Solution
Return to Noon
Show the problem again
A clock hand points at 12. Each second it moves to one of the two adjacent hour marks, forward or backward with probability 1/2 each. What is the expected number of moves until it first points at 12 again? Give an exact number.
Worked solution
The answer is 12. The walk is symmetric on a cycle of 12 positions, so its stationary distribution is uniform, 1/12 per position. For an irreducible Markov chain the expected return time to a state equals 1/π of that state, giving 12. Directly: after the first move the hand is at distance 1, and the hitting time from distance k on a cycle of n is k(n − k) = 11, so 1 + 11 = 12.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.