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Math Help - Probability of wining a series

  1. #1
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    Probability of wining a series

    A and B play a series of game, the probability of A wining is p and the probability of B wining is 1-p. The winner of the series is the player that wins 2 more games.

    A) Find the probability that A is the final winner.
    B) Compute the expectation of games played.

    A) in order for A to be the winner, one of these outcomes would have to occur:

    AA = p^2
    ABA = (p^2)(1-p)
    BAA = (p^2)(1-p)

    I think I would have to sum up the values, which would give:

    p^2+2(p^2)(1-p)= p^2(3-2p)

    B) I would think that would be around 3, considering that there are a total of 6 possible outcomes, I would think it would look something like this:

    E[N] = E[N=2|Y=1]P(Y=1)+E[N=3|Y=1]P(Y=1)+ E[N=2|Y=0]P(Y=0)+E[N=3|Y=0]P(Y=0)

    E[N] = E[N=2|Y=1]p^2+E[N=3|Y=1](p^2(3-p)+ E[N=2|Y=0](1-p)^2+E[N=3|Y=0](1-p)^2(1+2p)
    Last edited by lllll; September 28th 2008 at 09:50 PM.
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  2. #2
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    making some modifications to part B), would it be:

    Y = \left\{ \begin{array}{rcl}<br />
1 & \mbox{if} & \mbox{A wins the series} \\ 0 & \mbox{if} & \mbox{B wins the series}<br />
\end{array}\right.

    so then I would have:

    E[N] = E[N|Y=1]P(Y=1)+E[N|Y=0]P(Y=0)

    E[N] = E[N|Y=0](p^2(3-2p))+E[N|Y=1](1-p^2)(1+2p)

    at which point I'm stuck.
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  3. #3
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    Quote Originally Posted by lllll View Post
    A and B play a series of game, the probability of A wining is p and the probability of B wining is 1-p. The winner of the series is the player that wins 2 more games.

    A) Find the probability that A is the final winner.
    B) Compute the expectation of games played.

    A) in order for A to be the winner, one of these outcomes would have to occur:

    AA = p^2
    ABA = (p^2)(1-p)
    BAA = (p^2)(1-p)

    I think I would have to sum up the values, which would give:

    p^2+2(p^2)(1-p)= p^2(3-2p)

    B) I would think that would be around 3, considering that there are a total of 6 possible outcomes, I would think it would look something like this:

    E[N] = E[N=2|Y=1]P(Y=1)+E[N=3|Y=1]P(Y=1)+ E[N=2|Y=0]P(Y=0)+E[N=3|Y=0]P(Y=0)

    E[N] = E[N=2|Y=1]p^2+E[N=3|Y=1](p^2(3-p)+ E[N=2|Y=0](1-p)^2+E[N=3|Y=0](1-p)^2(1+2p)
    Do you mean 'The winner of the series is the player that first wins 2 games' ....? That's what your answer implies .... Except for AA, none of the others have A winning two more games than B ..... ABA and BAA both have A winning only one more game than B.
    Last edited by mr fantastic; September 29th 2008 at 01:07 AM.
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  4. #4
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    it's 2 more games then the other
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  5. #5
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    Quote Originally Posted by lllll View Post
    it's 2 more games then the other
    Then for reasons I've implied in my earlier post, your calculation is not correct. Consider the following cases:

    Case 1:

    AA
    ABAA
    ABABAA
    ABABABAA
    .
    .
    .

    Case 2:

    BAAA
    BABAAA
    BABABAAA
    .
    .
    .

    And there are obviously more cases .... I don't have time now to go into it further.
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