Rolls until two sixes in a row

Question

You roll a fair six-sided die until you get two 6s in a row. Any non-six resets your current streak of sixes to zero. On average, how many rolls does that take?

Accepted Answer

Let ej be the expected remaining rolls with a current streak of j sixes (j = 0, 1), and e2 = 0. Each roll shows a six with probability p = 1/6 (advance) or something else with probability 5/6 (crash to 0): e0 = 1 + 	frac16 e1 + 	frac56 e0, \qquad e1 = 1 + 	frac16 e2 + 	frac56 e0. From the second line, e1 = 1 + 	frac56 e0. From the first, 	frac16 e0 = 1 + 	frac16 e1, so e0 = 6 + e1 = 6 + 1 + 	frac56 e0, giving 	frac16 e0 = 7 and e0 = 42. The general rule for a run of k successes each with probability p is E = \frac{1}{p} + \frac{1}{p^2} + \cdots + \frac{1}{p^k}. Here p = 1/6, k = 2: E…

Rolls until two sixes in a row

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