Nice question! Do you have a source? Where did you see it?
Here is my approach:
A cube has 8 vertices.
1 is already taken (original vertex)
6 are neighbors.
1 is the most distant.
Here I can claim that long-run probability of being in 6-neighbor vertices are equal. (Why?)
This should reduce you equations to only 3 (I think) and it should be solvable. Mine is just a guess. Hope this helps.