I stumbled upon the following question: A company has a manager and $n \geq 3$ employees. The manager flips a coin $2n$ times: each time he flips heads he gives one task to one of his employees, chosen randomly and uniformly. We denote $X_i$ as the number of tasks given to employee number $i$. Find $Cov(X_1,X_2)$.
My attempt:
At first I calculated $\forall_{1\leq i \leq n}.\mathbb{E}[X_i]$.
$\mathbb{E}[X_i] = \sum_\limits{k=0}^{2n}\mathbb{E}[X_i|Number ~of~heads=k] \mathbb{P}(Number ~of~heads=k) = \sum_\limits{k=0}^{2n} \frac{k}{n} {2n \choose k}(\frac{1}{2})^{2n} = \frac{1}{n}\sum_\limits{k=0}^{2n} k {2n \choose k} (\frac{1}{2})^{2n} = \frac{1}{n} \mathbb{E}[Number~of~heads] = 1$
So we have $Cov(X_1, X_2) = \mathbb{E}[X_1X_2]-\mathbb{E}[X_1]\mathbb{E}[X_2]=\mathbb{E}[X_1X_2]-1$. I got stuck from here, failing to calculate $\mathbb{E}[X_1X_2]$.
Any help would be appreciated.
Let's define the random variables $$I_k = \begin{cases} 1, \quad \text{if employee 1 got the $k$-th task} \\ 0, \quad \text{if he didn't} \end{cases} $$ And similarly we define $J_\ell$ for employee $2$. Notice that $$ X_1 = \sum_{k=1}^{2n} I_k \quad \text{and} \quad X_2 = \sum_{\ell=1}^{2n} J_\ell. $$ By linearity of the covariance, we have $$ \rm{Cov}(X_1, X_2) = \rm{Cov}\left( \sum_{k=1}^{2n} I_k, \sum_{\ell=1}^{2n} J_\ell \right) = \sum_{k=1}^{2n} \sum_{\ell=1}^{2n} \rm{Cov}(I_k, J_\ell). \tag{1}\label{full cov} $$
Observe that, because the coin flips are independent, $I_k$ is independent of $J_\ell$ if $k \neq \ell$ and, hence, $\rm{Cov}(I_k, J_\ell) = 0$ in that case. Therefore, (\ref{full cov}) reduces to $$ \sum_{k=1}^{2n} \rm{Cov} (I_k, J_k). $$ Because they are identically distributed for every $k$, this is the same as $$ 2n \cdot \rm{Cov} (I_1, J_1). $$ Now all we have to do is compute this covariance. For the individual expectations, we have $$ \mathbb{E}[J_1] = \mathbb{E}[I_1] = \frac{1}{2} \cdot \frac{1}{n} = \frac{1}{2n}. $$ For the expectation of the product, we know that $I_1 J_1$ is only equal to $1$ if both of them are $1$, otherwise it's $0$. But the probability of that happening is $0$. So $\mathbb{E}[I_1 J_1] = 0$.
Hence, $$ \rm{Cov} (I_1, J_1) = \mathbb{E}[I_1 J_1] - \mathbb{E}[I_1]\mathbb{E}[J_1] = - \frac{1}{4 n^{2}}. $$ So, $$ \rm{Cov}(X_1, X_2) = 2n \cdot \left( - \frac{1}{4n^2} \right) = - \frac{1}{2n}. $$