Andy normally strolls across the hexagons of a truncated tetrahedron. Every hexagon that isn’t his home hexagon is adjacent to it, so as soon as he moves back to what he thinks is his home hexagon on the tiled floor, he finds out he’s not on the truncated hexagon unless he also happens to return to the home hexagon there (the moments when Andy encounters a red hexagon instead of his home green hexagon, in this image we assume without loss of generality that Andy’s first move was moving ‘southeast’). A nice system of equations can be set up, reduced to 4 states using the symmetry of the 6-cycle of hexagons he moves into (which he interprets to be the 3-cycle of blue nonhome hexagons on the truncated tetrahedron).

Let p_i be the probability that, starting i hexagons away from the tile adjacent to his home hexagon, Andy leaves the 6-cycle from the tile adjacent to his home hexagon. The blue 6-cycle hexagons are labeled with their value of i on the tiled floor. So p₀ is the probability he doesn’t notice he’s not on the truncated tetrahedron anymore. Since every move has probability 1/3, we can deduce

p₀ = 1/3 + 2/3 * p₁

p₁ = 1/3 * p₀ + 1/3 * p₂

p₂ = 1/3 * p₁ + 1/3 * p₃

p₃ = 2/3 * p₂

These can be solved to determine that p₀ = 9/20. We were asked for the probability that Andy discovers he isn’t on the truncated tetrahedron, which is the complement of this probability, 11/20.

Congrats to this month’s solvers!