I realized that an earlier question I'd posted was actually different than what I was actually asking.
My question is: say we have a game where you win 2 dollars, or lose 1 dollar, both with probability 1/2. What is the probability that you end up with exactly the same amount of money that you started with?
Is this equivalent to a ruin problem against an infinitely rich adversary?
To stop on your original position you need a multiple of three steps (twice as many $-1$ as $+2$ steps).
The probability of stopping after the first three steps is $\frac{3}{8} = 0.375$: eight possibilities for the steps, of which three end up at a net zero gain. In general the probability of first stopping after $3n$ steps is
$$\frac{{3n \choose n}}{2^{3n-1}(3n-1)}. $$
So the probability of ever stopping, rather than heading off to $+\infty$, is
$$\sum_{n=1}^{\infty} \frac{{3n \choose n}}{2^{3n-1}(3n-1)} =\frac{3}{4} \left(3-\sqrt{5}\right) \approx 0.572949\ldots$$