Binomial probability for sporting event where the order of the first 3 outcomes does not matter There is a sporting event where team A has $1/3$ chance of winning and team B has a $2/3$ chance of winning. In order to win a team needs to be the first to win 4 matches. There are no ties. 
The question: What is the probability that team A will win? 
My thoughts:
It will take no more than $7$ matches for team A or B to win.
Neither team will win within $3$ matches, so the order for the first $3$ matches does not matter. 
The possible scores where A wins are 
$(4 - 0)$
$(4 - 1)$
$(4 - 2)$
$(4 - 3)$
So I need to find the binomial probability for each possibility and sum them.
However, the order of the first 3 matches does not affect the outcome, so they should not be included in the $nCr$ part. 
That's where I'm stuck as to how I need to write out $nCr$ for each possibility whilst taking into account that the order of the first $3$ matches does not matter. 
 A: We describe two different approaches. Assume independence (this is not reasonable, in many games there is a home rink/field advantage).
Way $1$: Imagine that whatever happens, the teams  keep on playing until $7$ games have been played played. Then Team A wins the real Stanley Cup finals if and only if in the modified series A wins $4$ or more games of the $7$ games. The probability A wins $4$ games in the modified series is $\binom{7}{4}(1/3)^4(2/3)^3$. The probability it wins $5$ is $\binom{7}{5}(1/3)^5(2/3)^2$. And so on. Add up.
Way $2$: We look at the real series, and calculate the probability the series lasts $4$ games and A wins it, the probability the series lasts $5$ games and A wins it, and so on up to $7$.
The probability A wins in $4$ is $(1/3)^4$.
For winning in $5$ games, A must win $3$ of the first $4$, and then win the fifth. The probability of winning $3$ of the first $4$ games is $\binom{4}{3}(1/3)^3(2/3)^1$. Multiply by $1/3$ to find the probability of winning in $5$ games.
To win in $6$ games, A must win $3$ of the first $5$, and then win. The probability is $\binom{5}{3}(1/3)^3(2/3)^2(1/3)$. 
Now write down the probability of winning in $7$ games, and add up.
A: If $A$ wins after exactly $k\in\left\{ 4,5,6,7\right\} $ matches
then $A$ must win match $k$ and must win $3$ of the $k-1$ matches
that precede match $k$.
The probability that $A$ wins the $k$-th match is $\frac13$.
The probability that $A$ wins $3$ of the $k-1$ matches
that precede match $k$ is $\binom{k-1}{3}\left(\frac{1}{3}\right)^{3}\left(\frac{2}{3}\right)^{k-4}$.
So if $p_{k}$ denotes the probability that $A$ wins after exactly
$k\in\left\{ 4,5,6,7\right\} $ matches then: $$p_{k}=\binom{k-1}{3}\left(\frac{1}{3}\right)^{4}\left(\frac{2}{3}\right)^{k-4}$$
The probability that $A$ wins is: $$\sum_{k=4}^{7}p_{k}$$
