Three dice whose range is exactly four

Three fair six-sided dice are rolled. What is the probability that the largest value minus the smallest value equals $4$?

$\tfrac29$. The extremes must be $\{1,5\}$ or $\{2,6\}$. For each case, the middle die can repeat an extreme (two multisets with $3$ orderings each) or take one of three interior values ($6$ orderings each), giving $24$ outcomes. So $\tfrac{48}{216}=\tfrac29$.