Consider the regular octahedron as shown. Each edge of the octahedron has a length of 1. An ant starts at the vertex A and crawls a total distance of 3 units along the edges of the octahedron. Any time the ant reaches a vertex of the octahedron, it randomly chooses an edge to next crawl on that is different from the edge it just left. One such path the ant may take is shown. What is the probability that the ant will end up back at point A ? Express your answer as a common fraction.
+2666 There are \(4 \times 3 \times 3 = 24\) paths you can take.
Note that the only way to end up at point A is if you go along 1 of the triangles.
There are 4 triangles that contain point A, but there are 2 ways to go around each, making for \(4 \times 2 = 8\) successful paths.
This means that the probability is \({8 \over 24} = \color{brown}\boxed{1 \over 3}\)
