3. A fair cube die has the numbers $1, 2, 3, 4, 5, 6$ on its six faces. If the die is rolled three times, the probability that the sum of the numbers on the top faces is a multiple of 3 is
3. $\frac{1}{3}$.

Let the numbers on the top faces of the dice after three rolls be $x$, $y$, and $z$.
Notice that the multiples of 3 are
$$
\begin{array}{l}
3=1+1+1, \\
6=1+1+4=1+2+3=2+2+2, \\
9=1+2+6=1+3+5=1+4+4 \\
=2+2+5=2+3+4=3+3+3, \\
12=1+5+6=2+4+6=2+5+5 \\
=3+3+6=3+4+5=4+4+4, \\
15=3+6+6=4+5+6=5+5+5, \\
18=6+6+6 .
\end{array}
$$

Among the $(x, y, z)$ that make $x+y+z$ a multiple of 3, there are 8 groups where all three are different, 6 groups where two are the same, and 6 groups where all three are the same.
Therefore, the required probability is
$$
\left(8 A_{3}^{3}+6 A_{3}^{1}+6\right) \times\left(\frac{1}{6}\right)^{3}=\frac{1}{3} .
$$