Let's generalize the description of the problem slightly, so that item $R_i$ has weight $w_i$, and is moved to the front with probability $p_i = \frac{w_i}{w_1 + w_2 + \dots + w_N}$. Let $f(w_1, w_2, \dots, w_N)$ be the stationary probability of state $(R_1,R_2,\dots,R_N)$: by symmetry, that's the only probability we need to find.
Start the Markov chain in the stationary distribution and then make one more step from there. This will keep the distribution stationary, but make it easy to see that item $R_1$ is in front with probability $\frac{w_1}{w_1 + w_2 + \dots + w_n}$.
Moreover, given that item $R_1$ is in front, the order of $R_2, \dots, R_N$ is distributed exactly as it was in the stationary distribution, because it did not change. If we forget about $R_1$ entirely, we get a size-$(N-1)$ problem with weights $w_2, \dots, w_N$; therefore the probability that $R_2, \dots, R_N$ appear in that order is $f(w_2, \dots, w_N)$.
Multiplying, we get
$$
f(w_1, w_2, \dots, w_N) = \frac{w_1}{w_1 + \dots + w_n} f(w_2, \dots, w_N)
$$
from which the formula
$$
f(w_1, w_2, \dots, w_N) = \prod_{i=1}^N \frac{w_i}{w_i + \dots + w_N}
$$
immediately follows.
