Thread: Number of outcomes possible for best-of-nine series

1. Number of outcomes possible for best-of-nine series

How many different outcomes are there in a best-of-nine series between two teams A and B? Generalize to a best-of-n series where n is odd.
xxxxA xxxxxA xxxxxxA xxxxxxxA xxxxxxxxA
$\displaystyle \left({4\atop 4}\right)+\left({5\atop 4}\right)+\left({6\atop 4}\right)+\left({7\atop 4}\right)+\left({8\atop 4}\right)=$the number of outcomes where A wins the series. Multiply this by two to get number of outcomes where A or B wins the series. For best-of-n where n is odd
$\displaystyle 2*\sum_{k=\frac{n-1}{2}}^{n-1}\left({k\atop \frac{n-1}{2}}\right)$

Does this look correct?

2. Hello, oldguynewstudent!

How many different outcomes are there in a best-of-nine series
between two teams A and B? Generalize to a best-of-n series where n is odd.

$\displaystyle \text{xxxxA} \quad \text{xxxxxA} \quad \text{xxxxxxA} \quad \text{xxxxxxxA} \quad \text{xxxxxxxxA}$
.$\displaystyle {4\choose4} \;\;+\;\; {5\choose4} \;\;+\;\; {6\choose4} \quad+\quad {7\choose4} \quad+\quad {8\choose4}$ . . = .the number of outcomes where $\displaystyle A$ wins the series.

Multiply this by two to get number of outcomes where $\displaystyle A$ or $\displaystyle B$ wins the series.

$\displaystyle \text{For best-of-}n\text{ where }n\text{ is odd: }\;\;2\cdot\!\!\sum_{k=\frac{n-1}{2}}^{n-1}{k\choose\frac{n-1}{2}}$

Does this look correct? . Yes!

It all looks great! . . . Nice work!