I'm trying to solve this problem. The accepted answer is correct, but I'm not sure why my answer differs from it.
I first use the same approach and compute the probability of winning from parity (following the same notation as in the anwswer):
$$p_0=p^2+2pq\cdot p_0$$
I get $p_0 = \frac{4}{5}$.
There are two ways for the tournament to reach parity: $P_1$ to win first four games in a row in the next six games (the probability of which is $p^6$), or $P_2$ to win one game out of the next six games (the probability of which is $6p^5q$).
Then my answer would be
$$[p^4 + 6 p^5 q] p_0 =\frac{448}{1215}$$
My answer turns out to be different. Can I get a hint on what went wrong?
Thank you.
Your calculation $p_0 = 4/5$ is correct. The subsequent calculation is not, because the parity situation for which $p_0 = 4/5$ corresponds to the state of the tournament when both players each have five wins, not four. The reason why we choose this point to calculate the probability of winning from parity, is that we require at least six wins from a player in order to stop the tournament.
To illustrate more concretely, let the ordered pair $(P_1, P_2)$ represent the number of wins for each player. So $P_1 + P_2 = n$ is the number of games played so far. The starting state for the calculation is $(P_1, P_2) = (0,4)$. We enumerate a table of potential ending states for the tournament as follows:
$$\begin{array}{c|cccccccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \cdots \\ \hline 4 & (0,4) & (1,4) & (2,4) & (3,4) & (4,4) & (5,4) & \color{red}{(6,4)} & & & \\ 5 & (0,5) & (1,5) & (2,5) & (3,5) & (4,5) & (5,5) & (6,5) & \color{red}{(7,5)} & & \\ 6 & \color{blue}{(0,6)} & \color{blue}{(1,6)} & \color{blue}{(2,6)} & \color{blue}{(3,6)} & \color{blue}{(4,6)} & (5,6) & (6,6) & (7,6) & \color{red}{(8,6)} & \\ 7 & & & & & & \color{blue}{(5,7)} & (6,7) & (7,7) & (8,7) & \cdots\\ 8 & & & & & & & \color{blue}{(6,8)} & (7,8) & (8,8) & \cdots \\ \vdots & & & & & & & & \vdots & \vdots & \ddots \end{array}$$
Note we have omitted subsequent rows and columns as the pattern should be obvious. States in red correspond to victory for $P_1$, and states in blue are victory for $P_2$. States in black are unresolved. Cells that are empty are unattainable because the tournament ends before such a state can be reached. Each game played corresponds to a movement from the upper left cell $(0,4)$ either horizontally to the right (win for $P_1$) or vertically down (win for $P_2$).
As you can see from the fact that the only cell in the table from which the tournament can continue beyond $n = 10$ games played is $(5,5)$, as this is the only black cell in the diagonal with $P_1 + P_2 = 10$. So this is the proper point of reference for the parity state.
Why can't we use $(4,4)$ as the parity state? The reason is because the state $(4,4)$ can only be reached in one way: four consecutive wins by $P_1$ from $(0,4)$. But in fact, we can reach parity at $5$ wins each in other ways; e.g., $(0,4) \to (1,5) \to (2,5) \to (3,5) \to (4,5) \to (5,5)$ and this is not counted in the path that assumes we go straight from $(0,4)$ to $(4,4)$. So by taking $(4,4)$ to be the parity state, we are not accounting for the cases where $P_2$ wins exactly one more time before $P_1$ gets to $4$ wins.