Waiting for tail then head

Idea. The pattern TH appears exactly at the first head that comes after the first tail. So the wait is two geometric waits added together.

Step 1: the first tail. Each flip is a tail with probability $\frac{1}{3}$, so the number of flips up to and including the first tail is geometric with mean $3$.

Step 2: then a head. After that tail, every further tail keeps you in the same situation (the last flip is still a tail), and the first head completes TH. So this wait is geometric with success probability $\frac{2}{3}$, with mean $\frac{3}{2}$.

Step 3. $$\mathbb{E}[\text{flips}]=3+\frac{3}{2}=\frac{9}{2}.$$

In general. With $\mathbb{P}(\text{head})=p$ and $q=1-p$, the wait for TH is $\frac{1}{q}+\frac{1}{p}=\frac{1}{pq}$. The same holds for HT. For a fair coin it is 4, while HH takes 6 on average: HH can overlap with itself, and a failed attempt at HH throws away the head you had.