Tennis match probability - is my logic incorrect? A tennis match consists of sets, until one player wins three sets. Player $A$ has a two-thirds chance of winning. Player $B$ has a one-third chance of winning. If they play a match, what is the probability of $A$ winning?
My attempt:
The minimum games for $A$ victory is $3$, and the maximum is $5$ (both win $2$, then next round is decider)
I have probability = win in three + win in four + win in five
=$\left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
Which is more than $1$!
I don't understand why this is wrong though

Your working: $\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
Please see the second term. You are allowing any $1$ of the $4$ sets to be won by player B. But if there are $4$ sets, player A  cannot win first three sets. Same logic for the next term.
So it should be,
$\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {3\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{4\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
