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Math Help - Solving for n in binomial distribution

  1. #1
    Member Mollier's Avatar
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    Solving for n in binomial distribution

    Hi,

    I just came over a little problem that has me puzzled.

    If we have two teams, where one of the teams has a 55% chance of winning a match, how many times must the two teams meet such that we can be 95% certain that the best team wins?

    What pops into my mind is the binomial distribution, but then I would have to solve the following equation for n,

    \sum^{(n/2+1)}_{k=0}\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k} > 0.95,

    which looks daunting. Perhaps I could use Markov chains in some way?
    Any hints are appreciated!
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  2. #2
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    Hello, Mollier!

    The wording of the problem is quite fuzzy.
    I'll give you my interpretation.


    If we have two teams, where one of the teams has a 55% chance of winning,
    how many times must the two teams meet so that we can be 95% certain
    that the better team wins at least one game?

    Let \,A be the stronger team.

    Then P(\text{A wins}) \:=\:0.55\quad\Rightarrow\quad P(\text{A loses}) \:=\:0.45


    The probability that \,A loses \,n games is: . (0.45)^n
    . . and we want this to be less than 5%.

    So we have: . (0.45)^n \:<\:0.05

    Take logs: . \ln(0.45^n) \:<\:\ln(0.05) \quad\Rightarrow\quad n\ln(0.45) \:<\:\ln(0.05)

    Divide by \ln(0.45), a negative quantity: . n \;>\;\dfrac{\ln(0.05)}{\ln(0.45)}

    . . and we have: . n \:>\:3.751663619 \quad\Rightarrow\quad n \;\ge 4


    Therefore, if they play 4 or more matches,
    . . we can be 95% certain that \,A will win a game.

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  3. #3
    Member Mollier's Avatar
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    I'm so sorry, I translated it wrong.
    The problem asks for the number of matches that need to be played such that we can be 95% certain that the best team has won most matches. Is that better?
    Last edited by Mollier; December 6th 2010 at 06:56 AM.
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