smc

Call this cube $ABCDEFGH$, with $A$ being the starting point. Because in $7$ moves the bug has to visit the other vertices, the bug cannot revisit any vertex. Therefore, starting at $A$, the bug has a $\frac{3}{3}$ chance of finding a good path to the next vertex, and call it $B$. Then, the bug has a $\frac{2}{3}$ chance of reaching a new vertex next. Call this $C$. $A,B,$ and $C$ are always on the same plane. Now, we split cases. Case $1$: The bug goes to the vertex $E$ opposite $A$ on the space diagonal with probability $\frac{1}{3}$. Then, the bug has to visit $D$ on the plane of $ABC$ last, as there is no way in and out from $D$. Therefore, there is only $1$ way out of $81$ to get to $D$ last. Therefore, there is a $\frac{1}{243}\cdot \frac{6}{9}=\frac{6}{2187}$ chance of finding a good path in this case. Case $2$: The bug goes to vertex $D$ on plane $ABC$ with a chance of $\frac{1}{3}$. The bug then has only $1$ way to go to a point $E$ on the opposite face, therefore having a $\frac{1}{3}$ probability. Then, the bug has a choice of two vertices on the face opposite to $ABCD$. This results in a $\frac{2}{3}$ probability of finding a good path to a point $F$. Then, there is only $1$ way out of $9$ to visit both other vertices on that face in $2$ moves. Multiply the probabilities for this case to get $\frac{2}{243}\cdot \frac{6}{9}=\frac{12}{2187}$. Add the probabilities of these two cases together to get $\frac{18}{2187}=\boxed{\frac{2}{243}}.$