I recently encountered the following riddle:
Let a fair coin be flipped $N > 2$ times. Player A gets a point every time there are two heads in a row, and player B gets a point every time a tail is followed by a head, the player with more points wins. Thus, for example, for the sequence THHH, player A wins with 2 vs. 1 points while in the sequence THHTHH, neither wins, as they both score two points.
Clearly, the expected number of points is equal, as the sequences HH and TH are both equally likely. Now the riddle itself:
Are player A's chances of winning higher, lower or equal to that of player B?
I found it surprising that player A's chances of winning are lower than that of player B's.
Though there are intuitive explanations, I'd like to rigorously prove this.
This is not a complete answer, but a partial attempt too long for a comment. I hope someone can turn it into a full answer, as I have to give up thinking about it for the rest of the day:
The natural sample space is $\Omega:=\{H,T\}^{N}$, equipped with the counting measure. For convenience we assume $N$ is even. Now define, for $i\in\{1,\dots,N-1\}$ the random variables $X_i^{HH}$ and $X_i^{TH}$ $$X_i^{HH}(\omega)=\begin{cases} 1 & \omega_i=H=\omega_{i+1} \\ 0 &\text{otherwise}\end{cases},~ \text{and}\quad X_i^{TH}(\omega)=\begin{cases} 1 & \omega_i=T,~\text{and}~\omega_{i+1}=H \\ 0 &\text{otherwise}\end{cases}.$$ Thus the scores of players $A$ and $B$ are given by the random variables $$X_A=\sum_{i=1}^{N-1}X_i^{HH},~\text{and}~X_B=\sum_{i=1}^{N-1}X_i^{TH}.$$ As you noted, linearity of expectation and independence of different coin flips means that $\mathbb E(X_A)=\mathbb E(X_B)$. We can now use the fact that $$\mathbb E(X_A-X_B)=\sum_{\omega\in\Omega}(X_A-X_B)(\omega)\frac{1}{2^N}$$ to see that $$\sum_{\substack{\omega\in \Omega\\X_A(\omega)>X_B(\omega)}}(X_A-X_B)(\omega)=\sum_{\substack{\omega\in \Omega\\X_B(\omega)>X_A(\omega)}}(X_B-X_A)(\omega)$$ However, $\max_\Omega(X_A)=N-1$, but $\max_\Omega(X_B)=N/2$, so we can rewrite $$\sum_{n=1}^{N-1}n\mathbb P\{X_A-X_B=n\}=\sum_{n=1}^{N/2}n\mathbb P\{X_B-X_A=n\},$$ where every term in the sums are strictly positive. Given this, the only way that we could have $\mathbb P\{X_A>X_B\}>\mathbb P\{X_B>X_A\}$ is if, for enough $n$ close enough to $N/2$, $\mathbb P\{X_B-X_A=n\}$ is sufficiently larger than $\mathbb P\{X_A-X_B=n\}$. However, we know that $\mathbb P\{X_B-X_A=N/2\}=\frac{1}{2^N}$, and by a quick inspection $\mathbb P\{X_A-X_B=N/2\}\geq\frac{N}{2^N}$, which suggests that this is very unlikely (in the intuitive, not probabilistic, meaning of the word). Thus what is left to do is to find some suitable bounds on $\mathbb P\{X_B-X_A=n\}$ and $\mathbb P\{X_A-X_B=n\}$ in terms of $n$.