Winning chance in crooked best-of-2n+1

Question

I've got this question for a uni assignment where I have to calculate my winning change in a version of best-of-2n+1. It goes like this:

• I have a 50/50 chance of winning each game
• My opponent has to win n+1 games to win the best-of-2n+1
• I have to win only n games to win the best-of-2n+1

What I have so far is:

$P($I win$) \ = \sum_{k=n}^{2n} P($I win in k games$)$

note that I cant win in 2n+1 games

$= \sum_{k=n}^{2n} (\frac{1}{2})^k \cdot$ [Nr of ways where I win after k games]

$= \sum_{k=n}^{2n} (\frac{1}{2})^k \cdot {k-1 \choose n-1}$

here I win the k-th game and n-1 games before that, for k between n and 2n

Im requird to rewrite this so that there is no sigma-summation. If there is anyone who would know how to do so, any help would be appreciated!!

Answer 1

After you have played $2n$ games (pretending to keep playing if somebody wins the series before this) exactly one of you will have won enough games to win the series. The chances you win at least $n+1$ match the chances your opponent wins $n+1$. In addition, you win the even splits. That should suggest how to eliminate the summation.