Expectation and Variance of Card Pairs A 26 card deck containing 13 "hearts" and 13 "spades" is well-shuffled. Thirteen pairs are then dealt from the deck. A pair is a "match" if it contains 1 heart and 1 spade. Let N be the number of matches. The deck will always be exhausted in each dealing, so you will always have 13 pairs. What is the expectation and variance of N? 
N will always have be an odd number because you have 13 of each card, and the median of N is 7, so that's my educated guess for expectation. Beyond that, however, I'm not sure of how to approach this problem.
 A: Expected Value
The Expected Value can be solved easily using Linearity of Expectation. Define 
$$X_i = \begin{cases}
1, & \text{ if the $i^{th}$ pair is a match} \\
0, & \text{ otherwise}
\end{cases}$$
It should be clear that, with this set up, we have $N = \sum_{i=1}^{13} X_i$. 
Next we find $P(X_i = 1)$. Without loss of generality, assume that the first card is a heart which implies that there are $12$ hearts and $13$ spades remaining. Thus $P(X_i = 1) = \frac{13}{12+13} = 0.52$. In turn, we have $E(X_i) = 0(0.48) + 1(0.52) = 0.52$.
Now via Linearity of expectation,
$$E(N) = E\left(\sum_{i=1}^{13}X_i\right) = \sum_{i=1}^{13}E(X_i) = \sum_{i=1}^{13}0.52  = 6.76$$
Variance
The calculation for variance is somewhat more involved. Since the $X_i$ are not independent we have,
$$Var(N) = Var\left(\sum_{i=1}^{13}X_i\right) = \sum_{i=1}^{13}Var(X_i) + 2\sum_{i\neq j}Cov(X_i, X_j)$$
First note that $Var(X_i) = (0.52)(0.48)$ (by recognizing that $X_i \sim Bern(0.52)$). To find the covariance of term, we start with
$$Cov(X_i, X_j) = E(X_iX_j) - E(X_i)E(X_j)$$
The only new term here is $E(X_iX_j)$. With a little bit of thought, you should be able to convince yourself that 
$$E(X_iX_j) = P(X_i=1, X_j=1) = P(X_i=1|X_j=1)P(X_j=1)$$
We now focus on the term $P(X_i=1|X_j=1)$. In simpler terms (wlog), we are looking for "the probability the second pair is a match given that the first pair is a match." By similar logic to above, we have,
$$P(X_i=1|X_j=1) = \frac{12}{11+12} = \frac{12}{23}$$.
Finally, we put all of the pieces together.
\begin{align*}
Var(N) &= \sum_{i=1}^{13}Var(X_i) + 2\sum_{i\neq j}Cov(X_i, X_j) \\
&= 13(.52)(.48) + 2\binom{13}{2}\left[\frac{12}{23}(0.52) - (0.52)^2\right] \\
&= 3.385878
\end{align*}
Verification via Simulation
These answers can be easily checked in R
M <- 10000
count <- rep(0, M)
for(m in 1:M){
  deck <- c(rep(0, 13), rep(1,13))
  shuffle <- sample(deck, replace=F)
  for(i in seq(1,26,by=2)){
    if(shuffle[i]!=shuffle[i+1])
      count[m] <- count[m] + 1
  }
}

Running this simulation, mean(count) returns $6.76004$ and var(count) returns $3.396853$.
