Suppose we are tossing a fair coin. Then the expected waiting time for heads-heads is 6 throws, but the expected waiting time for tails-heads is 4 throws. This is very counterintuitive to me because the events heads-heads and tails-heads has the same probability, namely $\tfrac{1}{4}$. The general result is the following:

Suppose we are throwing a coin that has probability $p$ for heads and probability $q=1-p$ for tails. Let $V_{\text{HH}}$ be first time we encounter two heads in a row and $V_{\text{TH}}$ be the first time we encounter heads and tails in a row, i.e. $$ V_{\text{HH}}(\omega)=\min\{n\geq 2\mid \omega\in H_{n-1}\cap H_n\},\\ V_{\text{TH}}(\omega)=\min\{n\geq 2\mid \omega\in H_{n-1}^c\cap H_n\}, $$ where $H_n$ is the event that we see heads in the $n$'th throw. Then $$ E[V_{\text{HH}}]=\frac{1+p}{p^2},\\ E[V_{\text{TH}}]=\frac{1}{pq}. $$ Putting $p=q=\tfrac{1}{2}$ we see that if our coin is a fair coin then $E[V_{\text{HH}}]=6$ and $E[V_{\text{TH}}]=4$.