This is a pretty solution, and it gives you the probability generating function for free: by some power series manipulation, the pgf of the number of draws need

This is a pretty solution, and it gives you the probability generating function for free: by some power series manipulation, the pgf of the number of draws needed is g(z) = 1 + (z-1)*exp(z). So you can show fairly easily that Var(draws needed) = 3e - e^2, for example ...