In a regular octahedron, define the distance $d(A,B)$ between two vertices to be the minimum number of edges one must traverse to get from $A$ to $B$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular octahedron. What is the probability that $d(Q,R)>d(R,S)$?

1. Every vertex $R$ of the octahedron is adjacent to $4$ of the other $5$ vertices, and is at distance $2$ from the remaining one (its antipode).
2. So $d(Q,R)>d(R,S)$ happens only if $d(Q,R)=2$ and $d(R,S)=1$, i.e. $Q$ is the antipode of $R$ and $S$ is any other vertex.
3. Given $R$, the vertex $Q$ is the antipode with probability $\frac15$, and then every possible $S$ is adjacent to $R$. The probability is $\boxed{\dfrac15}$.