Keep track of each player's "state" after each game with two numbers. Let $r_i$ be the length of player $i$'s current run of wins, and let $g_i$ be player $i$'s number of games won. Let the state of the match be $(r_1,g_1,r_2,g_2)$. Initially, the match is in state $(0,0,0,0)$.
Let $w(a,b,c,d)$ be the generating function for the probability that a match that has reached state $(a,b,c,d)$ ends after a total of $i$ games. In other words, let $w(a,b,c,d)=\sum_i p_i x^i$, where $p_i$ is the probability that a match that has reached state $(a,b,c,d)$ ends after exactly $i$ games. Note that if $a=3$, $b=4$, $c=3$, or $d=4$, the match is over and took exactly $b+d$ games. In terms of the generating function, this means that $w(a,b,c,d)=x^{b+d}$.
Now suppose the state of a particular match is $(a,b,c,d)$.
If player 1 wins the next game, the state moves from $(a,b,c,d)$ to $(a+1,b+1,0,d)$.
If player 2 wins the next game, the state moves from $(a,b,c,d)$ to $(0,b,c+1,d+1)$.
If $p$ is the probability of player 1 winning a single game, the state of an ongoing match moves from state $(a,b,c,d)$ to either state $w(a+1,b+1,0,d)$ (with probability $p$) or state (with probability $1-p$). This gives a recurrence for the generating function. $w(a,b,c,d) = p w(a+1,b+1,0,d) + (1-p)w(0,b,c+1,d+1)$.
The generating function for the match-length probabilities of a fresh match is $w(0,0,0,0)$.
The recurrence and initial conditions fully characterize the generating function $w$.
- $w(a,b,c,d)=x^{b+d}$ if $a=3$, $b=4$, $c=3$, or $d=4$
- $w(a,b,c,d) = pw(a+1,b+1,0,d) + (1-p)w(0,b,c+1,d+1)$
$w(0,0,0,0)$ can be worked out by hand, but it's much easier to enlist the help of Mathematica or a similar tool.
$w(0,0,0,0)=\\(1-3 p+3 p^2) x^3+(p-3 p^2+4 p^3-2 p^4) x^4+(2 p-7 p^2+10 p^3-5 p^4) x^5\\+(7 p^2-28 p^3+49 p^4-42 p^5+14 p^6) x^6+(14 p^3-42 p^4+42 p^5-14 p^6) x^7$
If $p=\frac{1}{2}$, $w(0,0,0,0)=\frac{1}{4}x^3+\frac{1}{8}x^4+\frac{3 }{16}x^5+\frac{7 }{32}x^6+\frac{7}{32} x^7$.
The answer to many questions about this game can be deduced from $w$. For example, the probability that the match takes exactly $6$ games is the coefficient of $x^6$, $7 p^2-28 p^3+49 p^4-42 p^5+14$, which equals the value satish found, $7(p^4(1-p)^2 +p^2(1-p)^4)$.
If you want to answer questions about a particular player, you can use two variables in the generating function. Let the coefficient of $x^i$ represent the probability that player 1 wins and the match takes exactly $i$ games, and let the coefficient of $y^i$ represent the probability that player 2 wins and the match takes exactly $i$ games. This changes two of the initial condition values to $y^{b+d}$, but the recurrence is the same.
I feel like the chess players should have real names... Archibald and Bartholomew, perhaps? Anyway...
a) A gets $3$ wins, $2$ draws, $2$ losses
This is a question similar to "how many anagrams of "REFEREE" are there?" - the answer is to divide the total arrangements by the arrangements of each repeated element, so
$$N_a = \frac{7!}{3!2!2!} = \frac{5040}{24} = 210$$
We agree :-) - if you calculate out your binomials as factorials, you'll see they cancel down to the same form.
b) A gets exactly $4$ points
Of course we can calculate this by taking cases as above; let's try that:
$$N_b = \frac{7!}{4!0!3!}+\frac{7!}{3!2!2!}+\frac{7!}{2!4!1!}+\frac{7!}{1!6!0!} = 35+210+105+7 = 357$$
c) A gets exactly $4$ points and doesn't lose game $7$
As you can see I've interpreted the condition; basically our final answer can be obtaining by adding the number of ways, after $6$ games, to get to $3$ points, and to get to $3\frac 12$ points - followed by a win and a tie respectively.
Let's just see what that looks like...
$$\begin{align}N_c &= \left( \frac{6!}{3!0!3!}+\frac{6!}{2!2!2!}+\frac{6!}{1!4!1!}+\frac{6!}{0!6!0!} \right )\\
&\quad +\left( \frac{6!}{3!1!2!}+\frac{6!}{2!3!1!}+\frac{6!}{1!5!0!} \right ) \\
&= 20+90+30+1\:+\:60+60+6 = 141+126 \\
&= 267
\end{align}$$
Feel free to check my numbers, as this was done in my head and without the aid of a safety net.
Best Answer
I don't see a really elegant way to do this.
One simplification is to calculate the ways in which X wins and multiply by 2 because when you interchange all the Xs and Ys in all the "X wins" scenarios you get all the possible "Y wins" scenarios.
X wins in 3 - XXX - 1 way
X wins in 4 - YXXX - 1 way
X wins in 5 - YYXXX - XYXXX - XXYXX - 3 ways
for 6 and 7 organize yourself by considering the six possible results for the first 3 games (it must have been 2-1 after 3 games if no one wins in 3)
e.g. for X to win in 6 there are 2 scenarios starting with "XXY" but no scenarios starting with "YYX" .
For X to win in 7 all six 3 game starts are possible leading to either 2 or 3 scenarios.
I count 7 ways for X to win in 6 and 14 ways for X to win in 7
total ways for X to win $= 14 + 7 + 3+ 1+ 1 = 26$
So there must be 52 possible ways to play out the series, of which 7 start with XYY ( using the "X wins" list count sequences starting with either XYY or YXX )
An interesting result is that given you win in exactly 7 games the conditional probability that you were 2-1 down after 3 games is exactly 50% !