In this note (MIT Course: Mathematics for Computer Science), which you can see the pdf link, there is the following term on page 9:
In this case (probability is more than 0.5) there is a non-zero chance that you can play forever and never go broke.
The text is finished here and no further explanation is given. How can that probability be calculated and proven?
One can use a sort of inductive argument. I assume that "probability is more than $\tfrac12$" means that you are more likely to go up than down. For simplicity, let's suppose you go up with probability $\tfrac23$ and down with probability $\tfrac13$.
Let $p_{i \to j}$ denote the probability that at some point you ever hit $i$ starting from $j$ with $i,j \in \{0, 1, 2, ...\}$ and $i \ge j$. Note that $p_{k \to k} = 1$ for all $k$.
Start from $1$. Consider what happens in a single step: either you go down (probability $\tfrac13$) to $0$ and end or go up (probability $\tfrac23$) to $2$ and continue. We thus have $$ p_{1\to0} = \tfrac13 \cdot p_{0\to0} + \tfrac23 \cdot p_{2\to0} = \tfrac13 + \tfrac23 p_{2\to0}.$$
Now, how do we evaluate $p_{2\to0}$? Well, to go from $2$ to $0$ you have to pass via $1$, so $p_{2\to0} = p_{2\to1\to0}$. But by the Strong Markov Property applied at the time at which you hit $1$ (if you ever do), the evolution of the chain before hitting $1$ and after is independent! So $p_{2\to0} = p_{2\to1} p_{1\to0}$. By symmetry of the chain, $p_{2\to1} = p_{1\to0}$. Thus $p_{2\to0} = p_{1\to0}^2$.
Plugging this in and abbreviating $p := p_{1\to0}$ gives $$p = \tfrac13 + \tfrac23 p^2, \quad\text{or equivalently}\quad 2 p^2 - 3 p + 1 = 0.$$ But $2 p^2 - 3 p + 1 = (2 p - 1)(p - 1),$ and so the solutions to this are $1$ and $\tfrac12$.
In a Markov chains course, you should've had proved a theorem about this stuff which says that the correct solution is the minimal one. Thus $p = \tfrac12$.
In words, the chance $p_{1\to0}$ that you "go broke" at some point is non-zero.