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Place $N\ge 2$ books in a row and distinguish one of them. At each step select the distinguished book with Probability $p\in (0,1)$ and each of the other $N-1$ books with Probability $q=(1-p)/(N-1)$. Swap the selected book with its left neighbour; selecting the leftmost book leaves the ordering unchanged.

The distinguished book's position $I\in \{1,\dots ,N\}$ is a birth-death chain. From an interior position it moves left with Probability $p$, moves right with Probability $q$ when the book immediately to its right is selected, and otherwise holds. Missing outward moves at the endpoints become holding. Detailed balance therefore gives ${\pi }_{i}p={\pi }_{i-1}q$ for $i\ge 2$. With $r=q/p$,This finite Truncated geometric distribution is valid even when $r>1$: both numerator and denominator then have the same negative sign. The uniform case is $p=q=1/N$, when every book is selected equally often. Large selection Probability for the distinguished book biases it toward the left end. The strict range of $p$ ensures positive motion in both directions and avoids division by zero in the ratio.