This answer addresses the question as originally posted; I made it a community wiki because it is really an extended comment that would not have fit as a comment.
The previous question asked, inter alia, the following: given a two-sided coin with probability $p$ of showing heads, and the coin is tossed until either HTHTH or HTHH appears, how to compute the probability that the pattern we got was HTHTH.
Mike Spivey's answer cast the problem as a two-player game, in which one player wins if the first pattern to show up is HTHTH, and the second player wins if the pattern that shows up first is HTHH, and showed how to obtain the probability that either the first player wins, or the second player wins.
The current problem has a fair coin, and asks that the coin be tossed until a particular pattern, HTHT, shows up, and then asks what is the probability that another pattern, TTH, actually occurs before HTHT shows up.
Think of this problem as a two-player game again: the coin is tossed, until either HTHT or TTH show up. The first player wins if HTHT shows up first, the second if TTH shows up first. The exact same approach as used in the previous question works here to obtain the probability that the second player wins, i.e., the probability that TTH will show up first.
Now, you state in comments that you don't see this problem as the same as that game, because in this problem you keep tossing until HTHT appears, instead of stopping if TTH shows up. Well, take the game described above, and then imagine that the two players are utterly bored and so, if player two wins, they keep tossing the coin until HTHT shows up. Will that change the probability that the second player wins? No. The second player still wins if TTH shows up first, regardless of whether you keep tossing the coin or not after he wins. So the probability that TTH showed up at least once before you hit HTHT is the same, whether you keep tossing after TTH shows up or not, because the number of sequences in which TTH shows up but HTHT never does is negligible, so it will not affect the probability. (The probability that a particular finite pattern never shows up is $0$).
So in the end, the answer to your question is still "the probability that TTH showed up before HTHT shows up is equal to the probability that Player Two wins the game".