate the probability that 2 persons do not have the same birthday.
With 2 persons, the probability that they do not have the same birthday is .
With 3 persons the probability, that 2 persons do not have the same birthday, is ⋅.
With 4 persons the probability, that 2 persons do not have the same birthday, is ⋅ ⋅. Do you see the pattern now?
Within a group of n persons the probability, that 2 persons do not have the same birthday, is therefore:
This formula can be transformed to:
The initial problem that 2 persons out of n persons do have the same birthday is then P(n) = 1 -P′(n). Figure 10 is a graph of this function P(n). The graph shown that the probability that 2 persons in a group of 30 persons have the same birthday is more than 70%. In a party with 23 persons the probability is already larger than 50%.
The birthday paradox problem has a very useful application in computer science for calculating the probability that 2 entries clashes in a hash table: Given a hash table of size 365, what is the probability that 2 entries clashes on the same memory location after n entries. This is a good example for the fact that the same model can have very different interpretations (or applications) in completely different domains.
Another aspect makes this problem interesting from a modeling point of view. The calculation of the expression for P′(n) above is laborious. Isn’t there a simpler model? Indeed we know that:
Let x = -1∕365, then x = -2∕365, ... then P′(n) can be expressed as:
Simplifying and replacing by another approximation gives:
Hence, the birthday problem can be approximated by a much simple model that is a very good approximation. One can easily calculate the probabilities using a hand calculator.