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Gambling mathematics

  1. Jan 30, 2009 #1
    Hi,

    Let's say I have $1000 in a bookmaker, and this bookmaker would only allow bets of $500 at a time, each betting having a probability of 0.5 to win (this means odds of 1:1 or 2.0, so if you win you win 500, if you lose you lose 500 staked). How do I calculate the probability of going bankrupt over N bets, i.e. P(N)?

    I know when number of bets is 2, the outcomes are:

    WW (+500+500. leaving 2000 balance) 25%
    LW,WL =>balance of 1000 still, 50%
    LL=> balance of 0, bankrupt, 25%

    then for 3 bets, the outcomes are(remembering we would of stopped playing if LL had happened and bankrupt us after 2 bets)

    WWW(2500)
    LWW(1500), WWL(1500),WLW(1500)
    LWL(500),WLL(500)

    so no chance of going bust here except if we had already done it after 2 bets, so still 25%.

    After 4 bets we could have (excluding the bets we went bust after 2 times)


    L=0::WWWW(3000)(6.25%)
    L=1::WLWW(2000),WWLW(2000),LWWW(2000),WWWL(2000)(prob is 25%) (4!/3!1!=4 combos with L equals 1, and therefore balance of 2k)
    L=2::LWLW(1000),WLLW(1000),WLWL(1000),WWLL(1000),LWWL(1000)(prob is 31.25%) (4!/2!2!=6 combos with L=2, but one is LLWW, which is bust after two so excluded)
    L=3::LWLL(0),WLLL(0) (prob: 2*(0.5^4)=12.5%.....4!/3!=4 with L=3 , but two are LLLW,LLWL,which are excluded as they bust after two)
    (the other 25% is for times we went bust on first two, i.e LL...)


    Therefore the TOTAL prob of going bust after 4 moves is P(4)=P(2)+12.5%=37.5%


    How do you generalise this to get the probability of busting for any number of bets N?
     
  2. jcsd
  3. Jan 30, 2009 #2
    Is the game allowed to finish, even if you're accumulating debts?
     
  4. Jan 30, 2009 #3
    well you would only stop betting when you went bankrupt, there's no way to build debt, if you keep winning you're making profit on the original 1k. If it makes the problem simpler though I'd still be interested in seeing the answer, when you stop playing and withdraw when you'd won say 10k(not sure if that would actually make it simpler though, since still infinite paths)

    This problem seems like it should be simple, but I cant seem to get the answer for the life of me, haha, feel like it should converge to 100% as N->infinity...
     
    Last edited: Jan 30, 2009
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