How to figure out the relation of probability?

Question

There is a championship between $A$ and $B$. The champion will be the first to win $n$ matches in a series of $2n-1$ matches. For any given match there is a fixed probability $p$ that $A$ will win, and hence a probability $q=1-p$ that $B$ will win. Let $P_{ij}$ be the probability that $A$ will win the series given that they still need $i$ more victories, whereas $B$ needs $j$ more victories for the championship. $P_{0j}=1,1≤j≤n$, because $A$ needs no more victories to win. $P_{i0}=0,1≤i≤n$, as $A$ cannot possibly win if $B$ already has.
The professor asked me to figure out what is the relationship between $P_{ij}, p$ and $q$. The answer is $P_{ij}=pP_{i-1,j}+qP_{i,j-1}$. Can anyone tell me how to get this answer?

Answer 1

You are at the state where $A$ needs $i$ more wins and $B$ needs $j$ more wins. The chance that $A$ wins the whole championship is $P_{i,j}$. Now you play the next game. With probability $p\ A$ wins the game, leaving him needing $i-1$ more to win overall. $A$'s chance of overall victory is now $P_{i-1j}.$ With probability $q\ B$ wins the game, making $A$'s chance of winning $P_{ij-1}$. We must have that $P_ij$ is the mix of those, weighted by the chance $A$ wins the game, leading to your equation $P_{ij}=pP_{i-1j}+qP_{ij-1}$

Answer 2

Well the probability that A will win the series given they still need $i$ more victories while B still needs $j$ victories, i.e., $P_{i,j}$ is $$ (some probability)\times P_{i-1,j} + (some other probability) \times P_{i,j-1} $$ since you can get to that stage only by either (a) A winning from the state with probability $P_{i-1,j}$; or (b) B winning from the state with probability $P_{i,j-1}$. Now fill in the gaps. Also, think of going to $P_{1,j}$ and $P_{i,1}$ in the same vein to complete the argument.