2006 AMC 10A Problem 25

After $7$ moves there are $3^7 = 2187$ equally likely walks. A successful walk visits every vertex exactly once. Label the cube’s vertices by binary triples, with adjacent vertices differing in one coordinate. There are $3$ choices for the first move and $2$ for the second move if the bug is not to return to its starting point. By symmetry, fix these first moves as $000\to100\to110.$ The successful continuations are exactly $$\begin{aligned} 110&\to111\to101\\ &\to001\to011\to010, \end{aligned}$$ $$\begin{aligned} 110&\to010\to011\\ &\to001\to101\to111, \end{aligned}$$ and $$\begin{aligned} 110&\to010\to011\\ &\to111\to101\to001. \end{aligned}$$ Thus each allowed pair of first moves has $3$ successful continuations, giving $3\cdot2\cdot3=18$ successful walks. The probability is $\frac{18}{2187} = \frac{2}{243}.$ Thus, the correct answer is C.