Three. (50 points) Nine small balls numbered $1,2, \cdots, 9$ are randomly placed on nine equally spaced points on a circle, with one ball on each point. Let $S$ be the sum of the absolute differences of the numbers on all adjacent pairs of balls. Find the probability of the arrangement that minimizes $S$.

Note: If one arrangement can be rotated or reflected to coincide with another arrangement, they are considered the same.
Three, nine differently numbered small balls are placed on nine equally divided points on a circle, with one ball on each point, equivalent to a circular permutation of nine different elements, with a total of $8!$ ways of placement. Considering the factor of flipping, the number of essentially different ways is $\frac{8!}{2}$.

Next, we seek the number of ways to achieve the minimum value of $S$: On the circle, from 1 to 9, there are two paths, the major arc and the minor arc. For any one of these paths, let $x_{1}, x_{2}, \cdots, x_{k}$ be the numbers of the balls sequentially arranged on this arc.
$$
\begin{array}{l}
\text { Then }\left|1-x_{1}\right|+\left|x_{1}-x_{2}\right|+\cdots+\left|x_{k}-9\right| \\
\geqslant\left|\left(1-x_{1}\right)+\left(x_{1}-x_{2}\right)+\cdots+\left(x_{k}-9\right)\right| \\
=|1-9|=8 \text {. } \\
\end{array}
$$

The equality holds if and only if $1<x_{1}<x_{2}<\cdots<x_{k}<9$, meaning the numbers of the balls on each arc are in increasing order from 1 to 9.
Therefore, $S_{\text {min }}=2 \times 8=16$.
From the above, we know that when the set of ball numbers $\left\{1, x_{1}, x_{2}\right.$, $\left.\cdots, x_{k}, 9\right\}$ on each arc is determined, the sorting scheme that achieves the minimum value is uniquely determined.

Among $1,2, \cdots, 9$, except for 1 and 9, the remaining seven ball numbers $2,3, \cdots, 8$ can be divided into two subsets, with the subset having fewer elements having $\mathrm{C}_{7}^{0}+\mathrm{C}_{7}^{1}+\mathrm{C}_{7}^{2}+\mathrm{C}_{7}^{3}=2^{6}$ cases. Each case corresponds to a unique arrangement on the circle that makes the value of $S$ reach its minimum, i.e., the total number of favorable events is $2^{6}$. Therefore, the required probability $p=\frac{2^{6}}{\frac{8!}{2}}=\frac{1}{315}$.