[Solution] The arrays that meet the requirements are $(1,2,3),(2,3,4),(3,4,5),(4,5,6)$. Each array can have 3! different permutations.
Furthermore, the total number of different permutations when rolling three dice is $6^{3}$. Therefore, the required probability is
$$
\frac{4 \cdot 3!}{6^{3}}=\frac{1}{9} \text {. }
$$

Thus, the correct choice is $(B)$.
14-22 If three dice are rolled (this means all faces have the same probability). The probability that the three numbers facing up can be arranged into an arithmetic sequence with a common difference of 1 is
(A) $\frac{1}{6}$.
(B) $\frac{1}{9}$.
(C) $\frac{1}{27}$.
(D) $\frac{1}{54}$.
(E) $\frac{7}{36}$.
(28th American High School Mathematics Examination, 1977)