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Math Help - A "Random Walk" Problem

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    A "Random Walk" Problem

    Peter and Paul bet one dollar each on each game. Each is willing to allow the other unlimited credit.Use a calculator to make a table showing, to four decimal places, for each of p =1/10, 1/3, 0.49, .499, 0.501, 0.51, 2/3, 9/10 the probabilities that Peter is ever ahead by $10, by $100, and by $1000.

    Im not too sure what to do with this problem. Can someone help me get started with this?

    Attempt. For P = 1/10 and Peter is ahead by $10.
    Q = 9/10

    I tried using this formula
    P* = (1-r^s)/(1 - r^t)
    where
    P* is the probability that Peter wins all the money.
    r = q/p (q not equal p)
    Peter starts with s dollars
    Paul starts with t - s dollars.

    But the problem does not give the initial amount for each person.

    Okay, another attempt. The probability that peter is ever ahead by k games is
    {1 if p>=q
    { (p/q)^k if p < q


    Im doing the case where Peter is ahead by $10.
    p =1/10, 1/3, 0.49, .499, 0.501, 0.51, 2/3, 9/10

    by above since p >=q 0.501, 0.51, 2/3, 9/10.
    The probability that Peter is ever ahead is 1.

    For p = 1/10 , q = 9/10 , k =10
    I got 2.867E-10

    For p = 1/3 , q = 2/3 , k =10
    I got 1/1024

    For p = 0.49 , q = 0.51 , k =10
    I got 0.6703

    For p = .499 , q = .501 , k =10
    0.960789

    I definitely not too sure about this method. Please leave a comment if you have any ideas.
    Last edited by jjbrian; April 25th 2010 at 04:45 PM.
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