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Math Help - chess exercise

  1. #1
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    chess exercise

    Hi all,
    I am in the beginning of learning statistics and I am trying to solve problems from the internet.
    I found a problem with chess but I don't know how to solve it.
    Could you help me ??

    The problem is :
    Two equals (equal possibilities for win) players are playing chess.
    a. Is more possible to win someone 4 from 6 games or 5 from 9 ?
    b. Is more possible to win at least 3 from 4 or at least 5 from 8?
    c. How many games have to play a player so as the probability to win one of them to be > 0.9.


    Thanks a lot for your help
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  2. #2
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    Quote Originally Posted by peter20 View Post
    Hi all,
    I am in the beginning of learning statistics and I am trying to solve problems from the internet.
    I found a problem with chess but I don't know how to solve it.
    Could you help me ??

    The problem is :
    Two equals (equal possibilities for win) players are playing chess.
    a. Is more possible to win someone 4 from 6 games or 5 from 9 ?
    b. Is more possible to win at least 3 from 4 or at least 5 from 8?
    c. How many games have to play a player so as the probability to win one of them to be > 0.9.


    Thanks a lot for your help
    Hi Peter20,

    Although the problem doesn't say so, I think we have to assume that each game is won by one of the players, i.e. a draw is not a possibility (unlike in the Real World, where draws are quite common).

    So let's assume each player has a 1/2 probability of winning each game, and let's also assume the outcomes are independent (the outcome of one game does not influence the outcome of the next). With these assumptions, the number of games a player wins in n games has a Binomial distribution with p=1/2 and n=n (what else?). Then the probability of x wins in n games is

    \binom{n}{x} (1/2)^x (1/2)^{n-x} = \binom{n}{x} (1/2)^n
    where
    \binom{n}{x} = \frac{n!}{x! \; (n-x)!}

    So the probability of winning exactly 4 games out of 6 is

    \binom{6}{4} (1/2)^6

    and the probability of winning exactly 5 games out of 9 is

    \binom{9}{5} (1/2)^9.

    I'll leave you to do the arithmetic, and maybe with this start you can work the other problems too.
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  3. #3
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    thanks a lot awkward for your answer!
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