Problem statement:
Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.
It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.
My approach:
First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.
Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $\binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = \binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = \binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Hence, the final probability: $$ p' = p_{(a, a - b)} + p_{(a - b, a)} = \binom{2a - b}{a} (1-p)^{a} p^{a - b} + \binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$
Is my approach correct? I've tried to simulate the game in Python and got different result ($\approx 0.06$ by the simulation vs. $\approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).
It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?
Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).