What is the probability the team will win If a team 1 has a probability of p of winning against team 2. What is the probability "formula" that team one will win 7 games first. 
There are no ties and the teams play until one t am wins 7 games 
 A: Imagine that exactly $13$ games are played out, even though it is likely that the series will have been settled prior to the last game.  The advantage here is that we know that exactly one of the teams will have won  $7$ or more games and that determines the winner.  To finish, we remark that for Team $1$ to win the series, they must win between $7$ and $13$ games. Of course the probability that they win exactly $i$ games out of $13$ is $\binom {13}i p^i(1-p)^{13-i}$. Thus the answer is $$\sum_{i=7}^{13} \binom {13}i p^i(1-p)^{13-i}$$
A: A state $(a,b)$ where $a$ is the number of victories from team1 and $b$ from team2.
Team 1 wins if they reach the states $(7,0), (7,1),...,(7,6)$.
Each state $(7,k)$ are tied to $\binom{6+k}{k}$ possible scenarios of same probability of occuring, that is $p^7(1-p)^{k}$. Indeed, if the result is $(7,k)$ , it means that we played $7+k$ games, the $(7+k)^{th}$ game was won by team1, and we pick $k$ losses out of the $6+k$ remaining games.
Therefore $P((7,k))=\binom{6+k}{k}p^7(1-p)^{k}$.
Finally you probability is $$\sum_{k=0}^{6}P((7,k))=\sum_{k=0}^{6}{\binom{6+k}{k}p^7(1-p)^{k}}$$
