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?
 A: The first question is best done by enumeration, as suggested in the comments.
To handle the second one:
Let's introduce the relevant states of the game.  We will index these by the number of games by which $A$ is ahead.  Thus we have $5$ states: $(0),(1),(-1),Win,Loss$.  Let $p(state)$ denote the probability that $A$ will win if the teams are in the given state.  Thus the answer we seek is $p(0)$.  Obviously $p(Win)=1$ and $p(Loss)=0$.  By considering the possible outcomes of the next game, we get some basic relations between the probabilities as: $$p(0)=p(1)p+p(-1)(1-p)$$ $$p(1)=1^*p+p(0)(1-p)$$  $$p(-1)=p(0)p+0^*(1-p)$$  This system is easily solved to get $$p(0)=\frac {p^2}{1-2p(1-p)}$$  As a sanity check observe that this correctly gives $\frac 12$ if $p=\frac 12$.
A: The first is easy.
To handle the second, as you remarked, the game can go on indefinitely, so here's another way to solve using the sum of an infinite G.P. For compactness, I will call chances of $B$ winning a game $q$ instead of $(1-p)$
The earliest win for $A$ will be at $2-0$ for with probability $p^2$, but if they reach $1-1$, it will become like "deuce" in tennis, and A will need to create a lead of $2$.
Now draw a $2D$ lattice path with paths to reach possible $A$ wins counted on the lattice.
For $A$ to get 2 ahead before B does at any stage, you will find that it follows the pattern
$p^2(1+2pq+4p^2q^2+8p^3q^3+16p^4q^4+......)$
which is an infinite G.P. with $a=p^2,r=2pq$, sum $= \dfrac{a}{1−r}$
Thus the probability that ($A$ wins) $= \dfrac{p^2}{1−2pq}$
A: 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.
A: 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 : $p_A = \frac{p^2}{1 - 2pq}$
