A standard 52-car deck is shuffled, and cards are turned over one-at-a-time starting with the top card. What is the expected number of cards that will be turned over before we see the first Ace? (Recall that there are 4 Aces in the deck.)
There's a very clever way to do this. The $4$ aces partition the deck into $5$ components, with size on average ${{52 - 4}\over5} = 9.6$. We then have to draw the first Ace, so the expected number of cards that'll be turned over before we see it is $9.6 + 1 = 10.6$.
However, for those out there who are stupid like myself (or more generously put, want to practice our computational fortitude), let's do it by brute force. We want to calculate$$1\left({4\over{52}}\right) + 2\left({{48}\over{52}}\right)\left({4\over{51}}\right) + 3 \left({{48}\over{52}}\right)\left({{47}\over{51}}\right)\left({{4}\over{50}}\right) + 4 \left({{48}\over{52}}\right)\left({{47}\over{51}}\right)\left({{46}\over{50}}\right)\left({{4}\over{49}}\right) + \ldots + 48\left({{48}\over{52}}\right)\left({{47}\over{51}}\right)\ldots\left({{2}\over{6}}\right)\left({{4}\over{5}}\right) + 49\left({{48}\over{52}}\right)\left({{47}\over{51}}\right)\ldots\left({{2}\over{6}}\right)\left({{1}\over{5}}\right)\left({{4}\over{4}}\right) = {4\over{52}} + \sum_{n = 2}^{49}\left(n \left(\prod_{i=1}^{n-1} {{49 - i}\over{53 - i}}\right) \left({4\over{53 - n}} \right)\right)$$However, I'm not sure how to proceed with evaluating that expression. How on Earth can I get it to evaluate to $10.6$?