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Math Help - Best-of-Seven Probability Question

  1. #1
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    Best-of-Seven Probability Question

    Hi again,

    I am struggling once again (no surprise) with calculating probabilities...

    Two baseball games play a best-of-seven series in which the series ends as soon as one team wins four games. The first two games are played on A's field, the next 3 on B's and the last 2 on A's. Given that the probability A wins on their home field is 0.7, and on B's field is 0.5, what is the probability the series does not go to 6 games?

    I have that the probability they in in 4 is 0.1225, and 5 games is 0.175... Are these in any way related to the answer?

    I have thought about calculating the probability that the game goes to and taking the complement of that, but that would not include the series that went to 7 games. Kind of unsure where to piece this info together. Any insight would be much appreciated.

    Thanks in advance.
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  2. #2
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    Re: Best-of-Seven Probability Question

    Hello, KelvinScale!

    Two baseball teams play a best-of-seven series
    in which the series ends as soon as one team wins four games.
    The first two games are played on A's field, the next 3 on B's and the last 2 on A's.
    Given that the probability A wins on their home field is 0.7, and on B's field is 0.5,
    what is the probability the series does not go to 6 games?

    There is no formula for this problem.
    I had to make a brute-force LIST of the possible outcomes.


    Suppose A wins the series.

    A could win the first four games: . AA|AA \:=\:(0.7)^2(0.5)^2 \:=\:0.12250

    A could win 3 of the first 4 games, then win the 5th game.
    . . \begin{array}{cccccc}AA|ABA &=& (0.7)^2)0.5)^3 &=& 0.06125 \\ AA|BAA &=& (0.7)^2(0.5)^3 &=& 0.06125 \\ AB|AAA &=& (0.7)(0.3)(0.5)^3 &=& 0.02625 \\ BA|AAA &=& (0.3)(0.7)(0.5)^3 &=& 0.02625 \end{array}

    Hence: . P(A\text{ wins in less than 6 games})
    . . . . . . . . . . =\:0.12250 + 0.06125 + 0.06125 + 0.02625 + 0.02625 \:=\:0.2975


    Suppose B wins the series.

    B could win the first four games: . BB|BB \:=\:(0.3)^2(0.5)^2 \:=\:0.02250

    B could win 3 of the first 4 games, then win the 5th game.
    . . \begin{array}{cccccc}BB|BAB &=& (0.3)^2(0.5)^3 &=& 0.01125 \\ BB|ABB &=& (0.3)^2(0.5)^3 &=& 0.01125 \\ BA|BBB &=& (0.3)(0.7)(0.5)^3 &=& 0.02625 \\ AB|BBB &=& (0.7)(0.3)(0.5)^3 &=& 0.02625 \end{array}

    Hence: . P(B\text{ wins in less than 6 games})
    . . . . . . . . . . =\:0.02250 + 0.01125 + 0.01125 + 0.02625 + 0.02625 \:=\:0.0975


    Therefore: . P(\text{Series ends in 5 games or less}) \;=\; 0.2975 + 0.0975 \;=\;\boxed{0.395}

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  3. #3
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    Re: Best-of-Seven Probability Question

    Thanks a lot, it is clear now, I forgot to consider the case where B won in 4 and 5 games.
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