Expected draws to get two spades in a row

Question

You draw a card from a well-shuffled deck, note it, and return it, reshuffling each time. A spade appears with probability $\tfrac14$ on each draw. You continue until you draw spades on two consecutive draws.

What is the expected number of draws until you stop?

Accepted Answer

Let A be the state "last draw was not a spade" (or the start) and B be "last draw was a spade." With p = 	frac14, eA = 1 + 	frac14 eB + 	frac34 eA, \qquad eB = 1 + 	frac14(0) + 	frac34 eA. From the first, 	frac14 eA = 1 + 	frac14 eB, so eA = 4 + eB. Substitute: eB = 1 + 	frac34(4 + eB) = 4 + 	frac34 eB, giving 	frac14 eB = 4, eB = 16, and eA = 20. From a fresh start: eA = 20. Takeaway. With p = 	frac14, two in a row costs \frac{1+p}{p^2} = \frac{5/4}{1/16} = 20 draws.

Expected draws to get two spades in a row

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