Card Distribution: Expected Value Percy shuffles a standard $52$-card deck and starts turning over cards one at a time, stopping as soon as the first spade is revealed.
What is the expected number of cards that Percy turns over before stopping (including the spade)?
(Note: There are $13$ spades in a deck.)
I am completely stuck, and I can't formulate an expression. Can I receive a solution?

The expected number of cards is the sum over all cards $k$ of the probability $p_k$ that card $k$ will be turned. Card $k$ will be turned if no previous card was a spade. Thus
$$
p_k=\frac{\binom{39}k}{\binom{52}k}=\frac{39!}{52!}\frac{(52-k)!}{(39-k)!}
$$
and
$$
\sum_{k=0}^{39} p_k= \frac{39!}{52!}\sum_{k=0}^{39}\frac{(52-k)!}{(39-k)!}=\frac{53}{14}=\frac{52+1}{13+1}\;.
$$
The simple result suggests that there should be a more elegant proof.
P.S.: Here's the more elegant proof. Add a marked extra spade and arrange the $53$ cards uniformly randomly on a circle. Now start turning them clockwise, beginning after the marked card. A non-spade is turned if and only if it is in the first of the $14$ segments seperated by the $14$ spades, with probability $1/14$, so the expected number of turned non-spades is $39/14$. Add $1$ for the turned spade to arrive at $53/14$.
