Flips until four heads in a row

Question

You flip a fair coin until you see 4 heads in a row. Any tail resets your current head streak to zero. On average, how many flips does that take?

Accepted Answer

Let ej be the expected remaining flips when your current streak is j heads, for j = 0,1,2,3, with e4 = 0. Each flip advances the streak by one with probability 1/2 or crashes it to 0 with probability 1/2: ej = 1 + 	frac12 e{j+1} + 	frac12 e0. The pattern from the j=0 line always gives e0 = 2 + e1, and stepping down each level adds another factor. The tidy closed form for a run of k heads with a fair coin is E = 2^{k+1} - 2. For k = 4: E = 2^{5} - 2 = 32 - 2 = 30. You can also read it as 2 + 4 + 8 + 16 = 30, the reciprocals of the chances of holding streaks of length 1,2,3,4. Takeaway. For k…

Flips until four heads in a row

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