HHT and HTH in tossing a coin A coin is flipped infinitely until you or I win. If at any point, the last three tosses in the sequence are $HHT$, I win. If at any point, the last three tosses in the sequence are $HTH$, you win. Which sequence is more likely?
Unfortunately, this configuration does not seem like the ones as "$HHT$ versus $THH$" (where clearly only $HHT$ wins iff the first two occuring $H$ are consecutive). Of course, here we can still assume that $TT$ does not occur (as after such a thing the game restarts), but it does not seem to help me enough.
Any help appreciated!
 A: 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.
A: Hm, I think I actually have an answer. Assuming there is no TT, consider the first occurence of HTH and suppose it is winning. Then just before it we can't have HH (else HHHTH has HHT in the beginning), we can't have TT by above, we can't have HT (else we get TT) and if we have TH, then in THHTH we have HHT in the middle. So HTH to win it must occur among the first four tosses. The possible ones are HTH, THTH, with total probability $\frac{1}{8} + \frac{1}{16} = \frac{3}{16}$. So HHT to be winning has $\frac{13}{16}$ probability.
A: (1) If you started by $HH$ then the 1st person wins.
(2) If you started by $HT$ you have 2 equal probabilties that the game restarts (or it is a tie) if you got a $T$, and the 2nd wins if you got an $H$
(3) If you started by $TH$ then you have equal probabilities that the 1st person wins if you got an $H$, and if you got a $T$ then part (2) happens.
(4) If you started by $TT$ then the game restarts (or it is a tie).
Now you may have completed your tree diagram
