A and B play until one has 2 more points than the other... Question
A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A will win?
My attempt for the solution:
what is the probability they will play a total of 2n points?
For the first question, A and B will be "exchanging wins" until $|A-B| = 2$, if one wins twice in a row then its terminal state. Also, winning guarantees even total points,
$A = B + 2\\
A+B=B+B+2=2B+2$
A win always means even points then, making the job simpler. Probability prior to double winning streak shown in the following,

Let $E_{i}$ be a set of game from start till achieving $2*i$ points.

$P(E_{1}, E_{2}, \cdots, E_{n-1}) = P(E_{1}) + P(E_{2}) + \cdots + P(E_{n-1})\\
P(E_{1}, E_{2}, \cdots, E_{n-1}) = 1 + p*(p-1) + \cdots + (p*(p-1))^{2i-2}\\
P(E_{1}, E_{2}, \cdots, E_{n-1}) = \frac{1}{1-(p*(p-1))^{2}} - (p*(p-1))^{2i-1} - (p*(p-1))^{2i}$
*Note that $P(E_{1}) = P(\emptyset)$ since this is not the two winning streak.
The winning streaks can be a union of either $A$ wins twice in a row or $B$ wins twice in a row:

Let $S$ be the probability of 2n final points.

$P(S) = P(E_{1}, E_{2}, \cdots, E_{n-1})*p^{2} + P(E_{1}, E_{2}, \cdots, E_{n-1})*(p-1)^{2}$

What is the probability that A will win?
I didn't really understand this question, does it mean from a tie points to a double winning streak of A? Anyway, in that case, I just use the result from first question, 
$P(A_{win}) = P(E_{1}, E_{2}, \cdots, E_{n-1})*p^{2}$
 A: The probability $a$ that A (ultimately) wins is easy to compute. She wins if she wins the first two games, or if the players are tied after $2$ games, but A ultimately wins. Thus, conditioning on the outcome of the first two games, we have
$$a=p^2+2p(1-p)a.$$
Solve this linear equation for $a$. 
For the probability the game lasts for a total of $2n$ points, we need that there is a tie at $2n-2$ points, and no one has won by then, and then one of the two players gets $2$ in a row.
For tie at $2n-2$ points, with no one having won, we need $n-1$ occurrences of the pattern AB or BA (we can "mix" these).  This sort of pattern has probability $2p(1-p)$, so the required probability is
$$[2p(1-p)]^{n-1}(p^2+(1-p)^2).$$
Remark: We can alternately use the formula for winning in $2n$ games to find A's probability of winning. Her probability of winning in $2$ is $p^2$. Her probability of winning in $4$ is $(2p-2p^2)p^2$. Her probability of winning in $6$ is $(2p-2p^2)^2p^2$. And so on. So her probability of winning is
$$p^2+p^2(2p-2p^2)+p^2(2p-2p^2)^2+p^2(2p-2p^2)^3+\cdots.$$
This is an infinite geometric series, first term $p^2$, common ratio $2p-2p^2$. Now use the formula for the sum of such a series.
More complicated for sure than the simple linear equation approach we gave in the answer!
A: If the players are currently tied, and then one wins twice in a row, that player wins and the game ends. If one player is currently leading, then the other player has to win the next game to reach a tie. Therefore, in order for the result of the first $2n - 2$ games to be a tie, the wins have to be in the form of $n - 1$ pairs, each of the form $AB$ or $BA$ (for example: $ABBABAAB$). Thus, the wins are in the pattern $\underbrace{(P_1)(P_2)\ldots(P_n)}_{n-1}$, where each pair $P_i$ is either $AB$ or $BA$ (independently of the other pairs). If $q = 1 - p$, each of the pairs $AB$ and $BA$ has probability $pq$ (for $qp = pq$). As there are two possibilities for each of the $n - 1$ pairs $P_i$, there are $2^{n-1}$ possible configurations. Thus the probability of a tie after $2n - 2$ games is $2^{n-1}(pq)^{n-1}$.
Now, the last two wins can be either $AA$ with probability $p^2$, or $BB$ with probability $q^2$. So the probability of there being exactly $n$ games is $\boxed{2^{n-1}(pq)^{n-1}(p^2 + q^2)}$.
If $A$ wins, then the only possibility for the last two wins is $AA$, so the probability that $A$ wins in $2n$ games is $2^{n-1}(pq)^{n-1}p^2$.
The probability that $A$ wins after an indefinite number of games can be obtained by summing the probabilities for $n = 1, 2, \ldots$
$\displaystyle\sum\limits_{n=1}^{\infty}(2pq)^{n-1}p^2 = \boxed{\dfrac{p^2}{1 - 2pq}}$.
