Asymmetric ruin probability

I have $50$ dollars and I’m gambling on a series of coin flips. For each head I win $2$ dollars and for each tail I lose $1$ dollar. What’s the probability that I will run out of money?

Hint: Suppose we have $x$ dollars, then the probability of ruin satisfies the recursive equation

$$p(x+2) - p(x) = p(x) - p(x-1)$$

Find function $p(x)$.

Solution 1:

As indicated, the following recursion holds $$p(n+2)-2p(n)+p(n-1)=0.$$ It only has a solutions on the form $r^n$, where $r$ is a root of the characteristic polynomial $$r^3-2r^1+r^0=0.$$ Finding the roots one has that a general solution of $p$ is given by $$p(n)=A1^n+B\left(\frac{-1+\sqrt{5}}{2}\right)^n+C\left(\frac{-1-\sqrt{5}}{2}\right)^n.$$ Since the underlying random walk has increments with positive mean, $p(n)\rightarrow 0$, as $n\rightarrow\infty$. This gives $A=C=0$. Further, $p(0)=1$, so $B=1$. In conclusion, $$p(n)=\left(\frac{\sqrt{5}-1}{2}\right)^n.$$ $p(50)$ is now easy to calculate.

Solution 2:

We only need find the probability $r$ that you lose 1 dollar since the probability that you lose 50 dollars is $r^{50}$ since you must lose 1 dollar 50 times. We can write $$ r = 0.5 + 0.5r^3 $$

since half the time we lose on the first flip, and half the time we win on the first flip, thereby increasing our 1 dollar bankroll to 3 dollars, at which point our risk of ruin becomes $r^3$ since we must now lose 1 dollar 3 times. This has solutions $r = (\sqrt{5}-1)/2$ and $r = 1$. We obviously want the first of these since we have the advantage, and our opponent's $r$ is 1. So $$ P(lose \: 50) = \left(\frac{\sqrt{5}-1}{2}\right)^{50} $$ or about $3.55e^-11$.