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Thread: Probability of winning an n-game series

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    Probability of winning an n-game series




    With the baseball playoffs coming up, I was fiddling around with how to calculate the probability that one team would win a playoff series, given P, the probability that it would win any one game. The playoffs can be 1, 3, 5, or 7 games.

    A 1-game playoff is easy. It's just P.

    A 3-game series (first team to win 2 games) is:
    Code: 
         P(win) = PP + PQP + QPP = P^2 + 2*Q*P^2
    Note that Q = 1-P.

    A 5-game series is:
    Code: 
         P(win) = PPP + PPQP + PQPP + QPPP + PPQQP + PQPQP + PQQPP + QPPQP + QPQPP + QQPPP
                = P^3 + 3*Q*P^3 + 6*Q^2*P^3
    For a 7-game series (World Series), Team A must get to 4 wins before Team B does. That is, Team A could win 4-0, 4-1, 4-2, or 4-3. There is 1 way to win 4-0, 4 ways to win 4-1, 20 ways to win 4-2, and 20 ways to win 4-3:
    Code: 
         P(win) = P^4*(1 + 4Q + 10Q^2 + 20Q^3)
    Those coefficients look like half of a binomial series. It's not a full series because the powers of P are fixed at 4.

    Is there a general formula for an n-game series?

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    Re: Probability of winning an n-game series

    The probabilities are just the negative Binomial CDF - the probability that you have the first \frac {n+1} {2} wins on or before the n-th game.

    Also negative Binomial distribution has nice duality with the Binomial distribution. You may think that you always need to play all of the n games, and you win if and only if the number of wins (which has a Binomial distribution) is greater than or equal to \frac {n+1} {2}

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    Re: Probability of winning an n-game series

    Quote Originally Posted by BGM View Post
    The probabilities are just the negative Binomial CDF - the probability that you have the first \frac {n+1} {2} wins on or before the n-th game.

    Also negative Binomial distribution has nice duality with the Binomial distribution. You may think that you always need to play all of the n games, and you win if and only if the number of wins (which has a Binomial distribution) is greater than or equal to \frac {n+1} {2}
    If I understand, you are saying that the negative binomial distribution will provide the terms. For a 5-game series (first team to win 3 games), the series is:

    \left[{2\choose 0} P^2 + {2\choose 1} P^2Q + {2\choose 2} P^2Q^2\right]\times P

    Edit: This formula is wrong. The correct formula is (I think):

    \left[{2\choose 0} P^2 + {3\choose 1} P^2Q + {4\choose 2} P^2Q^2\right]\times P

    I know how to do that, but it requires iteration. I was asking if there might be a closed form equation for the general case?

    PS: How do I get larger parentheses around the series?
    Last edited by Jennifer Murphy; 09-29-2014 at 01:20 AM.

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    Re: Probability of winning an n-game series


    There should be no nice closed form formula for that.

    P.S. use \left( \right) to adjust the size of the brackets (or any other objects) to fit the size of the included objects.

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