Suppose we have a normal deck of $52$ cards. We shuffle them well and then turn over the first $13$ cards one-by-one. If the first card is one of the four aces we say that a match has occurred; similarly, if the second card is one of the twos; the third card is one of the threes, etc.; until the $13$th (one of the kings).
What is the expected number of matches?
What is the variance?
Partial answer.
This is similar to the 'hat check' problem in which 13 people put their hats in a pile, each selects a hat at random, and $X$ is the number of of people who retrieve their own hats. In that problem, $E(X) = V(X) = 1$.
Let $Y$ be the number of matches in your problem. Let $U_i = 1$ if the $i$th card is a match and $U_i = 0$ if not, for $i = 1, \dots, 13.$ It is easy to see that $P(U_i = 1) = 4/52 = 1/13$ and thus that $E(U_i) = 1/13.$ The $U_i$ are not independent because the deck is getting 'used up' as you go along. However, $$E(Y) = E(U_1 + \cdots + U_{13}) = E(U_1) + \cdots + E(U_{13}) = 13(1/13) = 1.$$
A similar equation for the variance would require independence of the $U_i$, which we do not have. You can look at the 'hat check' problem to see how one proves that $V(X) = 1$ there, even without independence. Maybe a modification of that argument works here.
Based on a simulation of a million performances of this experiment, it is easy to verify that $E(Y) = 1$ and it seems that $V(Y) \approx 0.942.$ (I would not bet on the third decimal place and I'd be happy to post the code in a comment if asked.) As in the hat check problem the PDF of $Y$ is reasonably approximated by POIS(1), with the worst errors at 0 and 1 (and of course $P(Y > 13) = 0$).