Product of Heads and Tails

Since every toss is either a head or a tail, $Y = n - X$ , where $X \sim \mathrm{Bin}(n,\tfrac12)$ . Hence

\mathbb{E}[X] = \frac{n}{2}, \qquad \operatorname{Var}(X) = \frac{n}{4}, \qquad \mathbb{E}[X^2] = \operatorname{Var}(X) + \mathbb{E}[X]^2 = \frac{n}{4} + \frac{n^2}{4}.

By linearity,

\mathbb{E}[XY] = \mathbb{E}[X(n-X)] = n\,\mathbb{E}[X] - \mathbb{E}[X^2] = \frac{n^2}{2} - \frac{n^2+n}{4} = \frac{n^2-n}{4}.

\boxed{\mathbb{E}[XY] = \frac{n^2-n}{4} = \frac{n(n-1)}{4}}

Alternative. Let $H_i$ and $T_j$ be the indicators that toss $i$ is a head and that toss $j$ is a tail. Then

XY = \sum_{i,j} H_i T_j .

When $i=j$ we have $H_iT_i = 0$ . When $i \ne j$ the tosses are independent, so $\mathbb{E}[H_iT_j] = \tfrac14$ . There are $n(n-1)$ ordered pairs with $i \ne j$ , which gives $\mathbb{E}[XY] = \tfrac{n(n-1)}{4}$ again.

Key idea. Since $\mathbb{E}[X]\,\mathbb{E}[Y] = n^2/4$ , the answer says $\operatorname{Cov}(X,Y) = -\operatorname{Var}(X) = -n/4$ . This is the price of the constraint $X+Y=n$ .