Frog tour of eight pads with a single backward hop

Eight lily pads in a row are numbered $1$ to $8$ from left to right, and a frog starts on pad $1$. On every hop it jumps to a pad it has never visited, chosen uniformly from those still unvisited, until all eight pads have been visited. What is the probability that exactly one of its hops goes to the left?

$\tfrac{1}{42}$. The visiting order of pads $2,\dots,8$ is a uniformly random permutation of seven items, and leftward hops are its descents (the first hop is always rightward). Permutations of $7$ items with exactly one descent number $2^7-8=120$ (the Eulerian number), so the probability is $\tfrac{120}{5040}=\tfrac1{42}$.