Two players, 1 and 2, are playing a chess game which consists of a series of matches. The probability that Player 1 wins is $2/3$, and that Player 2 wins is $1/3$. The winner will be the one who is ahead of the other by 2 games and has won at least 6 games.
If Player 2 wins the first four matches, find the probability that Player 1 wins the championship.
The answer is given to be $1088/3645$.
Player $1$ wins by winning the next $6$ games, or by winning $5$ out of the next $6$ games and then winning a pair of games before Player $2$ wins a pair of games (where we don't have to worry about any series of alternating wins in between since these just extend the match without changing the winning probabilities).
The probability to win $6$ games is $\left(\frac23\right)^6=\frac{64}{729}$. The probability to win $5$ out of $6$ games is $6\cdot\frac13\cdot\left(\frac23\right)^5=\frac{64}{243}$. The probability to win a pair of games before Player $2$ wins a pair of games is
$$ \frac{\left(\frac23\right)^2}{\left(\frac23\right)^2+\left(\frac13\right)^2}=\frac45\;. $$
Thus the total probability is
$$ \frac{64}{729}+\frac{64}{243}\cdot\frac45=\frac{1088}{3645}\;. $$