winning the match (which is first to win n games), if i know the probability of them winning a game? Say i have two players who each have a certain probability of winning a game,
for example for player $1$: $p_1=0.8$ and player $2$: $p_2=0.2$.
There are $n$ games in a match, what is the probability of player $1$ being the first to win $n$ games,and therefore win the match?
Thanks.
 A: Calculus for a fixed number of games 
If both play a total of $m$ games, then $n$ games must be won for the first player with probability $p$ each one, being them independent:
$$p^n$$
The second player must won $m-n$ games. That with probability $1-p$ each one.
$$(1-p)^{m-n}$$
Finally to conclude the game the last one must be won by the first player, so we must choose from $m-1$ games which $n-1$ to be won.
$${m-1 \choose n-1}$$
So the probability is:
$$P(x=m)={m-1 \choose n-1} p^n(1-p)^{m-n}$$
Calculus for the total of games:
The constraint in the number of games is that the second player must win no game or at most $n-1$ so:
$$0\leq m-n \leq n-1$$
$$n\leq m \leq 2n-1$$
So adding up:
$$\sum_{k=n}^{2n-1}P(x=k)=\sum_{k=n}^{2n-1} {k-1 \choose n-1} p^n(1-p)^{k-n}=\sum_{k=0}^{n-1} {k+n-1 \choose n-1} p^n(1-p)^{k}$$
If $n=2$:
$$\sum_{k=2}^{3}P(x=k)=\sum_{k=2}^{3} {k-1 \choose 1} p^{2}(1-p)^{k-2}$$
$$=\sum_{k=2}^{3} (k-1) p^{2}(1-p)^{k-2}=(3-2p)p^2$$
As you can see that is not greater than one.
