- #1

h0dgey84bc

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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?