Expected value - random sets cardinality I've got a set $|A|=n$ and two random subsets $B,C\subseteq A$ and $|B|=|C|=k$. What is the expected value of $|B\cap C|$? My approach: I consider $B$ given and try to fill $C$. The probability of picking no common terms ($l=0$) is $\dfrac{n-k\choose k}{n\choose k}$, because there are $n-k\choose k$ ways to pick one of the terms that hasn't been picked already and $n\choose k$ possible scenarios. For $l=1$, I pick one term from $B$ into $C$ and calculate the probability of the remaining $C$-terms not being in $B$. There are $k\choose1$ ways for choosing the common term, and $k-1$ remaining terms: ${k\choose1}\dfrac{n-k\choose k-1}{n\choose k-1}$. So, the expected value is expected to be $$0\cdot\dfrac{{{n-k}\choose k}}{{n\choose k}}+1\cdot {k\choose1}\dfrac{{{n-k}\choose {k-1}}}{{n\choose {k-1}}}+\ldots+k{k\choose k}\dfrac{{{n-k}\choose 0}}{{n\choose0}}$$But the last term is apparently equal to $k$ and other terms are non-negative, so the expected value is at least $k$, and that seems kind of unbelievable. Where did I make a mistake?
 A: There’s no harm in taking $B$ to be a fixed $k$-element subset of $A$ and picking $C$ at random. To choose $C$ so that $|B\cap C|=j$, we must choose $j$ elements of $B$ and $k-j$ of $A\setminus B$; this can be done in $\binom{k}j\binom{n-k}{k-j}$ ways. There are altogether $\binom{n}k$ ways to choose $C$, so the desired expectation is
$$\begin{align*}
\binom{n}k^{-1}\sum_{j=0}^kj\binom{k}j\binom{n-k}{k-j}&=k\binom{n}k^{-1}\sum_{j=0}^k\binom{k-1}{j-1}\binom{n-k}{k-j}\\\\
&=k\binom{n}k^{-1}\sum_{j=1}^k\binom{k-1}{j-1}\binom{n-k}{k-j}\\\\
&=k\binom{n}k^{-1}\binom{n-1}{k-1}\\\\
&=\frac{k\binom{n-1}{k-1}}{\frac{n}k\cdot\binom{n-1}{k-1}}\\\\
&=\frac{k^2}n\;.
\end{align*}$$
The step that gets rid of the summation uses Vandermonde’s identity.
As a quick sanity check, note that if $k=n$ this gives an expected value of $n$, which is certainly correct, as in that case $B=C=A$, and if $k=1$, it give an expected value of $\frac1n$, which can also be verified directly, since in that case the expected value is evidently
$$\frac1n\cdot1+\frac{n-1}n\cdot0=\frac1n\;.$$
A: Without loss of generality we may assume that $B$ consists of the integers from $1$ to $k$. For $i=1$ to $k$, let $X_i=1$ if $i\in C$, and $0$ otherwise. Then the cardinality of $B\cap C$ is $X_1+X_2+\cdots +X_k$.
By the linearity of expectation, we want to find $\sum_1^k E(X_i)$.
The probability that $i\in C$ is $\frac{k}{n}$. It follows that $\sum_1^k E(X_i)=\frac{k^2}{n}$.
Remarks: $1.$ Finding the distribution of a random variable may be difficult. Even if one does, finding a simplified form for the resulting complicated expression for the mean can be painful. Surprisingly often, the method of indicator functions provides a simple alternate path to the calculation of the expectation.  
$2.$ In the calculation of the probabilities, the denominators do not appear to be correct. Fix $B$. There are $\binom{n}{k}$ equally likely ways to choose $C$. Thus $\binom{n}{k}$ should be in all the denominators. For the numerators, for any amount of overlap $l$, one chooses the elements of $B$ that will be contained in $C$. There are $\binom{k}{l}$ ways to choose these, and for each such way there are $\binom{n-l}{k-l}$ ways to choose the remaining elements of $C$. 
