Probability to win. Two players $A$ and $B$ are playing the final of chess championship which contains a series of matches. 
The probability that $A$ wins is ${2\over 3}$.
The probability that $B$ wins is ${1\over 3}$.
The winner will be the one who is 
ahead by 2 games as compared to the
other and wins at least 6 games.
If the player $B$ wins the first 4 matches, 
find the probability that $A$ wins the championship.
My attempt:
We should find the following probabilities:
$A$ wins 6 matches given $B$ has won the first 4.
$A$ wins 7 matches given $B$ has won the 
first 4 and wins one out of the next 8.
But I am not getting any patterns.
 A: Go ahead and play another $6$ games, even if the ultimate winner is decided before all six games have been played.  If player A wins all six of these games, which happens with probability $(2/3)^6$, he is the champion at that point (having won two more games than player B's four wins).  If player A wins five of the six, which happens with probability $6(2/3)^5(1/3)$, the contest at this point is a tie, to which we'll return.  Finally, if player A wins four or fewer of the next six games, then player B is the champion (because she will have won at least two games, bringer her to six wins, which will be at least two more than player A).
So let's see what player A's chances become when starting from a tie.  Call this $p$.  We know we have to play at least two more games.  Either player A wins them both, which happens with probability $(2/3)^2$, or they split wins, leaving them tied again, which happens with probability $2(2/3)(1/3)$, or else A loses.  Thus
$$p=\left(2\over3\right)^2+2\left(2\over3\right)\left(1\over3\right)p={4\over9}(1+p)$$
from which we find $p=4/5$.  In total, the probability that player A wins the championship is
$$\left(2\over3\right)^6+6\left(2\over3\right)^5\left(1\over3\right)\left(4\over5\right)=\left(2\over3\right)^5{34\over15}={1088\over3645}\approx0.2985$$
Remark:  A key point to understand in this approach is that allowing "meaningless" games to be played does not affect the answer, but simplifies some of the analysis.
A: You can make a Markov chain with states $0-4,1-4,2-4,3-4,0-5,1-5,2-5,3-5,0,-1,+1$ where $0$ represents a tie in the number of games with each having won at least $4$ so two wins will win the match.  $+1$ is the case that B has won one more game and has won at least $5$ and $-1$ is the case that A has won one more game and has won at least $5$.  You can start to find B's chance to win from $0,-1,+1$ by the usual technique, then work backwards for the other cases.
A: First consider the next 5 games, for B to fail to win, A must win 4 or more of the next 5 games, leaving the A-B score at 5-4 or 4-5
ok, so you should be able to work out the probability of after 5 games
1) A wins all to lead 5-4
2) A wins 4 of 5 to trail 4-5
3) the compliment of 1 & 2 is that neither of those happened, i.e. P3  = 1 - P1 - P2 and in that case B won - but you don't need all the combinations, 1 & 2 are easier to calculate
can you calculate 1 & 2?
so then the 6 win rule ceases to matter, let's say A is winning 5-4, we might as well call that +1 
let's say what are the chances of A winning at -1, 0 and +1
P(-1) = (2/3)P(0) + 0 = (2/3)P(0) 
A has a 2/3 chance of getting to a situation 0, and a 1/3 chance of losing completely
P(0) = (2/3)P(1) + (1/3)P(-1) 
P(1) = 2/3 + (1/3)P(0)
so then you can change this all into terms of P(0) by substitutiong into the middle expression - you can then get P(0), P(-1) and P(1)
you can then multiply P(-1) and P(1) by the probabilities we worked out for the lead of 1 or -1 after 5 games
