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Math Help - a probability problem about chess

  1. #1
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    a probability problem about chess

    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?
    ----the answer is 7808/19683, i don't understand the chess thing basicly, i only can guess out what is the 19683, 3^9, but i don't know how to explain it. Can somebody explain me the chesses, and how to get the answer.
    thanks a lot
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ives View Post
    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?
    ----the answer is 7808/19683, i don't understand the chess thing basicly, i only can guess out what is the 19683, 3^9, but i don't know how to explain it. Can somebody explain me the chesses, and how to get the answer.
    thanks a lot
    what's an "upset victory"?
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  3. #3
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    i don't know either..........
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  4. #4
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    that's the key to the answer. i think
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ives View Post
    that's the key to the answer. i think
    not really

    i think an upset victory just means the person who is least likely to win, wins.

    thus you need to find the probability of Elena winning the best of 5 games, which means she wins three

    so you want the probability of her winning 3 games

    that should give you the answer, but don't quote me on that. as i said, i am a noob with probability
    Last edited by Jhevon; May 7th 2008 at 07:31 PM.
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  6. #6
    Behold, the power of SARDINES!
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    This is a binomial probability(the formula is set up below)

    p(x)=\binom{5}{x} \left( \frac{4}{9}\right)^x \left( \frac{5}{9}\right)^{5-x}

    We want the prob of her winning at least 3 games so we want
    P(3)+P(4)+P(5)

    p(3)=\binom{5}{3} \left( \frac{4}{9}\right)^3 \left( \frac{5}{9}\right)^{5-3}=\frac{16000}{59409}

    p(4)=\binom{5}{4} \left( \frac{4}{9}\right)^4 \left( \frac{5}{9}\right)^{5-4}=\frac{6400}{59409}

    p(5)=\binom{5}{5} \left( \frac{4}{9}\right)^5 \left( \frac{5}{9}\right)^{5-5}=\frac{1024}{59409}

    so we add all these up to get

    \frac{1024}{59409} +\frac{6400}{59409}+\frac{16000}{59409}=\frac{2342  4}{59409}=\frac{7808}{19803}
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  7. #7
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    thanks...
    i haven't leaned the binomial things
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