A truck transports goods among $10$ points located on a circular route. These goods are carried only from one point to the next with probability $p$, or to the preceding point with probability $q=1-p$.
The transition matrix is the circulant matrix $M = q \cdot P + p \cdot P^T$, where $P$ is the permutation matrix in the link. Computing the stationary distribution can be done by computing the solution to the system $(M - I)x = 0$.
However, rather than solving this system of equations, we can more easily prove that your guess of the stationary distribution $\pi = (1/10,\dots,1/10)$ is correct by verifying that $\pi M = M$. To see that this holds, note that $\pi = \frac 1{10} (1,\dots,1)$, and that the entries of $(1,\dots,1)M$ are the column-sums of $M$. The only non-zero entries of a given column of $M$ are $p$ and $q$, which means that every entry of $(1,\dots,1)M$ will be $p+q = 1$, which means that we have $$ (1,\dots,1)M = (1,\dots,1)M \implies \pi M = \pi. $$ So, $\pi$ is indeed the stationary distribution.
