 Solution

Treat one visit to a level as a whole gambler's ruin excursion, starting at its chamber $1$ and ending at local chamber $0$ or $4$. The room choices are fair, so the chance of reaching $4$ before $0$ is $1/4$, and the chance of reaching $0$ first is $3/4$. The one-way hatches reset the next excursion to chamber $1$ of the adjacent level. Thus the embedded Markov chain at level exits moves upward with probability $1/4$ and downward with probability $3/4$, even though individual room choices are unbiased.

Let $u$ be the food probability from the lower starting chamber, and $v$ that from the upper chamber $1$. The lower level either exits unsuccessfully or reaches the upper level, while the upper level either finds food or returns to the lower start. First-step analysis at these level exits givesSubstituting yields $u=1/16+3u/16$, soThis also follows from biased gambler's ruin probability on levels $0,1,2,3$, starting at $1$, with ratio $(1-p)/p=3$: $(1-3)/(1-3^3)=1/13$. Absorption occurs almost surely, since from each active level a fixed positive chance of terminal absorption occurs within two excursions; the original chamber excursions are also almost surely finite.