We have $3$ Aces left, and $52-k$ cards left. We will find the expected number of additional draws until we get an Ace.
We change the notation, and solve a more general problem. There are $b$ boys and $g$ girls. People are chosen, without replacement, until we get a boy. Let $X$ be the number of draws. We want to find $E(X)$. In our card problem, we have $b=3$ and $g=52-k-3$.
Label the girls $G_1, G_2, \dots, G_g$. For $i=1$ to $g$, define the random variable $Y_i$ by
$$Y_i=1 \quad\text{if $G_i$ is drawn before a boy},\quad \text{$Y_i=0$ otherwise}.$$
Then
$$X=1 +(Y_1+Y_2+\cdots +Y_g),$$
since $\sum E(Y_i)$ is the mean number of girls before the first boy.
We will be finished if we can find the $E(Y_i)$. These are all the same, so we find $E(Y_1)$.
But note that
$$E(Y_1)=P(Y_1=1)=\frac{1}{b+1}.$$
This is because in the group of $b+1$ people consisting of $G_1$ and the $b$ boys, $G_1$ must be chosen first, and all possible orders among these $b+1$ people are equally likely. To see this, it is necessary to think of the selections going on until everybody is chosen.
Thus
$$E(X)=1+\frac{g}{b+1}=\frac{b+g+1}{b+1}.$$
Comment:
There are less pleasant, but more combinatorial ways to solve the problem. We can without much difficulty find an expression for the probability that the first boy is chosen on the $i$-th choice, and use the ordinary expression for expectation. It is a bit of a mess, and simplifying it is tricky. But it is easier now that we know, from the simpler calculation above, what the answer should be!