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 values of the differences between the numbers of all adjacent balls on the circle. Find the probability of the arrangement that minimizes $S$.
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. 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.
$$
\begin{array}{l}
\text { Then }\left|1-x_{1}\right|+\left|x_{1}-x_{2}\right|+\cdots+\left|x_{k}-9\right| \\
\geqslant\left|\left(1-x_{1}\right)+\left(x_{1}-x_{2}\right)+\cdots+\left(x_{k}-9\right)\right| \\
=|1-9|=8 .
\end{array}
$$

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

Among the numbers $1,2, \cdots, 9$, excluding 1 and 9, the remaining seven ball numbers $2,3, \cdots, 8$ can be divided into two subsets. The subset with fewer elements has $\mathrm{C}_{7}^{0}+\mathrm{C}_{7}^{1}+\mathrm{C}_{7}^{2}+\mathrm{C}_{7}^{3}=2^{6}$ cases, each corresponding to a unique arrangement on the circle that minimizes the value of $S$, i.e., the number of favorable events is $2^{6}$. Therefore, the required probability $p=\frac{2^{6}}{\frac{8!}{2}}=\frac{1}{315}$.