2006 AMC 12A Problem 20

From the start there are $3^7$ equally likely $7$-move walks. For a walk visiting all $8$ vertices, there are $3$ choices for the first move and $2$ for the second, since it cannot return to the starting vertex. Label cube vertices by three-bit strings. By symmetry, after fixing those first two moves we may take the first three vertices to be $000,001,011.$ A branch check gives exactly these three completions: $$\begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010. \end{aligned}$$ Thus there are $3 \cdot 2 \cdot 3 = 18$ such walks. The probability is $\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.$ Thus, the correct answer is C.