Gambling Math: Probability of Bankruptcy Over N Bets

AI Thread Summary
The discussion focuses on calculating the probability of going bankrupt over a series of bets with a starting balance of $1000 and a betting limit of $500 per bet, each with a 50% chance of winning. Initial calculations show that after two bets, the probability of bankruptcy is 25%, and this probability remains the same after three bets since any bankruptcy would have already occurred. By the fourth bet, the probability of going bust increases to 37.5%, factoring in various outcomes while excluding those that lead to bankruptcy after two bets. The poster seeks a general formula for calculating bankruptcy probability over any number of bets, questioning whether the game continues even with accumulated debts. The complexity of the problem suggests that as the number of bets increases, the probability of bankruptcy may converge to 100%.
h0dgey84bc
Messages
159
Reaction score
0
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?
 
Mathematics news on Phys.org
Is the game allowed to finish, even if you're accumulating debts?
 
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 can't seem to get the answer for the life of me, haha, feel like it should converge to 100% as N->infinity...
 
Last edited:
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...

Similar threads

Replies
11
Views
4K
Replies
1
Views
2K
Replies
53
Views
8K
Replies
1
Views
1K
Replies
1
Views
3K
Replies
48
Views
11K
Back
Top