4. If 4 dice are thrown at random, the probability that the sum of the numbers on two of them is 9 is $\qquad$ .
4. $\frac{145}{324}$.

Let $A_{i}=\{4$ dice do not show $i\}(i=1,2, \cdots, 6)$.
Then $\left|\overline{A_{3}} \cap \overline{A_{6}}\right|=\left|\overline{A_{3} \cup A_{6}}\right|=6^{4}-\left|A_{3} \cup A_{6}\right|$
$$
\begin{array}{l}
=6^{4}-\left(\left|A_{3}\right|+\left|A_{6}\right|-\left|A_{3} \cap A_{6}\right|\right) \\
=6^{4}-\left(5^{4}+5^{4}-4^{4}\right)=302 .
\end{array}
$$

Similarly, $\left|\overline{A_{4} \cup A_{5}}\right|=302$.
$$
\begin{array}{l}
\text { Therefore, }\left|\left(\overline{A_{3}} \cap \overline{A_{6}}\right) \cup\left(\overline{A_{4}} \cap \overline{A_{5}}\right)\right| \\
=\left|\overline{A_{3} \cup A_{6}} \cup \overline{A_{4} \cup A_{5}}\right| \\
=\left|\overline{A_{3} \cup A_{6}}\right|+\left|\overline{A_{4} \cup A_{5}}\right|-\left|\overline{A_{3}} \cap \overline{A_{6}} \cap \overline{A_{4}} \cap \overline{A_{5}}\right| \\
=604-4!=580 .
\end{array}
$$

Therefore, the required probability is $P=\frac{580}{6^{4}}=\frac{145}{324}$.