Probability of a team winning the series? Suppose that two sports teams were to play each other in a series, and the winning team is the one that wins two games. One of the teams, Team A, has a probability of 80% chance of winning any one game. What is the probability of Team A winning the series?

The answer that I reached is 64%, and my reasoning is that for Team A to win, there must be two conditions:


Team A must win its first game
Team A must win its second game


With that in mind, I multiplied 80% (the chance of Team A winning its first game) with 80% again (because it has 80% chance of winning its second game). Is this correct? I'm getting a bit tripped up because the probability of Team A winning its second game is not dependent on its probability of winning the first game, so that's why I'm sort of leaning on the total probability being 80% again (because if we can assume Team A wins its first game then everything else is irrelevant; the chances of them winning again is 80%).
i will solve a general case , say probability of $A$ winning is $p$ and for a team to win it needs to win $n$ matches. now say $k$ matches have been played , and $A$ won the series , it means $A$ won the last match , thus out of the rest $k-1$ matches $A$ won $n-1$ games. thus for $A$ to win the series in $k$ matches probability is $$\binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$

now $k&gt;=n$ and $k&lt;=2n-1$ ($A$ needs to win before $B$) thus our total probability is $$\sum_{k=n}^{2n-1} \binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$
I stumbled upon this when I hit the same dilemma - probability of winning game 1 and game 2 are independent , its like picking a red ball from a jar that has 3 red and 2 blue with replacement - a general example problem. 

However, on thinking  more, even though the individual game result is independent, the overall series win is dependent of the result of game 1 and game 2. 

Thus to get the probability of winning the series, the P(winning game 1) has to be multiplied with P(winning game 2), which is:

0.8 * 0.8 = 0.64.