The five strings of non-aces can be permuted amongst each other, conserving probability. Note that some of the strings may be empty and where there are multiple empty strings, the 120 possible "abstract" permutations will produce fewer than 120 physical permutations of the cards.
The symmetry implies that the expected length each string of non-aces is the same, or 48/5 for a standard deck of cards.
The expected waiting time to the first ace, or from the first to the second ace, or ace $k$ to ace $k+1$, is the preceding number $+1$. That is the expected length of a full interval of non-aces followed by an ace. So 53/5 is the answer for a standard deck.
The general answer is (cards+1)/(aces+1), by the same argument.
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
At each step, four cards are removed from the deck, so a deck is exhausted in $13$ steps. Let's call that a 'round'.
The game you propose can be restated as follows. At each round, we draw $13$ ordered cards from the deck, and add up the values of the cards that came before the first ace in our $13$ drawn cards. If no ace is drawn, we add up the values of all $13$ drawn cards, shuffle them into the remaining $39$ cards (the ones which were not drawn), and repeat the process.
There are $\binom{48}{13}$ sets of $13$ cards which do not contain an Ace, and so there are $\binom{52}{13}-\binom{48}{13}$ sets of $13$ cards which contain at least one Ace. Hence, the probability that the game ends in a given round is
$$p=1-\frac{\binom{48}{13}}{\binom{52}{13}}=\frac{14498}{20825}\simeq69.62\%$$
Now, if the game has not ended in a given round, then the expected sum of the cards drawn is $13$ times the expected value of a card drawn (by linearity of the expectation). Since no card drawn was an ace, the expected value of a card drawn is
$$\frac{2+3+4+5+6+7+8+9+10+11+12+13}{12}=\frac{15}2,$$
and hence the expected sum is $l=13\cdot\frac{15}2=\frac{195}2=97.5$.
Now, we need to calculate the expected sum of the cards drawn before the first Ace in a round where the game ends. Here, we will break the thing into cases.
$\qquad$Number of Aces in cards drawn: $1$
There are $\binom{48}{12}\cdot\binom{4}{1}$ sets of $13$ cards which contain exactly one Ace. Therefore, given that the $13$ cards drawn contain at least one ace, the probability that we fall in this case (exactly one Ace drawn) is
$$a_1=\frac{\binom{48}{12}\cdot\binom{4}{1}}{\binom{52}{13}-\binom{48}{13}}=\frac{9139}{14498}\simeq 63.04\%$$
The expected position of the lone ace is $\frac{1+2+3+4+5+6+7+8+9+10+11+12+13}{13}=7$, so on average $6$ non-Ace cards will be drawn before it. The expected sum for this case is hence $s_1=6\cdot\frac{15}2+1=46$.
$\qquad$Number of Aces in cards drawn: $2$
There are $\binom{48}{11}\cdot\binom{4}{2}$ sets of $13$ cards which contain exactly two Aces. Like before, the probability that we fall in this case is
$$a_2=\frac{\binom{48}{11}\cdot\binom{4}{2}}{\binom{52}{13}-\binom{48}{13}}=\frac{2223}{7249}\simeq 30.67\%$$
Now, things get trickier. The position of the pairs of aces is a subset of size $2$ on $S=\{1,2,\dots,13\}$, and we are interested in the minimimum of this subset. Let $X$ denote this random variable.
There are $\binom{13}2$ $2$-subsets of $S$, and only $12$ of them contain the number $1$, which is a guaranteed minimum. Therefore
$$\mathbb{P}(X=1)=\frac{12}{\binom{13}2}$$
Similarly, there are $11$ $2$-subsets of $S$ whose minimum is $2$.
More generally, for each $k\in\{1,2,\dots,12\}$, $\binom{13-k}1$ of the $2$-subsets of $S$ have minimum $k$, and we find that
$$\mathbb{P}(X=k)=\frac{\binom{13-k}1}{\binom{13}2}.$$
As a sanity check, notice that they add up to $1$. The expected sum for this case is hence:
$$s_2=\sum_{k=1}^{12}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \mathbb{P}(X=k)=\frac{57}2$$
$\qquad$Number of Aces in cards drawn: $3$
Now we've got most of the work done. We have that
$$a_3=\frac{\binom{48}{10}\cdot\binom{4}{3}}{\binom{52}{13}-\binom{48}{13}}=\frac{39}{659}\simeq 5.92\%$$
For this case, let $Y$ be the random variable which denotes the minimum of a uniformly sampled $3$-subset of $S$. Notice that there are $\binom{13}{3}$ $3$-subsets of $S$.
We will have that for each $k\in\{1,2,\dots,11\}$, $\binom{13-k}2$ of the $3$-subsets of $S$ have minimum $k$. Hence:
$$\mathbb{P}(Y=k)=\frac{\binom{13-k}2}{\binom{13}3}$$
Finally, the expected sum for this case is
$$s_3=\sum_{k=1}^{11}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \mathbb{P}(Y=k)=\frac{79}4$$
$\qquad$Number of Aces in cards drawn: $4$
For this final case we have
$$a_4=\frac{\binom{48}{9}\cdot\binom{4}{4}}{\binom{52}{13}-\binom{48}{13}}=\frac{5}{1318}\simeq 0.38\%$$
and expected sum
$$s_4=\sum_{k=1}^{10}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \frac{\binom{13-k}3}{\binom{13}4}=\frac{29}2$$
Let's put it all together. Supposing the game ends on a given round, the expected sum for that round will be $($and you can check that the $a_i$ add up to $1)$
$$w=\sum_{i=1}^4a_is_i=\frac{282424}{7249}\simeq38.96$$
Finally, the expected sum for the game will be given by
$$\sum_{n=1}^\infty\, \underbrace{\mathbb{P}(\text{Game ends on round $n$})}_{(1-p)^{n-1}\cdot p} \cdot \underbrace{\mathbb{E}(\text{Value of sum of cards of a game ending on round $n$})}_{(n-1)\,l+w}\\ =\sum_{n=1}^\infty\,(1-p)^{n-1}\cdot p \cdot \Big((n-1)\,l+w\Big)=(w-l)+\frac{l}p $$
This last step is standard manipulation of series and term-by-term differentiation, but I can further explain if it's not clear. Therefore, the final answer is