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Thread: Looking for an elegant mathematical expression

  1. #1
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    Looking for an elegant mathematical expression




    Hi to you all,

    i am new at this forum and i am glad there is such.
    Problem description
    suppose i am playing two seperate games (independently!) with the following rules:
    game #1- i have 4 empty cells in a row and i am trying to fill each cell (1 at the time) with red balls, the probability of filling each cell is p1 and (1-p1) for the cell to remain empty - -> only 4 attempts.
    game #2- i have 4 empty cells in a row and i am trying to fill each cell (1 at the time) with black balls, the probability of filling each cell is p2 and (1-p2) for the cell to remain empty - -> only 4 attempts.
    Question: what is the probabilty of having a total of n=1,2,3,...,8 filled cells? after a total of 4+4 attempts.
    i have already solved the problem but i am looking for an elegant mathematical expression. right now my solution looks like:
    p(n=0)=(1-p1)^4* (1-p2)^4
    p(n=1)=4*(1-p1)^4*p2*(1-p2)^3+4*(1-p2)^4*p1*(1-p1)^3
    p(n=2)=...etc.
    Thanks in advanced.
    Yaron B.S.

  2. #2
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    Re: Looking for an elegant mathematical expression

    Try writing the expression with a summation sign. Also, try to figure out a way to write the powers in a cleaver way

    This is a well known expression!

  3. #3
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    Re: Looking for an elegant mathematical expression


    The answer should be no.

    Let the number of cells filled in each game be M_1, M_2 respectively.

    Then M_i \sim \text{Binomial}(k=4, p_i)

    and thus

    \Pr\{N = n\} = \Pr\{M_1 + M_2 = n\}

    = \sum_{m=\max\{0, n-4\}}^{\min\{4, n\}} \Pr\{M_1 = n - m\}\Pr\{M_2 = m\}]

    = \sum_{m=\max\{0, n-4\}}^{\min\{4, n\}} \binom {4} {n-m} p_1^{n-m}(1 - p_1)^{4-n+m}\binom {4} {m} p_2^m(1 - p_2)^{4-m}

    for n = 0, 1, \ldots, 8

    Unless p_1 = p_2, you cannot further simplify the expression.

  4. The Following User Says Thank You to BGM For This Useful Post:

    yaron (08-03-2014)

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