Find the probability of $P_1$ winning the championship 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 dont have thoughts as to how to tackle this problem.Because a player has to just ahead of 2 games as compared to another player,neither more than 2 nor less than 2.Please help me.
 A: Let $p=2/3$ and $q=1/3$ the single match winning probabilities of $P_1$ and $P_2$. For $P_1$ not to lose, he has to win at least $5$ out of the next $6$ matches: if he wins $6$ matches (prob. $p^6$) he wins the championship, if he wins $5$ matches (prob. $6p^5q$) the two players will be on parity.
Let $p_0$ be the probability $P_1$ to win the championship starting from a parity situation. It is easy to see that:
$$
p_0=p^2+2pq\cdot p_0,
$$
because to win the championship $P_1$ must win the following two matches, while in case he wins only one, parity will ensue again. From that one can solve for $p_0$:
$$
p_0={p^2\over1-2pq},
$$
so that the probability for $P_1$ to win is:
$$
p^6+6p^5q\cdot p_0=p^6\left(1+6{pq\over 1-2pq} \right)={1088\over3645}.
$$
A: Calling $P_1's$ probability  of winning $p$, and $P_2's$, $q$,  the earliest win for $P_1$ will be at $6-4$ for which she will have to firstly win 4 games, with probability $p^4$.
Now draw a $2D$ lattice path with further wins plotted on the lattice.
For $P_1$ to get 2 ahead at any stage, you will find that it follows the pattern
$p^2 + 2p^3q + 4p^4q^2 + 8p^5q^3 + 16p^6q^4 + ......$
which is an infinite G.P. with $a= p^2, r = 2pq,$ sum = $\dfrac{p^2}{1-2pq}$
Thus the probability that $(P_1$ wins) = $\dfrac {p^6}{1 -2pq}$ 
Something seems wrong, though, because putting $p = 2/3, q = 1/3,$ I get $Pr = \dfrac{64}{405}$
ps:
Aretino has found the flaw. I should have let $P_2$ win $1$ more game before declaring "deuce" !
