Solution for 2 variable recurrence for a problem similar to gambler's ruin [closed]

Question

Let $p_A,p_B$ with $p_A+p_B=1$, $p_A \geq p_B$ be the probabilities of a biased coin flip. Player $A$ gets 1 point if the coin flip gives heads, $B$ gets 1 if tails. The player who reaches $N$ points first wins.

I believe this is a problem similar to the gambler's ruin, but with a recurrence relation given by $$\boxed{P(i,j) = p_A P(i+1,j) + p_B P(i,j+1)}$$ where $P(i,j)$ is the probability $A$ wins if $A$ has $i$ points and $B$ has $j$ points. I want to calculate $P(0,0)$ (i.e. the probability $A$ wins in when the game begins), and my boundary conditions are (I think!) $$P(N,j) = 1\quad \forall j\in\{0,\ldots,N-1\}$$ (i.e. $A$ has) and $$P(i,N) = 0\quad\forall i\in\{0,\ldots,N-1\}.$$

I tried to solve the problem by brute force for some small $N$ cases and got the expression $$P(0,0) = p_A^N \sum_{k=N-1}^{2n-2} \binom{k}{N-1} p_B^{k-N+1}$$ which I think is correct.

I want to solve the recurrence relation and prove that this formula is correct, but don't know how to properly solve that equation.

Answer 1

See Example 8d of Section 4.8 of "A 1st course in Probability" by Sheldon Ross.

Here it is:

If independent trials, each resulting in a success with probability $p$, are performed, what is the probability of $r$ successes occurring before $m$ failures?

Solution. The solution will be arrived at by noting that $r$ successes will occur before $m$ failures if and only if the $r$th success occurs no later than the $(r + m − 1)$th trial. This follows because if the $r$th success occurs before or at the $(r + m − 1)$th trial, then it must have occurred before the $m$th failure, and conversely. Hence, from Equation (8.2) (that's the explicit form of the Negative Binomial distribution), the desired probability is $$ \sum_{n=r}^{r+m-1} \binom{n-1}{r-1}p^r(1-p)^{n-r}. $$