Expected rolls until two sixes in a row

Question

You roll a fair six-sided die repeatedly. What is the expected number of rolls until you see two sixes in a row?

Accepted Answer

Set up progress states. Let E0 be the expected rolls remaining with no active six, and E1 with exactly one six just rolled. From state 0: with probability 	frac16 move to state 1, else stay: E0 = 1 + 	frac16 E1 + 	frac56 E0. From state 1: with probability 	frac16 finish, else back to scratch: E1 = 1 + 	frac16\cdot 0 + 	frac56 E0. The first equation simplifies to 	frac16 E0 = 1 + 	frac16 E1, i.e. E0 = 6 + E1. Substituting the second: E0 = 6 + 1 + 	frac56 E0 \implies 	frac16 E0 = 7 \implies E0 = 42. The expected number of rolls is 42, neatly, 6 + 6^2. Why 6 + 36? For a "success" of pr…

Expected rolls until two sixes in a row

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