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Math Help - 2 player game rolling a die

  1. #1
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    2 player game rolling a die

    Homer and Marge are about to play a game with the following rules. They take turns rolling a standard fair die, starting with Homer. The game ends when someone rolls a 5 or 6. That person wins the game.
    (b)
    Suppose that the game ends in a draw if ever two 1's come up in a row. With this change to the rules, find the probability that Homer will win the game.

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  2. #2
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    Hmm, anyone able to help with how to treat the draw?

    I know that the probability of Homer winning (ignoring the new rule) is;
    define W as winning, and N as not winning. Then P(w)=1/3 and [tex]P(N)=2/3[/math.
    So for homer to win;
    P(w)=1/3
    P(NNW)=\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}=\frac{4}{9}\cdot \frac{1}{3}
    P(NNNNW)=\frac{2}{3}\cdot \frac{2}{3}\frac{2}{3}\cdot \frac{2}{3}\frac{1}{3}=(\frac{4}{9})^2\cdot \frac{1}{3}
    etc.
    So the probability for homer to win is just the sum of these;
    P(H)=P(w)+P(NNW)+P(NNNNW)+....
    which is the sum of a geometric process, so P(H)=\frac{1}{3} \cdot \frac{1}{1-\frac{4}{9} } =\frac{1}{3}\cdot \frac{9}{5}=\frac{3}{5}

    So now I just have no idea how to incorporate the draw into the game. I know that at the end of each players turn, there will be 3 states (1) they won, (2) they rolled a 1 (3) they rolled a 2, 3, 4. In state (1) the game ends, in state (2) The other player will roll, and if a 1 the game ends. and (3) the game continues if rolls a 1, and ends if roll a 6.
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  3. #3
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    If you consider a Markov-Chain with the following states:

    Homer won,
    Marge won,
    Homer rolled 1,
    Marge rolled 1,
    Homer rolled 2,3 or 4,
    Marge rolled 2,3 or 4,
    The game ended in a draw,


    You can represent the transition probabilities as a matrix A and may find the answer to your question, by finding the limit (A^n) and multiplying that by the initial vector (which would be 0's on all positions and a 1 in the "Marge rolled 2,3 or 4").

    Using Matlab and finding A^10000 (it converges), and then multiplying by the initial state vector I found that:

    P(Homer winning) = 0.5647
    P(Marge winning) = 0.3686
    P(Draw) = 0.0667
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  4. #4
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    Hmm, I think I understand this now.. we still want to know the probability of homer winning, so we dont want to include the possibility of a draw occuring;

    So for any roll;
     W \to \mbox{win (5,6)}; P(W)=\frac{1}{3}
     N \to \mbox{not win (1,2,3,4)}; P(N)=\frac{2}{3}
     D \to \mbox{draw(1)}; P(D)=\frac{1}{6}
     ND \to \mbox{not win,but not roll a 1(2,3,4)}; P(ND)=\frac{1}{2}

    So the probability of homer winning is:
    P(H)=P(W)+P((N \cdot ND) \cdot W) + P((N \cdot ND) ( N \cdot ND) \cdot W)+...
    P(H)=\frac{1}{3}+\left(\left(\frac{2}{3}\cdot \frac{1}{2}\right)\cdot \frac{1}{3}\right)+\left(\left(\frac{2}{3}\cdot \frac{1}{2}\right)^{2}\cdot \frac{1}{3}\right)+...=\frac{1}{3} \cdot \frac{1}{1-\frac{2}{6} }=\frac{1}{2}

    Thanks pedrosorio, but no matlab for this question.. :P
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  5. #5
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    I think you're missing something in your calculation because the second term for example, should be:

    P((ND * N) + (N * ND)) * P(W) , no?
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  6. #6
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    Hmm, yeah, i think you are on the right track.. I think what I have done is incorrect... as what i am saying is for the 2nd probablilty,
    Homer rolls a 1,2,3,4 and marge rolls a 2,3,4...

    The probability has to be less then what it was before the new rule was introduced.... How about;
    P(H)=P(w)+P(D*ND+ND*D)*P(w)+P(D*ND+ND*D)^2*P(w)+.. .
    P(H)=1/3+(1/6*1/2+1/2*1/6)*1/3+....=1/3*[1/(1-1/6)]=2/5

    ie. this is saying homer rolls a 1, marge rolls a 2,3,4 or homer rolls a 2,3,4 and marge rolls a 1. This way for every roll a 1,2,3, or 4 is covered so homer/marge wont win on that roll and the game cannot end in a draw.
    sorry for the lack of latex
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  7. #7
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    P(H)=P(w)+P(D*ND+ND*D)*P(w) ...

    This can't be, it must be

    P(w)+P(N*ND + ND*N)*P(w)...

    I agree that the probability of Homer winning should be under 3/5 for sure (because of the draw rule).

    The probability of homer winning on the 3rd try would be:

    (P(N)*P(ND)*P(ND)*P(N) + P(N)*P(ND)*P(N)*P(ND) + P(ND)*P(N)*P(ND)*P(N)) * P(w) = 3 * (P(N)*P(ND))^2 * P(w)

    In general I think you'll have probability of winning on nth try:

    n*(P(N)*P(ND))^(n-1) * P(w)
    Last edited by pedrosorio; September 6th 2009 at 03:31 AM.
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  8. #8
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    argh, still have no idea how to solve this;
    redifine X to be ND (ie. 2,3,4 is rolled), and D to be a 1 is rolled. Then homer can win if the following rolls happen:

    Homer wins on his first roll;
    W
    Homer wins on his second roll;
    XXW
    XDW
    DXW
    Homer wins on his third roll;
    XXXXW
    XXXDW
    XXDXW
    XDXXW
    XDXDW
    DXXXW
    DXXDW
    DXDXW
    Homer wins on his fourth roll;
    XXXXXXW
    XXXXXDW
    XXXXDXW
    XXDXXXW
    XXDXXDW
    XXDXDXW
    XXXDXXW
    XXXDXDW
    XDXXXXW
    XDXXXDW
    XDXXDXW
    XDXDXXW
    XDXDXDW
    DXXXXXW
    DXXXXDW
    DXXXDXW
    DXDXXXW
    DXDXXDW
    DXDXDXW
    DXXDXXW
    DXXDXDW

    etc. I cant work out how to calculate a recursive formula for this, or see if its a geometric series. I tried working out the probability of a draw but still cant work it out...
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  9. #9
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    Hi Robb, I'm doing the same problem as you, are you by any chance at ANU?
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