Next, we calculate the number of arrangements, which make $S$ the minimum. Along the circle there are two routes from $1$ to $9$, the major arc and the minor arc. For each of them, let $x_1, x_2, \dots, x_k$ be the numbers of the successive balls on the arc, then
$$\begin{aligned} & |1-x_1| + |x_1-x_2| + \dots + |x_k-9| \\ \ge & |(1-x_1) + (x_1-x_2) + \dots + (x_k-9)| \\ = & |1-9| = 8. \end{aligned}$$
The equality occurs if and only if $1 < x_1 < x_2 < \dots < x_k < 9$, i.e. the numbers of the balls on each route is increasing from $1$ to $9$.

Therefore, $S_{\min} = 2 \cdot 8 = 16$.

From the above analysis, when the numbers of the balls $\{1, x_1, x_2, \dots, x_k, 9\}$ on each arc are fixed, the arrangement which gets the minimum value is uniquely determined. Divide the set of $7$ balls $\{2, 3, \dots, 8\}$ into two subsets, then the subset which contains less elements has $C_9^0 + C_9^1 + C_9^2 + C_9^3 = 2^6$ cases. Each case corresponds to a unique arrangement, which achieves the minimum value of $S$.

Thus, the number of the arrangements when $S$ takes the minimum value is $2^6$ and the corresponding probability is $p = \frac{2^6}{\frac{8!}{2}} = \frac{1}{315}$.