Even coin tosses, 2 players The game consists of a sequence of independent plays for each of the two players. Each round I have probability $p$ of making a point and my opponent has probability $1-p$ of making a point. (so each round only involves one player taking their chances and each player alternates, there has to be an even number of games). What are my chances of winning?
So my chances of winning a game with $2n$ trials is: $\sum_{n+1}^{2n}  {{2n}\choose{x}} p^x (1-p)^{2n-x}$
In the case of $p=0.5$ for two rounds, my probability of winning is 1/4, then 4 rounds my probability is 5/16 which I obtained by manually counting outcomes. Can I obtain a general formula for 2n games which uses counting outcomes?
Thanks!
 A: From the fact that you have to have scored some number of points we have:
$$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} = 1-\sum_{i=0}^{n} \binom{2n}{i} p^i (1-p)^{2n-i}$$
Using the fact that there are just as many ways to distribute $i$ points as to distribute $2n-i$ points when there are $2n$ points available:
$$  1-\sum_{i=0}^{n} \binom{2n}{i} p^i (1-p)^{2n-i}=1-\sum_{i=0}^{n} \binom{2n}{2n-i} p^i (1-p)^{2n-i}$$
letting $j=2n-i$ to bring the expression closer to the thing we are solving for:
$$1-\sum_{i=0}^{n} \binom{2n}{2n-i} p^i (1-p)^{2n-i} =1-\sum_{j=n}^{2n} \binom{2n}{j} p^{2n-j} (1-p)^{j} $$
For $p=1/2$ we have that $p=1-p$ and thus they are exchangeable:
$$1-\sum_{j=n}^{2n} \binom{2n}{j} p^{2n-j} (1-p)^{j}=1-\sum_{j=n}^{2n} \binom{2n}{j} p^{j} (1-p)^{2n-j} $$
splitting out the $j=n$ term:
$$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} =1-\binom{2n}{n}p^n(1-p)^n-\sum_{j=n+1}^{2n} \binom{2n}{j} p^{j} (1-p)^{2n-j} $$
Converging to:
$$ \sum_{i=n+1}^{2n} \binom{2n}{i} p^i (1-p)^{2n-i} = \frac{1}{2} - \frac{1}{2^{2n+1}}\binom{2n}{n}$$
Essentially if the probability of winning a point is 1/2 then you are as likely as your opponent to score some number of points. Your probability of winning is thus $(1- P_{tie})/2$. The probability of a tie is $2n$ games pick $n$ points multiplied by the probability of getting exactly that number of points for both sides.
