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!

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.
