Random walk around a hexagon

Question

Six stones sit at the corners of a regular hexagon. A bug sits on one corner and each second steps to one of the two neighboring corners, choosing left or right with probability $\tfrac12$ each.

Over a long time, what fraction of its steps does the bug spend at any one chosen corner?

Accepted Answer

We do not need to solve six balance equations. Notice the symmetry in the arrivals: each corner is reached only from its two neighbors, and each neighbor sends the bug there with probability 	frac12. So the total probability of arriving at any corner is 	frac12 + 	frac12 = 1, the same for every corner. Rows sum to 1 (the bug always moves) and now columns sum to 1 too, so the chain is doubly stochastic. That forces the long-run distribution to be even across all six corners. With six corners each sharing the time equally, \pi = 	frac16 	ext{ per corner}. So the bug spends one sixth of its…

Random walk around a hexagon

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