Heads times tails

Idea. $HT=H(100-H)$ is a quadratic in $H$, so its mean needs the variance of $H$, not just its mean.

Step 1. $H$ is binomial with $n=100$, $p=\frac{1}{2}$: $\mathbb{E}[H]=50$ and $\mathrm{Var}(H)=25$, so $\mathbb{E}[H^2]=25+2500=2525$.

Step 2. $$\mathbb{E}[HT]=100\cdot 50-2525=2475.$$

The trap. $50\times 50=2500$ treats $H$ and $T$ as if they were independent. They are perfectly negatively related, and $\mathbb{E}[HT]=\mathbb{E}[H]\mathbb{E}[T]+\mathrm{Cov}(H,T)=2500-25$.

Another route. Count pairs: $HT$ is the number of ordered pairs (a head toss, a tail toss). There are $100\times 99$ ordered pairs of different tosses, and each is (head, tail) with probability $\frac{1}{4}$: $\frac{9900}{4}=2475$.