(14) Randomly throw 4 dice, then the probability that the sum of the numbers shown on two of the dice is 9 is $\qquad$ .
(14) For 4 dice, each roll's result is a basic event, denoted by $X$ as the set of all basic events. Clearly, $|X|=6^{4}$, where $|X|$ represents the number of elements in the set $X$. For each $i, i=1,2, \cdots, 6$, define the event $A_{i}$:
$A_{i}=\{4$ dice show numbers without $i$ in all basic events $\}$.
Since $9=3+6=4+5$, the event that the sum of the numbers shown on two of the dice is 9 in each roll is denoted as $B$, then
$$
B=\overline{A_{3} \cup A_{6}} \cup \overline{A_{4} \cup A_{5}} .
$$

Here $\bar{A}$ represents the event complementary to $A$.
$$
\begin{aligned}
\left|\overline{A_{3} \cup A_{6}}\right| & =|X|-\left|A_{3} \cup A_{6}\right| \\
& =6^{4}-\left(\left|A_{3}\right|+\left|A_{6}\right|-\left|A_{3} \cap A_{6}\right|\right) \\
& =6^{4}-\left(5^{4}+5^{4}-4^{4}\right)=302 .
\end{aligned}
$$

Similarly, $\left|\overline{A_{4} \cup A_{5}}\right|=302$, thus we have
$$
\begin{aligned}
|B| & =\left|\overline{A_{3} \cup A_{6}}\right|+\left|\overline{A_{4} \cup A_{5}}\right|-\left|\overline{A_{3} \cup A_{6}} \cap \overline{A_{4} \cup A_{5}}\right| \\
& =604-24=580 .
\end{aligned}
$$

The required probability $=\frac{|B|}{|X|}=\frac{580}{6^{4}}=\frac{145}{324}$.