13 cards are drawn from a deck one after another... What the expected number of...? 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 13th (one of the kings).
What is the expected number of matches?
what is the variance?
 A: 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 $P(Y > 13) = 0$).
