Bug on a square, steps to the opposite corner

Question

A bug sits at corner A of a square with corners A, B, C, D and edges A-B, B-C, C-D, D-A. Each minute it walks along an edge to one of the two neighboring corners, chosen at random. Corner C is diagonally opposite A.

On average, how many minutes until the bug first reaches corner C?

Accepted Answer

Let a, b, d be the expected minutes to reach C from corners A, B, D. By the mirror symmetry of the square, b = d. Each move costs 1 minute plus the expected time from wherever you land, and once at C nothing more is needed: a = 1 + 	frac12 b + 	frac12 d = 1 + b, \qquad b = 1 + 	frac12 a + 	frac12 \cdot 0 = 1 + 	frac a2. Substitute the second into the first: a = 1 + \left(1 + 	frac a2ight) = 2 + 	frac a2 \implies 	frac a2 = 2 \implies a = 4. Takeaway. Expected hitting times come from one equation per corner, "1 plus the average over neighbors"; symmetry shrinks the system, and on a s…

Bug on a square, steps to the opposite corner

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