Suppose, a snokker match is best of $2n-1$, so the player who wins $n$ frames wins the match.
Suppose, the probability for winning a frame is $p$ for player $1$.
- What is the probability (denoted by $q(n,p)$ ) that player $1$ wins the match ?
Trivial cases are $$n=1\ \ \ (q(1,p)=p)$$ $$p=0\ \ \ (q(n,p)=0\ for\ all\ n)$$ $$p=1\ \ \ (q(n,p)=1\ for\ all\ n)$$ $$p=\frac{1}{2}\ \ \ (q(n,p)=\frac{1}{2}\ for\ all\ n\ due\ to\ symmetry)$$
My idea is a recursive approach : If both players have won $n-1$ frames, the probability that player $1$ wins is $p$ and so on. But I do not know how to do this for arbitary many frames.