HHT and HTH in tossing a coin

An informal argument as to why HHT is more likely:

After starting the game, we keep tossing the coin until we see a head. The next toss puts one of the players closer to winning, H for you and T for me. Suppose it’s the latter. If the next toss is H, then I win, otherwise the game starts over and we’re both effectively two steps away from winning again. On the other hand, if we had HH, when the next toss is an H the game remains in the same state: you’re still one step closer to winning than I am. So, whenever a non-winning coin toss comes up that doesn’t favor you, you never lose any ground, whereas I get put back to square one whenever the coin toss goes against me. “On average,” I’m usually two steps from winning, but you’re only one step away. I would even hazard a guess that you’re twice as likely to win this game.

This is borne out by a calculation. The game can be modeled by an absorbing Markov chain with transition matrix $$P = \begin{bmatrix}\frac12&\frac12&0&0&0&0\\0&0&\frac12&\frac12&0&0\\0&0&\frac12&0&\frac12&0\\\frac12&0&0&0&0&\frac12\\0&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix}$$ with state 5 representing the HHT win and state 6 the HTH win. The absorption probabilities work out to be $2/3$ and $1/3$, respectively.