One step from the edge

Idea. Let $P$ be the probability of ever falling from one step away. Falling from two steps away means first coming back to one step away and then falling, which has probability $P^2$. Conditioning on the first step gives an equation for $P$.

Step 1: the equation. From one step away, he falls at once with probability $\frac{1}{3}$, or moves to two steps away with probability $\frac{2}{3}$, from where he falls with probability $P^2$: $$P=\frac{1}{3}+\frac{2}{3}{P}^2.$$

Step 2: solve. $2P^2-3P+1=0$, that is $(2P-1)(P-1)=0$, so $P=\frac{1}{2}$ or $P=1$.

Step 3: the right root. Put a far wall at distance $N$: he stops if he reaches it. This is gambler's ruin with ratio $r=\frac{1/3}{2/3}=\frac{1}{2}$, and the probability of reaching the edge before the wall is $$\frac{r-r^N}{1-r^N}\to r=\frac{1}{2}(N\to \infty ).$$ Since the event of falling is the increasing limit of falling before reaching distance $N$, the answer is $P=\frac{1}{2}$.

In general. If each step goes away from the edge with probability $p$, he falls with probability $\frac{1-p}{p}$ when $p>\frac{1}{2}$, and for sure when $p\le \frac{1}{2}$, even though for $p=\frac{1}{2}$ the expected time to fall is infinite.