共有 \(4^6\) 种等可能移动。把顶点标为 \(A, B, C, A', B', C'\)，其中带撇号的是对应对顶点。有效结果是一个排列 \(f\)，且 \(f(A) \notin \{A, A'\}\)，其他顶点同理。
有序对 \((f(A), f(A'))\) 有 \(4 \cdot 3 = 12\) 个选择。其中 \(f(A)\)、\(f(A')\) 互为对顶点的有 \(4\) 种，相邻的有 \(8\) 种。
若 \(f(A), f(A')\) 互为对顶点，例如 \(B, B'\)，则 \(\{f(C), f(C')\} = \{A, A'\}\)，且 \(\{f(B), f(B')\} = \{C, C'\}\)，给出 \(4 \cdot 2 \cdot 2 = 16\) 种。
若 \(f(A), f(A')\) 相邻，例如 \(B, C\)，则 \(f(B), f(B')\) 中必须有一个是 \(C'\)，有 \(4\) 个有序选择 \((f(B), f(B'))\)，每个给 \((f(C), f(C'))\) 留下 \(2\) 种，共 \(8 \cdot 4 \cdot 2 = 64\) 种。
概率为 \[ \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}\text{。}\]
所以正确答案是 A。
English
There are \(4^6\) equally likely combinations of moves. Label the vertices \(A, B, C, A', B', C',\) where primed vertices are opposite the corresponding unprimed ones. An ant cannot move to its own vertex or the opposite one, so a valid outcome is a permutation \(f\) with \(f(A) \notin \{A, A'\},\) and similarly for each pair.
There are \(4 \cdot 3 = 12\) ordered choices for \((f(A), f(A')).\) Of these, \(f(A)\) and \(f(A')\) are opposite in \(4\) cases and adjacent in \(8.\)
If \(f(A), f(A')\) are opposite, say \(B, B',\) then \(\{f(C), f(C')\} = \{A, A'\}\) and \(\{f(B), f(B')\} = \{C, C'\},\) giving \(4 \cdot 2 \cdot 2 = 16\) valid combinations.
If \(f(A), f(A')\) are adjacent, say \(B, C,\) then one of \(f(B), f(B')\) must be \(C'\) and there are \(4\) ordered choices for \((f(B), f(B')),\) each leaving \(2\) for \((f(C), f(C')):\) that is \(8 \cdot 4 \cdot 2 = 64\) valid combinations.
Hence the probability is \[ \dfrac{16 + 64}{4^6} = \dfrac{80}{4096} = \dfrac{5}{256}. \]
Thus, the correct answer is A.