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 can see that the upvoted answer is incorrect to the answer b, so I give a way to solve this type of problem recursively.
Let us call $p_A$ probability that team A wins  and $q=1-p$. First, $p_A \neq \frac{p^2}{p^1 + (1-p)^2}$.
In fact, A can win at first two stages, or B one one of the two games. In the latter case, the game restart. Translated mathematically :
$$
p_A = p^2 + 2pq \cdot p_A
$$
Solve it for $p_A$ we can find as in other correct answers : $p_A = \frac{p^2}{1 - 2pq}$
