What is the probability that a team (either A or B) wins the series in exactly 6 games, given that one team (either A or B) won both the first and second game?
Prob (A winning an individual game)= .68
Prob (B winning an individual game) = .32
Let's look at from the point of view of one of the teams whose probability of winning is P and the probablity of losing is P'. In order to meet your requirements, then the team must either a) win the first two games, win only one of the next two, then win the sixth game or b) lose the first two, then win the next four.
a) Prob of winning the first two = P^2
Prob of winning only one of the next three = 3 P P'^2
Prob of winning the last game = P
-- total probability of this scenario = 3 P^4 P'^2
b) Prob of losing two then winning four = P'^2 P^4
Compute for team A = 3 (.68)^4 (.32)^2 + (.32)^2 (.68)^4 = 8.76%
Compute for team B = 3 (.32)^4 (.68)^2 + (.68)^2 (.32)^4 = 1.94%
Total = 10.7%