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% !
There is a more general version how to calculate it via thinking about 'states'. Let me illustrate this for your example. Let state be $(a,b)$ where $a$ is the number of games won by $A$ and $b$ is the number of games won by $B$. Value of state $(a,b)$, $v_{a,b}$ is the probability of $A$ wining the series. Hence $v_{3,3}=0.55$ (you can think of $v_{4,3}=1$ and $v_{3,4}=0$ and about the transition probabilities in the next sentence). From this, you work backwards since any state $(a,b)$ transitions either to state $(a+1,b)$ with probability $0.55$ or to state $(a,b+1)$ with probability $0.45$. Hence $$\begin{aligned}
v_{3,2}&=0.55\cdot1+0.45\cdot v_{3,3}=0.7975\\
v_{3,1}&=0.55\cdot1+0.45\cdot v_{3,2}=0.908875\\
\end{aligned}$$ so that your answer is correct. By calculating the value of each state you figure out the probability of $A$ winning for any combination of games already won.
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