TITLE: Calculating probability of winning best-of-7-games tournament. Why is my method wrong?
QUESTION [10 upvotes]: The question is as follows:

A and B participate in a tournament of "best of 7 games". It is equally likely that either A wins the game or B wins the game, or the game ends in a draw. What is the probability that A wins the tournament?

So I tried an approach like this. I made a table like:
$$ \begin{array}{c|c}
\text{Ways of Winning} & \text{Probability} \\
\hline
\text{W W W W _ _ _} & (1/3)^4 \\
\color{red}{\text{W W W L }} \text{W _ _} & (1/3)^5\times 4 \\
\color{red}{\text{W W W L L }} \text{W _} & (1/3)^6\times \dfrac{5!}{3!\,2!} \\
\color{red}{\text{W W W L L L }} \text{W} & (1/3)^7\times \dfrac{6!}{3!\,3!} \\
\color{red}{\text{W W W D }} \text{W _ _} & (1/3)^5\times 4 \\
\color{red}{\text{W W W D L }} \text{W _} & (1/3)^6\times \dfrac{5!}{3!} \\
\color{red}{\text{W W W D L L }} \text{W} & (1/3)^7\times \dfrac{6!}{3!\,2!} \\
\vdots & \vdots \\
\color{red}{\text{W D D D D L }} \text{W} & (1/3)^7\times \dfrac{6!}{4!} \\
\color{red}{\text{W D D D D D }} \text{W} & (1/3)^7\times 6 \\
\color{red}{\text{W D D D D D D}} & (1/3)^7\times 7 \\
\end{array}
$$
Here the W represents a win for A, L represents a loss while D represents a draw. If a character is in red, it means that it can be exchanged (rearranged) with the other characters in red. If in black, it's position is fixed. Underscores represent any value can be taken at that point.
The procedure I followed is that, for zero draws, I kept adding an extra red L before the last W and still letting A win the tournament. Then I added a red D, then kept adding a red L, again letting A win the tournament. I wrote all such arrangements in this fashion and wrote the corresponding probabilities and added them together. There were 16 such rows for me.
The answer i got was $651/3^7$ or $217/729$, but the answer given is $299/729$. They calculated it by subtracting the probability of a draw from one and then dividing it by two. I understand why they did it, what I don't understand is why our answers don't match!
So, what is wrong with my approach? Am i missing some cases? Or is totally scrap?

REPLY [4 votes]: Alternative route
Let $N_{A}$ denote the number of times that $A$ wins, and $N_{B}$
the number of times that $B$ wins. 
Then:

$\Pr\left(\text{no winner}\right)=\sum_{k=0}^{3}\Pr\left(N_{A}=N_{B}=k\right)=3^{-7}\sum_{k=0}^{3}\frac{7!}{k!k!\left(7-k\right)!}$
$\Pr\left(A\text{ wins}\right)+\Pr\left(B\text{ wins}\right)+\Pr\left(\text{no winner}\right)=1$
$\Pr\left(A\text{ wins}\right)=\Pr\left(B\text{ wins}\right)$

This leads to:
$$\Pr\left(A\text{ wins}\right)=\frac{1}{2}\left[1-3^{-7}\sum_{k=0}^{3}\frac{7!}{k!k!\left(7-k\right)!}\right]=\frac{1794}{4374}=\frac{299}{729}$$