HH vs. THTH Coin Toss Alice tosses a fair coin until either she or Bob wins. Alice wins if she gets two consecutive heads (HH) and Bob wins if he gets Tails and Heads twice in a row (THTH). What is the probability that Bob wins the game?
Attempt: If the first toss is T, there will certainly be a TH before HH. If the first toss is H, the only way to get HH before TH is if the second toss is also H. Thus, the probability of getting TH before HH is 3/4. Given any sequence with TH before HH, there are three possibilities for the tosses following that TH:

*

*H, in which case Alice wins

*TT, in which case Alice will keep tossing

*TH, in which case Bob wins.

Hence, the probability that Bob wins must be greater than 3/4 * 1/4 = 3/16. But what is the exact probability? Thanks in advance!

Your work almost gets there.
Let's suppose $b$ is the probability that Bob wins given that nobody has won yet and the last two tosses were $TH$.  If the next two are $TT$ then neither wins immediately and eventually there will be another $H$ bringing the game back to $TH$.   So you have said in effect that $$b= \frac12\times 0 + \frac14 \times b +\frac14 \times 1$$ which has the solution $b=\dfrac13$.
But you also said Alice could have won at the start too with $HH$ before the first $T$ with probability $\dfrac14$, so overall the probability Bob wins is $$\frac14\times 0+\frac34 \times b = \frac14.$$
