Suppose that we toss balls into \(b\) bins until some bin contains two balls. Each toss is independent, and each ball is equally likely to end up in any bin. What is the expected number of ball tosses?

This is another way of describing the birthday paradox. The expected number of balls tossed \(T(B) = \sum\limits^B_{i=1} \frac{B!}{(B-i)!B^i}\). If we have 365 bins, \(T(b) \approx 24.6\) which is in line with our previously discussed expectations of people in a room sharing birthdays. (Note the difference between the average tosses to fill a duplicate bin and the calculation of when the probability of a matched birthday is greater than 50%)