7. A fair cube die, with six faces marked with the numbers $1,2,3,4,5,6$. Randomly toss the die three times (each toss is independent of the others), and the numbers obtained are $a_{1}, a_{2}, a_{3}$, then the probability of the event “ $\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{1}\right|=6$ ” is $\qquad$.
Assume $a_{1} \geqslant a_{2} \geqslant a_{3} \Rightarrow a_{1}-a_{2}+a_{2}-a_{3}=a_{1}-a_{3}$, let $x=a_{1}-a_{2}, y=a_{2}-a_{3} \Rightarrow x+y=3$, we get a total of 12 cases:
$$
6,6,3 ; 5,5,2 ; 4,4,1 ; 6,5,3 ; 5,4,2 ; 4,3,1 ; 6,4,3 ; 5,3,2 ; 4,2,1 ; 6,3,3 ; 5,2,2 ; 4,1,1 \text {, }
$$

Thus, the cyclic results of $a_{1}, a_{2}, a_{3}$ are $6 \times 3+6 \times 6=54$ cases.
Therefore, the probability of the event occurring is $P=\frac{54}{6^{3}}=\frac{1}{4}$.