There is a distribution to cover these waiting time problems without replacement. It is called the Negative Hypergeometric Distribution. It has an expected value of
$ \frac{N+1}{M+1} \cdot k = \frac{265}{14}$
where N ( 52 cards ) is the total number, M ( 13 spades ) is the total number of the target type and k ( 5 spades needed ) is the number wanted from M.
The indicator variable is, as you suspected, for the event that a given card is guessed correctly. I'll call it $X_i$ for convenience:
For $i=1,\ldots,52$, let
$$X_i =
\begin{cases}
1 & \qquad\text{if card $i$ is guessed correctly} \\
0 & \qquad\text{otherwise}
\end{cases}$$
Then, using linearity of expectation,
$$E(X) = E\left(\sum_{i=1}^{52}{X_i}\right) = \sum_{i=1}^{52}{E\left(X_i\right)} = \sum_{i=1}^{52}{P\left(X_i=1\right)}.$$
We seek $P(X_i = 1)$ now. If we denote by Y (N) a correct (incorrect) guess for any particular card then the event "$X_i=1$" can be represented by the set of all possible strings of length $i$ of Ys and Ns ending in Y.
Take such a string and consider its substrings of maximally consecutive Ns and the following Y. For example, YNNYNNNNYYNY has $5$ such substrings (because it has $5$ Ys) of lengths $1,3,5,1,2$. We can calculate the probability of this outcome, given the rules of the game, to be:
$$\frac{1}{52}\;\cdot\;\frac{50}{51}\cdot\frac{49}{50}\cdot\frac{1}{49}\;\cdot\;\frac{49}{50}\cdot\frac{48}{49}\cdot\frac{47}{48}\cdot\frac{46}{47}\cdot\frac{1}{46}\;\cdot\;\frac{1}{49}\;\cdot\;\frac{47}{48}\cdot\frac{1}{47}\;\; = \;\;\frac{1}{52}\cdot\frac{1}{51}\cdot\frac{1}{50}\cdot\frac{1}{49}\cdot\frac{1}{48}$$
$$\\$$
Hopefully, it's not hard to convince yourself of the pattern here that the probability of any given Y-N string is $\dfrac{(52-k)!}{52!}$ where $k$ is the number of Ys in the string.
Now, of all Y-N strings of length $i$ ending in Y, there are $\binom{i-1}{k-1}$ strings with exactly $k$ Ys since we need to choose $k-1$ positions from $i-1$ possibilities for all Ys but the last. And since this $k$ can range from $1$ to $i$, we have,
\begin{eqnarray*}
P(X_i = 1) &=& \sum_{k=1}^{i}{\binom{i-1}{k-1} \dfrac{(52-k)!}{52!}}.
\end{eqnarray*}
So,
\begin{eqnarray*}
E(X) &=& \sum_{i=1}^{52}{\left[ \sum_{k=1}^{i}{\binom{i-1}{k-1} \dfrac{(52-k)!}{52!}} \right]} \\
&& \\
&=& \sum_{k=1}^{52}{\left[ \dfrac{(52-k)!}{52!} \sum_{i=k}^{52}{\binom{i-1}{k-1}} \right]} \qquad\text{(swapping the order of summation)} \\
&& \\
&=& \sum_{k=1}^{52}{\left[ \dfrac{(52-k)!}{52!} \binom{52}{k} \right]} \qquad\text{(using identity $\sum\limits_{r=p}^{q}{\binom{r}{p}} = \binom{q+1}{p+1}$)} \\
&& \\
&=& \sum_{k=1}^{52}{\dfrac{1}{k!}} \\
&& \\
\end{eqnarray*}
Note now the Taylor expansion: $\;\;e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$.
So, $\;e-1 \;=\; \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots\; = \;\sum\limits_{k=1}^{\infty}{\frac{1}{k!}}$.
We can see $E(X)$ is a partial sum of this series and is extremely close to $e-1$.
Best Answer
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 $53/14$.