[jay_shark]

This is Gambler's Ruin .

Two Gamblers A and B bet on the outcomes of successive flips of a coin . For each flip , if the coin comes up heads , A collects 1 unit from B whereas if it comes up tails ,A pays 1 unit to B . They continue to do this until one runs out of money . We can make the assumption that the coin is bias with some probability p of landing heads . A question of interest is to determine the probability that A ends up with all the money if he starts with x units and B has y units .

First you should realize that there are 3 possible outcomes of this gambling game . Either A wins , B wins , or the game goes on forever with no one winning . It turns the last part has probability 0 of occurring and that either A or B must have all the money in the end .

I'll give you the formula without the derivation but if you want the derivation I'll have to charge you ...haha jk .

Here is the formula :

Px = [1-(q/p)^x]/(1-(q/p)^(x+y)) if p is not 1/2
Px = x/(x+y) if p =1/2

So as an example , if A were to start with 10 units and B were to start with 20 units then the probability A wins if p =60% is 98.26% which means A has a RoR of 100-98.26% =1.734% .

So if you are a heads up player and you have a bankroll of 10X your buy in and you play repeatedly against a player with 20X the buy in with p=0.6 , you'll wind up with all the money about 98.26% of the time .