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Math Help - Odds and probability question!

  1. #1
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    Odds and probability question!

    If the odds in favour of Boris beating Elena in a chess game are 5 to 4,
    what is the probability that Elena will win an upset victory in a best-of-five chess tournament?

    I found the probability of Elena winning to be 4/9. Then for winning the first 3 games of the series to be (4/9)^3= 0.088. However the textbook's answer key says the correct answer is 7808/19683 or .397. Can anyone help me please
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  2. #2
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    Hello, ibetan!

    f the odds in favour of Boris beating Elena in a chess game are 5 to 4,
    what is the probability that Elena will win an upset victory in a best-of-five tournament?

    I found the probability of Elena winning to be 4/9.
    Then for winning the first 3 games of the series to be (4/9)^3
    This is true, but it is not the only way Elena can win the tournament.

    However, the answer key says the correct answer is: 7808/19683

    A best-of-5 tournament means the first to win 3 games is the winner.


    Elena can win in three games.
    There is one way: . EEE
    Hence: . P(\text{Elena, 3 games}) \;=\;\left(\frac{4}{9}\right)^3 \;=\;\frac{64}{729}


    Elena can win in four games.
    There are three ways: . BEEE,\;EBEE,\;EEBE

    Hence: . P(\text{Elena, 4 games}) \;=\;3\cdot\left(\frac{4}{9}\right)^3\left(\frac{5  }{9}\right) \;=\;\frac{960}{6,\!561}


    Elena can win in five games.
    There are 6 ways: . BBEEE,\;BEBEE,\;BEEBE,\; EBBEE,\;EBEBE,\;EEBBE

    Hence: . P(\text{Elena, 5 games}) \;=\;6\cdot\left(\frac{4}{9}\right)^3\left(\frac{5  }{9}\right)^2 \;=\;\frac{9,\!600}{59,\!049}


    Therefore: . P(\text{Elena wins tourna{m}ent}) \;=\;\frac{64}{729} + \frac{960}{6,\!561} + \frac{9,\!600}{59,\!049} \;=\;\frac{23,\!424}{59,\!049} \;=\;\frac{7,\!808}{19,\!683}

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