Random walk hitting times comparison

Question

If we take $(S_n)_{n\geq0}$ a random walk st $S_0 = 0$, do we have a neat formula for $\mathbb{P}_0[\tau_0^+ > \tau_K]$ where $\tau_K = \inf\{n\geq0: S_n = K\}$ and $\tau_0^+ = \inf\{n\geq1: S_n = 0\}$ ? Or an equivalent as $K\to+\infty$? (My guess is that it is equivalent to $\tfrac{1}{K}$, but not sure how to prove it)

Answer 1

Assume $K>2$, otherwise it's easy. The first two moves must be to the right if $K$ is hit before $0$. $P_{0}[\tau_{0}^{+}>\tau_{K}]\\ =P_{0}[X_{1}=1,X_{2}=1, S_{n}-X_{1}-X_{2} \text{ will first hit $K-2$ before hitting $-2$ for n>2}]\\ =1/4*P_{0}[\text{the walk will first hit $K-2$ before hitting $-2$}]\\ =1/4*(2/K)\\ =1/(2K)$

where the second to last equality follows by a classical computation in the Gambler's ruin problem