In a group of 23 people, the probability that at least two have the same birthday is bigger than 50%. This result may be counterintuitive and probably this is why it is called  ‘paradox’, although it is a real and provable fact.

A more general formulation:  given a group of $n$ persons, what is the probability that at least two of them will have the same birthday?

To keep things simple, let’s suppose that each year have exactly $365$ days. The birthday of every person can be one of the $365$ days. This means that the total number of birthday possibilities for a group of $n$ people is $365^n$. Some of the possibilities will contain people having same birthday, and we are interested to find how many,  in order to compute the probability.

The total number of birthday possibilities such that each two birthday are different can be computed using k-permutations of n (‘aranjamente’ in Romanian).  For the birthday problem it is: $\frac{365!}{(365-n)!}=365(365-1)(365-2)\cdots(365-n+1)$.

Alternatively this can be thought as follows: first person can have any of the $365$ days as birthday, second person can have the remaining $365-1$ days as birthday, … ,the n-th person can have any of the remaining $365-(n-1)$ days as birthday. Multiplying the total number is: $365(365-1)(365-2)\cdots(365-(n-1))$.

Then the number of birthday possibilities for which at least two people have the same birthday is: $365^n-\frac{365!}{(365-n)!}$. Thus the probability of having two persons with the same birthday is:

$f(n)=\frac{365^n-\frac{365!}{(365-n)!}}{365^n}$.

For $n=23$ the result is $f(n)=0.5073$. Here is a plot of the probability based on the number of people in the group: