Number of outcomes possible for best-of-nine series

Oct 2009
255
20
St. Louis Area
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?
 

Soroban

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May 2006
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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!

 
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