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Variation of Gambler's ruin problem

  1. Dec 3, 2011 #1

    In the usual Gambler's ruin problem one calculates the probability of reaching some target balance N before going broke, given that one starts with holdings of 'h', and given that on each bet he either increases or decreases his balance by +/-1. See http://www.mathpages.com/home/kmath084/kmath084.htm for an excellent discussion.

    I'd like to do something similar to this but slightly different, instead of assuming the gambler plays indefinitely until he reaches target balance N or goes bankrupt, I'd like assume the gambler plays until he either goes bankrupt, or has placed 'n' bets (as opposed to reaching some target balance N).

    For example, the gambler starts with 100, he plans to play until he either goes bankrupt or has placed n=45 bets at which point he will stop. What's the probability the he will go bankupt before having placed these 45 bets? (each bet will raise or lower his balance by +/-1)
  2. jcsd
  3. Dec 3, 2011 #2
    I *think* if [itex] P_{hn} [/itex] is the probability of going bankrupt given current holdings of 'h' and given the gambler has 'n' bets to place. Must obey the 2-dimensional recursion relation:

    [itex] P_{hn}=\alpha P_{(h+1,n-1)}+(1-\alpha)P_{(h-1,n-1)} [/itex] where [itex] \alpha [/itex] is the probability of a bet winning. The boundary conditions would then be [itex] P_{0n}=1 [/itex] (already bankrupt), and [itex] P_{h0}= 0 [/itex] (for nonzero h, given that no bets left, no chance of bankruptcy).

    I have no clue how to obtain the solution of this though.
  4. Dec 3, 2011 #3
    I found the answer here if anyone is interested: http://www.math.ucdavis.edu/~blakehunter/Masters.pdf [Broken] it's called the "Gambler's ruin in finite time"...turned out to be a much more complicated calc than I had imagined..
    Last edited by a moderator: May 5, 2017
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