16. Four fair cubic dice are rolled once. The probability that at least three of the four dice show the same number is ( ).
(A) $\frac{1}{36}$
(B) $\frac{7}{72}$
(C) $\frac{1}{9}$
(D) $\frac{5}{36}$
(E) $\frac{1}{6}$
16. B.

In two scenarios:
(1) If the numbers on the top faces of the four dice are all the same, then the probability is
$$
1 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{6^{3}} ;
$$
(2) If three of the four dice have the same number on the top face, then the probability is
$$
\mathrm{C}_{4}^{1} \times 1 \times \frac{1}{6} \times \frac{1}{6} \times\left(1-\frac{1}{6}\right)=\frac{20}{6^{3}} .
$$

Therefore, the probability that at least three of the four dice have the same number on the top face is
$$
\frac{1}{6^{3}}+\frac{20}{6^{3}}=\frac{21}{216}=\frac{7}{72} .
$$