4. A fair die is rolled three times consecutively. The probability that the numbers facing up form an arithmetic sequence is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{12}$
(C) $\frac{1}{18}$
(D) $\frac{1}{36}$
4. B.

It is easy to know that the common differences are $0, 1, 2, -1, -2$, corresponding to the number of cases $6, 4, 2, 4, 2$, totaling 18 cases.
Therefore, the required probability is $\frac{18}{6 \times 6 \times 6}=\frac{1}{12}$.