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Math Help - Number of outcomes possible for best-of-nine series

  1. #1
    Member oldguynewstudent's Avatar
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    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
    \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
     <br />
2*\sum_{k=\frac{n-1}{2}}^{n-1}\left({k\atop \frac{n-1}{2}}\right)<br />

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  2. #2
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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.

    \text{xxxxA} \quad \text{xxxxxA} \quad \text{xxxxxxA} \quad \text{xxxxxxxA} \quad \text{xxxxxxxxA}
    .  {4\choose4} \;\;+\;\; {5\choose4} \;\;+\;\; {6\choose4} \quad+\quad {7\choose4} \quad+\quad {8\choose4} . . = .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.


    \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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