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Negative binomial, Poisson, or gambler's ruin?

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    . 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,
    .49, .499, .501, .51, 2/3, 9/10 the probabilities that Peter is ever
    ahead by $10, by $100, and by $1000. (p is the probability of Peter winning the game)

    2. Relevant equations

    3. The attempt at a solution
    (i) Negative binomial distribution:
    p= probability of winning on any given game
    r=number of successes wanted (10, 100, 1000)
    x=number of trials needed to obtain r
    E(X)= r/p
    (So, take r= 10, 100, 1000 and for each plug in the p for Peter’s winning)
    This is only if the problem was misstated when giving "unlimited credit".
    I could define a new variable Peter-Paul=10 (respectively 100 and 1000), but how would I transfer the probabilities for winning a game to the new variable?

    (ii) Poisson distribution (large n):
    P(X=r) = (e^-λ)(λ^r)/r!
    n = trials
    p= probability for each success
    r= number of successes
    λ= n*p

    (But here I have no idea what value I could give lambda, except infinite!!!)

    (iii) Gambler’s ruin:
    P(i,N) = [1-(q/p)^i]/[1-(q/p)^N] , for p not equal to q
    P(i,N) = i/N , for p=q = o.5
    P(i,N) = the "be ahead by N" probability
    p= Peter’s wining probability for a game
    q= Peter’s failure probability for a game
    i = amount of money Peter has overall
    N = amount of money he must be ahead by (10, 100, 1000)

    If "i" is infinite ("unlimited credit"), even taking limits wouldn't get me to any definable probability.

    Thank you very much for any help.
    Last edited: Dec 2, 2009
  2. jcsd
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