How can I find $\operatorname{Cov}(X,Y)$ $n=240$ trials with a $6$ sided dice
$X = \#5$'s
$Y = \#6$'s
How do I go about showing that $\operatorname{Cov}(X,Y) = -20/3$? I think I need to find $V(X+Y)$ but I'm not sure how. $V(X)=V(Y)=240* 1/6 * 5/6$. 
 A: That is the covariance if the die rolled 240 times is fair (or the 240 dice rolled are fair). What if the die/dice is/are not fair?
X and Y are binomially distributed:
$$P(X=x) = \binom{n}{x}(p_x)^x(1-p_x)^{n-x}$$
$$P(Y=y) = \binom{n}{y}(p_y)^y(1-p_y)^{n-y}$$
$p_x$ and $p_y$ represent the probability of rolling a 5 and a 6, resp.
$$E[X] = np_x = 240p_x \stackrel{if \ fair}{=} 40$$
$$E[Y] = np_y = 240p_y \stackrel{if \ fair}{=} 40$$
Now to compute $Cov(X,Y) = E[XY] - E[X]E[Y]$ we also need $E[XY]$.
$$E[XY] = \sum_{x=0}^{n}\sum_{y=0}^{n} xy P(X=x, Y=y)$$
I guess $X$ and $Y$ are not independent.
$$P(X=x, Y=y)$$
is what you call the bivariate binomial distribution:
$$P(X=x, Y=y) = \binom{n}{x}\binom{n-x}{y}(p_x)^x(p_y)^y(1-p_x-p_y)^{n-x-y}$$
Note that
$$\binom{n}{x}\binom{n-x}{y} = \frac{n}{(n-x-y)!x!y!}$$
agreeing with the form in the link above, which derives $E[XY]$ to be:
$$E[XY] = n(n-1)p_xp_y \stackrel{if \ fair}{=} (240)(239)(1/6)(1/6)$$
$$\therefore, Cov(X,Y) = E[XY] - E[X]E[Y] \stackrel{if \ fair}{=} -240(1/6)(1/6)$$
