Expected value of sums

It depends (slightly) on how one interprets before. There are two interpretations possible: (i) Before means not including the draw that got us the first Ace and (ii) We include in the count the draw that got us the first Ace.

There is no big difference between (i) and (ii): The count (and expectation) in (ii) is just $1$ more than the count, and expectation, in (i), We use interpretation (i). So let $W$ be the number of draws before the first Ace, not including the draw that got us the Ace. We want $E(W)$.

The argument is simple but a bit delicate, so the solution below is given in great detail. Luckily, the actual computation is almost formula-free. We use indicator random variables. Label the $48$ non-Aces $1$ to $48$. Don't bother to label the Aces.

Define random variable $X_i$ by $X_i=1$ if the card with label $i$ was drawn before any Ace, and let $X_i=0$ otherwise. Then $$W=X_1+X_2+\cdots+X_{48}.$$ By the linearity of expectation, which holds even when the random variables are not independent, we have $$E(W)=E(X_1+X_2+\cdots+X_{48})=E(X_1)+E(X_2)+\cdots+E(X_{48}).$$ By symmetry, all the $X_i$ have the same distribution. We find, for example, the probability that $X_1=1$. So we want the probability that card with label $1$ is drawn before any Ace.

Consider the $5$-card collection consisting of the $4$ Aces and the card labelled $1$. All orders of these cards in the deck are equally likely. It follows that the probability that card with label $1$ is in front of the $4$ Aces is $\frac{1}{5}$. Thus $E(X_1)=\frac{1}{5}$.

We conclude that $E(W)=\dfrac{48}{5}$.

If we want to take interpretation (ii), and include the draw that got us the Ace, our expectation is $1+\dfrac{48}{5}=\dfrac{53}{5}$.

Add a fifth ace, uniformly randomly arrange the $53$ cards in a circle, and break the circle into a line at the added ace. By symmetry, the expected number of cards between two of the $5$ aces is $\frac{48}5$, so this is the expected number of cards between the beginning of the line and the first ace.

Distribute the $52$ cards uniformly between $0$ and $1$, so on average they're at $k/53$ for $k=1,2,3,\ldots,52$. The four aces are on average at $1/5,\ 2/5,\ 3/5,\ 4/5$. So $$ 0.2 = \frac{k}{53} $$ implies $$ k = 10.6. $$ and $(0.8)\cdot53 = 42.4$. So on average the four aces are the $10.6$th, $21.2$th, $31.8$th, and $42.4$th cards.

I came across this in The Theory of Gambling and Statistical Logic By Richard A. Epstein so pasting a snippet here. I got this from Google Books (public domain) so hopefully there is no copyright violation here. It's amazing how Andre, Michael and Patrick derived the correct answer differently.

A good way to go about this is as follows. Let $T$ be the number of cards drawn at the time of the first ace. This is a $\mathbb{N}$ valued random variable. Therefore $$E(T) = \sum_{n=0}^\infty P(T > n).$$ You may find this helpful.