A and B play a series of games. Find the probability that a total of 4 games are played. A and B play a series of games. Each game is independently
won by A with probability $p$ and by B with probability $1 - p$. They stop when the
total number of wins of one of the players is two greater than that of the other player.
The player with the greater number of total wins is declared the winner of the series.  
(a) Find the probability that a total of 4 games are played.
(b) Find the probability that A is the winner of the series.  
I though I could do this using conditional probability and independent trials/bernoulli trials but I am really confused.  
let's say I want to find the probability that A wins three games and B wins 1 game and add that to the probability of B wins 3 games and A wins 1 game, but I am not sure of how to do that because B has to win at least 1 game before A wins 2.
If I set $n$ = number of games played would the sample space equal $p^n * (1-p)^n$?
How do I solve this? is there an easier way?

I would consider two games at a time.  Two games can result in AA (team A wins, probability $p^2$) or BB (team B wins, probability $(1-p)^2$) or AB, BA considered together (match is back to starting state, with probability $2p(1-p)$).
Then the probability that the match goes for four games (two pairs) is $(2p(1-p))(p^2+(1-p)^2)$, occurring when the first pair is split and the second pair is swept.
The probability that team A wins the match is $\frac{p^2}{p^2+(1-p)^2}$ since that is the ratio of team A winning to team B winning in any given round of two games.
