2005 AMC 12B Problem 25

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.