Math Is Fun Forum

My solution :
Suppose     n = number of balls
            b = number of bins

    The ith stage is the stage when the first bin contains 2 balls.
    So in ith stage there are (n-i) balls remain

    We have the expected number of balls (n) in a given bin (b) = n/b
    So expected number of ball tosses after at least one of the bins contains two balls = (n-i)/b
    So expected number of ball tosses before at least one of the bins contains two balls
            =1–[(n-i)/b] = (b-n+i)/b