Separation chain for two labels in a circular transposition shuffle

Place six positions on a cycle and distinguish two labels. At each step choose one of the fifteen unordered pairs of distinct positions uniformly and transpose their contents. This convention excludes identity moves. The shorter circular distance between the distinguished labels takes values $1,2,3$. Cycle symmetry and exchangeability of the other labels make this distance a Lumped Markov chain.

Counting transpositions gives the transition matrix and Stationary distributionThe stationary weights also follow by counting unordered position pairs: six pairs have distance one, six have distance two, and three are opposite. If $\Pi$ has every row equal to $\pi$, then $P=\frac{1}{3}I+\frac{2}{3}\Pi$ and $P^n=3^{-n}I+(1-3^{-n})\Pi$. Starting with adjacent labels, the Probability of opposite labels after $n$ steps is therefore $(1-3^{-n})/5$.

The parent Random transposition shuffle uses a different convention: two positions are sampled independently, allowing an identity move with Probability $1/6$. Under that convention the distance transition matrix is $\tilde{P}=\frac{1}{6}I+\frac{5}{6}P=\frac{4}{9}I+\frac{5}{9}\Pi$, and the opposite-label Probability is $(1-(4/9{)}^n)/5$. Explicitly stating the sampling convention prevents a spurious change in the convergence rate.