What is the probability that $HH$ occurs before $TH$ in an infinte sequence of coin flips? This is one of the questions of a set of exam review questions that don't have solutions to them. I can't get my head around this but it seems so simple.

By flipping a fair coin repeatedly and independently, we obtain a sequence of
  H's and T's. We stop flipping the coin as soon as the sequence contains either HH or TH.
  Two players play a game, in which Player 1 wins if the last two symbols in the sequence
  are HH. Otherwise, the last two symbols in the sequence are TH, in which case Player 2
  wins.
A = "Player 1 wins"
  and
  B = "Player 2 wins."
Determine Pr(A) and Pr(B)


The above answers are right.
The key point to understand here is that Player B wins for TH.
So the only chance for Player A to win is through HH in the first two flips.
Once you get a T it'll nullify Player A's chances of winning because you can never get 2 H's again. The game will become whether Player B wins or not.
So if we consider the unrolling of the probability tree for the first two levels,
we see that: 
P(HH occurring before TH) = P(HH in first two flips) = P(A winning) = 1/4 
P(TH occuring before HH) = P(B winning) = 3/4 
I too was initially confused with this problem but the above answers helped me articulate the solution.
