14. Randomly throwing four dice, the probability that the sum of the numbers shown on two of the dice is 9 is $\qquad$ .
14. $\frac{145}{324}$.

Let the result of each roll of four dice be a basic event, and let $X$ represent the set of all basic events. Clearly, $|X|=6^{4} (|X|$ represents the number of elements in set $X$).
For each $i (i=1,2, \cdots, 6)$, define the event $A_{i}$:
$A_{i}=\{$ all basic events where the number shown on the four dice does not include $i\}$.
Let the event where the sum of the numbers shown on two of the dice is 9 be denoted as $B$.
Since $9=3+6=4+5$, we have:
$$
B=\overline{A_{3} \cup A_{6}} \cup \overline{A_{4} \cup A_{5}},
$$

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

Similarly, $\left|\overline{A_{4} \cup A_{5}}\right|=302$.
$$
\begin{array}{l}
\text { Therefore, } |B|=\left|\overline{A_{3} \cup A_{6}}\right|+\left|\overline{A_{4} \cup A_{5}}\right|-\mid \overline{A_{3} \cup A_{6}} \cap \overline{A_{4} \cup A_{5} \mid} \\
=604-24=580 .
\end{array}
$$

Thus, the required probability is $\frac{|B|}{|X|}=\frac{580}{6^{4}}=\frac{145}{324}$.