I’m assuming that color equates to deck in your description. There are $6$ ways to choose the deck that gets picked three times and then $5$ ways to choose the deck that gets picked once, for a total of $6\cdot5=30$ different combinations of decks. The odd card can appear in any of $4$ places in the sequence, so there are $120$ different draws that result in $3$ cards from one deck and one from another. Since a deck cannot be exhausted in $3$ draws, there are $6^4$ possible sequences of $4$ draws. The desired probability is therefore $$\frac{120}{6^4}=\frac{120}{1296}=\frac{5}{54}=0.0\overline{925}\;.$$
I’m not sure that I understand your last paragraph. The probability of drawing three cards from one deck and one from another if the odd card is either the second or third drawn can be calculated similarly; the only difference is that thre are only $2$ places in the sequence in which it can appear instead of $4$, so there are only $60$ different draws that meet the conditions, and the probability is only half as much.
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
Suppose your deck has $n$ cards, and you are interested in the expected number of draws until a particular card is drawn. Call this $u(n)$. We condition on the result of the first draw. With probability $1/n$, it is your target, and you are done. With probability $1 - 1/n$, it is not your target: you are now in the same situation as before except that you have a deck of $n-1$ instead of $n$. Thus we have the recursion equation
$$ u(n) = 1 + \frac{n-1}{n} u(n-1) $$
The solution of this with initial condition $u(1)=1$ is $$ u(n) = \frac{n+1}{2}$$
Now consider case (2) where you are waiting for two specific cards to be drawn. Let $v(n)$ be the expected number of draws here. With probability $2/n$, the result of the first draw is one of your two target cards; the expected number of additional draws you need is then $u(n-1)$. With probability $1-2/n$, it is not one of your targets, and then your expected number of additional draws is $v(n-1)$. Thus we have the recursion
$$ v(n) = 1 + \frac{2}{n} u(n-1) + \frac{n-2}{n} v(n-1) $$ With initial condition $v(2)=2$, the solution is
$$ v(n) = \frac{2n+2}{3} $$
I'll leave case (3) to you to figure out.