Gambler going broke at an exact round

A player begins with $n\ge1$ dollars (an integer). Every round they gain one dollar with probability $p$ or lose one dollar with probability $1-p$, and they stop as soon as their wealth reaches zero. Let $f(p,n,k)$ be the probability that ruin happens at precisely round $n+2k$, for an integer $k\ge0$. Evaluate $f(1/3,5,3)$, rounded to four decimal places.

About 0.1084. Ruin at round $n+2k$ needs $k$ wins and $n+k$ losses, with probability $\binom{n+2k}{n+k}q^{n+k}p^{k}$ for the counts ($q=1-p$); by the ballot theorem (read backwards from ruin) a fraction $\tfrac{n}{n+2k}$ of those paths first hit zero at the last step. With $p=\tfrac13,n=5,k=3$: $\binom{11}{3}\left(\tfrac13\right)^3\left(\tfrac23\right)^8\cdot\tfrac5{11}\approx0.1084$.