If three dice are thrown, the probability that they show the numbers in A.P. is

$n\left(S\right)=6^3=216$

E= numbers in A.P  

$\text{ A.P. with common difference one are }=4=\left(123\right),\left(234\right),\left(345\right),\left(456\right)$

$\text{ A.P. with common difference two are }=2=\left(135\right),\left(246\right)$

$\text{ A.P. with common difference zero are }=6$

$n\left(E\right)=2+4+6=12$

$\text{ Required probability }P=\frac{12}{216}=\frac{1}{18}$