<s>[INST] Covariance of dice tosses that result in 1 or 2 (fake proof) Question: 
Consider n independent tosses of a $k$-sided fair dice. Let $X_i$ be the number of tosses that result in $i$.
What is the covariance $\mathrm{cov}(X_1,X_2)$ of $X_1$ and $X_2$.

\begin{align}
\mathrm{cov}(X_1,X_2) = \mathbf{E}[X_1X_2] - \mathbf{E}[X_1]\mathbf{E}[X_2]
\end{align}
I get a different $\mathbf{E}[X_1X_2]$ than the given solution.
The solution given is
Let $A_t$ (respectively, $B_t$) be a Bernoulli random variabe that is equal to 1 if and only if the $t$th toss resulted in 1 (respectively, 2). We have E$[A_tB_t] = 0$ (since $A_t \neq 0$ implies $B_t \neq 0$)
$$ \mathbf{E}[A_tB_s] = \mathbf{E}[A_t]\mathbf{E}[B_t] = \frac{1}{k} \cdot \frac{1}{k}   \mathrm{for}\  s \neq t.$$
Thus,
\begin{align}
\mathbf{E}[X_1X_2] &= \mathbf{E}[(A_1+\cdots + A_n)(B_1+\cdots B_n)]\\
 &=n\mathbf{E}[A_1(B_1+\cdots+B_n)] = n(n-1)\cdot   \frac{1}{k} \cdot \frac{1}{k} \\
&= \frac{n(n-1)}{k^2}
\end{align}
and

My solution that gives slightly off answer
My approach uses iterated expectations.
\begin{align}
\mathbf{E}[X_1X_2] = \mathbf{E}[\mathbf{E}[X_1X_2|X2]]
\end{align}
If I had $k$ instead of $k-1$ in the following equation, I would get an answer identical to given solution but if I already know $X_2=x_2$ then dice tosses should be identically distributed among k-1 remaining options, right?
\begin{align}
\mathbf{E}[X_1|X_2=x_2] = \frac{n-x_2}{k-1} 
\end{align}
Then 
\begin{align}
\mathbf{E}[X_1X_2|X_2] = \frac{n-X_2}{k-1} \cdot X_2
\end{align}
\begin{align}
\mathbf{E}[\mathbf{E}[X_1X_2|X_2]] = \mathbf{E}[\frac{nX_2-{X_2}^2}{k-1}]
\end{align}
given $\mathbf{E}[{X_2}^2] = \mathbf{E}[{X_2}] = \frac{n}{k}$
\begin{align}
\mathbf{E}[X_1X_2] = \frac{n(n-1)}{k(k-1)}
\end{align}
So my answer differs to the solution on the matter of $\mathbf{E}[X_1X_2]$
\begin{align}
\mathbf{E}[X_1X_2] = \frac{n(n-1)}{k^2} \neq \frac{n(n-1)}{k(k-1)}
\end{align}
Whats wrong with my logic? Or maybe MIT is wrong.
 [/INST] Unfortunately it turns out MIT was right and I was wrong.
Thanks to the help from JeanMarie I was able to come up with the following derivation. 
\begin{align}
\mathbf{E}[{X_2}^2] &= \mathrm{var}(\mathbf{E}[{X_2}^2]) + \mathbf{E}[X_2]^2 \\
\mathrm{var}(\mathbf{E}[{X_2}^2]) &= \frac{n(k-1)}{k^2} \\
\mathbf{E}[X_2]^2 &= \big(\frac{n}{k}\big)^2 \\
\mathbf{E}[{X_2}^2] &= \frac{n(k-1)}{k^2} + \frac{n^2}{k^2} \\
&= \frac{n^2+n(k-1)}{k^2}
\end{align}
Following from where my own solution left off, bunch of algebra.
\begin{align}
\mathbf{E}[X_1X_2] &= \mathbf{E}[\frac{nX_2-{X_2}^2}{k-1}] \\
&= \frac{1}{k-1}\big(n \cdot \frac{n}{k} - \frac{n^2+n(k-1)}{k^2}\big) \\
&= \frac{1}{k-1} \big( \frac{n^2k}{k^2} - \frac{n^2}{k^2}\big) - \frac{n}{k^2} \\
&= \frac{n^2}{k^2} - \frac{n}{k^2} \\
&= \frac{n(n-1)}{k^2}
\end{align}
Which agrees with the MIT solution.
</s>