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Math Help - probability of a die

  1. #1
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    probability of a die

    so I am doing some math that my friend asked me to do and for the life of me I cant figure it out... Here is the question. If anyone could solve this that would be amazing.

    Suppose you have a fair 6-sided die. If you roll the die repeatedly, with 1 worth 1 point, and 6
    worth 3 points, and all other outcomes (2, 3, 4 and 5) worth 2 points, and add up the points as you
    go along (starting at 0), what is the probability pn that you’ll hit n? Check your formula with a
    direct calculation of p
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  2. #2
    MHF Contributor matheagle's Avatar
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    P(X=1)=1/6=P(X=3) while P(X=2)=2/3
    where X is the points you accumulate in each play.
    NOW, I doubt you want n twice.
    I think you want

    P(X_1+\cdots +X_n=k) where each X_i are iid copies of that above distribution.

    Let S_n=X_1+\cdots +X_n the points you've accumulated after n plays.

    P(S_n=n)=(1/6)^n since you tossed a 1, n times

    P(S_n=n+1)=n(1/6)^{n-1}(2/3)
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  3. #3
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    well the way we have been taught in our lin alg class is that we want to set it as a recursion formula and use eigenvectors/values I believe, I may be wrong though. Then he would set it as a linear combination.
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  4. #4
    MHF Contributor matheagle's Avatar
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    you can look at the vector (S_{n-1},S_n)

    Maybe use http://en.wikipedia.org/wiki/Probabi...ating_function

    G_X(z)=E(z^X)=(1/6)z+(2/3)z^2+(1/6)z^3

    Then G_{S_n}(z)=\big(E(z^X)\bigr)^n=\big((1/6)z+(2/3)z^2+(1/6)z^3\big)^n

    So P(S_n=k)={G^{(k)}_{S_n}(0)\over k!}

    You can use the trinomial theorem to calculate the probabilities
    http://staff.spd.dcu.ie/breens/documents/BiTri2.pdf
    Last edited by matheagle; March 15th 2011 at 11:00 PM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    maybe you can use a markov chain here with an upper triangular matrix.
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  6. #6
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    I think I've solved this. No time tonight to verify and post the answer, but I'll follow up tomorrow probably. It's based on a recursion and simple matrix stuff. All that's left to do is diagonalize an appropriate matrix to get the answer in a nice form.
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  7. #7
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    Quote Originally Posted by theodds View Post
    I think I've solved this. No time tonight to verify and post the answer, but I'll follow up tomorrow probably. It's based on a recursion and simple matrix stuff. All that's left to do is diagonalize an appropriate matrix to get the answer in a nice form.
    It would be better to wait a while before posting and see what progress the OP has made ....
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    It would be better to wait a while before posting and see what progress the OP has made ....
    Okay. I'll at least post the answer so others can check it/against it, if they wish.

    The probability that n occurs in a given infinite sequence of rolls, if my work is correct, is

    \frac 1 2 + \left(\frac {-1} 2\right)^n - \frac 1 2 \left(\frac {-1} 3 \right)^n
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