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Math Help - Geometric distribution problem

  1. #1
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    Geometric distribution problem

    I am am trouble with the following homework problem.
    (Note:This is exactly how the problem is stated)

    Two players each put one dollar into a pot. They then decide to throw a pair of dice alternately. The first one who throws a sum of 5 wins the pot. How much should the player who starts add to the pot to make this a fair game?

    I think it is a geometric distribution.
    Which is defined as p(x)= p(1-p)^x and the mean is (1-p)/p.
    I also know the probability of 2 dice summing to 5 is 4/36 or approx. .1111.
    I can determine the probability that player A wins when P{X=1}, P{X=3}...
    Similarly, for player B P{X=2}, P{X=4}
    The professor also gave the following clues:
    P{A wins}=? P{B wins} =?
    E[A's reward]=P{A wins}*1
    E[B's reward]=P{B wins}*(1+$x)

    But I am lost.
    Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by kid funky fried View Post
    I am am trouble with the following homework problem.
    (Note:This is exactly how the problem is stated)

    Two players each put one dollar into a pot. They then decide to throw a pair of dice alternately. The first one who throws a sum of 5 wins the pot. How much should the player who starts add to the pot to make this a fair game?

    I think it is a geometric distribution.
    Which is defined as p(x)= p(1-p)^x and the mean is (1-p)/p.
    I also know the probability of 2 dice summing to 5 is 4/36 or approx. .1111.
    I can determine the probability that player A wins when P{X=1}, P{X=3}...
    Similarly, for player B P{X=2}, P{X=4}
    The professor also gave the following clues:
    P{A wins}=? P{B wins} =?
    E[A's reward]=P{A wins}*1
    E[B's reward]=P{B wins}*(1+$x)

    But I am lost.
    Any help would be greatly appreciated.
    Calculate Pr(A wins). Call it p.

    Pr(B wins) = 1 - p.

    For a fair game, the expected winnings of each player should be zero.

    If A wins s/he wins $1. If A loses s/he 'wins' -$(1 + x).

    Therefore p(1) - (1-p)(1 + x) = 0.

    Solve for x.
    Last edited by mr fantastic; October 21st 2008 at 10:07 PM. Reason: The last line was dopey
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Calculate Pr(A wins). Call it p.

    Pr(B wins) = 1 - p.

    For a fair game, the expected winnings of each player should be zero.

    If A wins s/he wins $1. If A loses s/he 'wins' -$(1 + x).

    Therefore p(1) - (1-p)(1 + x) = 0.

    Solve for x.


    I haven't done the calculation but symmetry tells me that p is probably equal to 2/3 ......
    Painfully, I must ask-
    Is the probability of player A winning 4/36???
    How do i determine this?

    p(1).. is this 4/36 * (32/36)^1 ???
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  4. #4
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    Quote Originally Posted by kid funky fried View Post
    Painfully, I must ask-
    Is the probability of player A winning 4/36???
    How do i determine this?

    p(1).. is this 4/36 * (32/36)^1 ???
    No. That would mean the proability of B winning was 32/36. But obviously the person who starts first must have the advantage (because they can win on the very first throw, before their opponent has even thrown). In fact, 4/36 = 1/9 is the probability of either player winning on a single throw.

    There are many different way of doing the calculation.

    Let the probability of A winning be p and the probability of B winning be q.

    Let the probability of success in a single throw be r.

    p = r + (1-r)^2 r + (1 - r)^4 r + .... = r (1 + [1-r]^2 + [1-r]^4 + ....)

    q = (1-r) r + (1 - r)^3 r + (1-r)^5 r + .... = r (1-r) (1 + [1-r]^2 + [1-r]^4 + ....) = r (1 - r) p.

    Therefore:

    p + q = 1 .... (1)

    q = r (1 - r) p .... (2)

    Solve equations (1) and (2) simultaneously:

    p = \frac{1}{2 - r} and q = \frac{1- r}{2 - r}.

    In your case r = \frac{4}{36} = \frac{1}{9}.

    Therefore p = \frac{1}{2 - \frac{1}{9}} = \frac{9}{17}, showing that A has (as expected) a slight advantage over B by starting first.

    Note that as r --> 1, p --> 1 and q --> 0 (as expected).
    Last edited by mr fantastic; October 21st 2008 at 11:58 PM.
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  5. #5
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    Thanks Mr Fantastic. I do not think I would have gotten the answer without your help. But, on the positive side, I do understand now how you got the answer.
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