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World Series, probability of victor with evenly matched teams

  1. Feb 25, 2009 #1
    1. The problem statement, all variables and given/known data

    When two teams meet in the World Series, they play a series of games which stops when one team has won four games (a "best-of -seven" series). The team in the World Series are A and B.
    (a) If the two teams are evently matched, what is the probability that the World Series ends in only four games?

    (b) Again assuming they are equally matched, what is the probability that team A wins games 1,2,4 and 7 (and thus the World Series) and the team B wins games 3, 5 and 6?

    (c) Again assuming they are equally matched, what is the probability that A wins the World Series in seven games?

    (d) if the probability that A wins any given game against B is 1/4, what is the probability that A wins the World Series in either four of five games?



    2. Relevant equations



    3. The attempt at a solution

    Ans for a) (1/2)^4

    I don't know how to solve the problem after a). Please help. Thank you.
     
    Last edited: Feb 25, 2009
  2. jcsd
  3. Feb 25, 2009 #2
    Re: probability

    I think that its like this:

    a) P(X = 0) + P(X = 4), X ~ Binomial(4, .5)
    b) This is an ordered sequence, so since p = .5 then it is just .5^7
    c) P(X = 4), X ~ Binomial(7, .5)
    d) your question is incomplete. However, if the probability that A wins is p, then it is (P(X = 4), (X ~ Bin(4, p))) + (P(X = 4), (X ~ Bin(5, p))
     
  4. Feb 25, 2009 #3
    Re: probability

    I don't think that I completly understand your answer, but Thank you.
     
  5. Feb 25, 2009 #4
    Re: probability

    Look up the binomial distribution on wikipedia. You need to use its probability function to calculate the probabilities P(X = x).

    the answers are:

    << answers deleted by berkeman -- giving answers to homework problems are not permitted on the PF >>
     
    Last edited by a moderator: Feb 26, 2009
  6. Feb 26, 2009 #5

    HallsofIvy

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    Re: probability

    (c) and (d) are a bit more complicated than just the binomial distribution. For example, if we write "A" for a game that A wins and "B" for a game B wins, the binomial coefficient the number of different ways A could win four games and B one could be written AAAAB, AAABA, AABAA, ABAAA, and BAAAA- but the first of those is invalid in this problem because then the series would end in four games.
     
  7. Feb 26, 2009 #6
    Re: probability

    it is not binomial distribution, it is like fliping coins:

    a) is (1/2)^4 x (1/2)^4 for the two teams

    b) is AABABBA which is 1,2, 4 and 7 win games for team A and 3,5 and 6 for team B
    which is (1/2)^4 x (1/2)^3

    c) is how many ways

    d) I don't know yet, I am trying :confused:
     
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