Two players $P_1$ and $P_2$ are playing the final of a chess championship,which consists of a series of matches.Probability of $P_1$ winning a match is $\frac{2}{3}$ and that of $P_2$ is $\frac{1}{3}$.The winner will be the one who is ahead by two games as compared to the other player and wins atleast $6$ games.Now if the player $P_2$ wins first 4 matches,prove that the probability of $P_1$ winning the championship is $\frac{1088}{3645}$
I have couple of doubts in this question after following the solution given by "Aretino" .
Find the probability of $P_1$ winning the championship
Its assumed that After a Parity occurs, Probability of winning $P1$ is $p_0$.
But there are different parity situations right?
For example we have
$P_2 P_2 P_2P_2 P_2 P_1P_1P_1P_1P_1$ also a parity situation.why cant we consider this case?
am i missing something here?