Let $W = \text{We get a Y}$.

Let $D = \text{The game is decided after the first two tosses}$

The crucial observation here is that if we don't finish after the first two tosses, we are facing the same game again.

Therefore, $P(W/D^c) = P(W) \implies P(W/D) = P(W)$.

This gives us that $P(W) = P(W \cap D)/P(D) = \frac{1/4}{1/4 + 2/4} = 0.33$