An individual has 50 dollars and is gambling on a series of coin flips. A head results in a win of 2 dollars and a tail results in a loss of 1 dollar. What’s the probability that the person will run out of money?
This is a Gambler's Ruin/Random Walk problem. However I'm not sure how to set up this variation.
I know that the probability of starting at i dollars and reaching a value of N dollars based on a 1 dollar win/loss fair game is:
$P_1 = \frac{i}{N}$
Any help would be much appreciated.
Best Answer
Define $P(n)$ as the chance that he eventually runs out of money if he starts with $n$.The recurrence is $$P(n)=\frac 12P(n-1)+\frac 12P(n+2)\\P(0)=1$$ We can rewrite the recurrence as $P(n+2)=2P(n)-P(n-1)$. The characteristic equation is $x^3=2x-1$ with roots $1, \frac 12(\sqrt 5-1), \frac 12(-1-\sqrt 5)$ or $1,\phi-1,-\phi$.
The general solution is $P(n)=a1^n+b(\phi-1)^n+c(-\phi)^n$ with $a+b+c=1$ to match the condition $P(0)=1.$ $c$ must be $0$. If $c \gt 0$ the probability will be negative for large odd $n$ and greater than $1$ for large even $n$. If $c \lt 0$ the same will be true reversing odd and even. If he starts with a lot of money the chance of going broke becomes small, so $\lim_{n \to \infty}P(n)=0$, which tells us that the coefficient on $1^n$ is zero. We get $$P(n)=(\phi-1)^n$$ and in particular if he starts with just one dollar he has about $0.382$ chance of playing forever.