The answer to any question of this form is: Unless you happen to be able to solve it quickly by doing mathematics, simulate! import random def one(): n

The answer to any question of this form is: Unless you happen to be able to solve it quickly by doing mathematics, simulate! import random def one(): n = 0 while True: k = random.randint(0,2) if k==0: return n n += k def many(k): return sum(one() for i in range(n)) / float(n) [many(1000) for i in range(10)] many(100000) will make it pretty easy to guess what the right answer is. The previous-but-one posting on the same blog has a distinctly more difficult question: repeatedly pick independent random numbers uniform on [0,1]; stop when your latest number is smaller than some previous one; what's the average number of numbers you have to pick? Exercise for the reader: provide the appropriate definition of one() to make the calculation above work. It's only a few lines. (Mathematically expert readers will be able to solve both these problems rigorously and quickly without simulating. But if such a problem comes up in a context where you actually care about the answer you'd do well to do the simulation anyway, since it's an easy way to check your reasoning.)