= 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 distribution>
$$
P=\frac1{15}\begin{pmatrix}9&4&2\\4&9&2\\4&4&7\end{pmatrix},
\qquad \pi=(2/5,2/5,1/5).
$$
The 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 $P=\tfrac13 I+\tfrac23\Pi$ 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 $\widetilde P=\tfrac16I+\tfrac56P=\tfrac49I+\tfrac59\Pi$, and the opposite-label <Probability> is $(1-(4/9)^n)/5$. Explicitly stating the sampling convention prevents a spurious change in the convergence rate.
Back to article page
