Expected spins to land red twice in a row

Question

A spinner lands on red with probability 	frac13 and on some other color with probability 	frac23, each spin independent. You spin until red appears on two consecutive spins. What is the expected number of spins until you stop?

Accepted Answer

Two states: A = last spin not red (or start), B = last spin red. Let eA, eB be the expected additional spins, with red probability p = 	frac13. eA = 1 + 	frac13 eB + 	frac23 eA, \qquad eB = 1 + 	frac13(0) + 	frac23 eA. The first gives 	frac13 eA = 1 + 	frac13 eB, so eA = 3 + eB. Substitute into the second: eB = 1 + 	frac23(3 + eB) = 3 + 	frac23 eB, hence 	frac13 eB = 3, eB = 9, and eA = 12. Starting fresh: eA = 12. Takeaway. Using \frac{1+p}{p^2} with p = 	frac13 gives \frac{4/3}{1/9} = 12 spins for two reds in a row.

Expected spins to land red twice in a row

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