China Mathematical Competition (Jiangxi)

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 $$\begin{array}{rl}& |1-x_1|+|x_1-x_2|+\cdots +|x_k-9|\\ \ge & |(1-x_1)+(x_1-x_2)+\cdots +(x_k-9)|\\ =& |1-9|=8.\end{array}$$ The equality occurs if and only if $1<x_1<x_2<\cdots <x_k<9$, 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 $p=\frac{2^6}{\frac{8!}{2}}=\frac{1}{315}$.