4. A fair die is rolled three times continuously, and the numbers facing up when it lands form an arithmetic sequence. The probability of this happening is
A. $\frac{1}{3}$
B. $\frac{1}{12}$
C. $\frac{1}{18}$
D. $\frac{1}{36}$
The number of arithmetic sequences is $6+2 A_{3}^{2}=18$, let $A=$ “the numbers facing up form an arithmetic sequence”, then $P(A)=\frac{18}{6^{3}}=\frac{1}{12}$, so the answer is $B$.