Variance of the number of adjacent head-head pairs

Question

A fair coin is flipped 10 times. Look at each of the 9 neighboring pairs of flips (positions 1-2, 2-3, and so on). Let X count how many of those pairs are both heads. What is the variance of X?

Accepted Answer

Write X as a sum of 9 little yes/no counters, one per neighboring pair: Ik = 1 if flips k and k+1 are both heads. The trap is that neighboring counters overlap and are not independent, so their covariances matter. Each counter's own variance. A specific pair is HH with chance 	frac14, so each counter has variance 	frac14 \cdot 	frac34 = 	frac{3}{16}. There are 9 of them, contributing 9 \cdot 	frac{3}{16}. Covariance between counters. Two counters that do not share a flip are independent, so their covariance is 0. The only linked pairs are neighbors that share a flip, like the pair at pos…

Variance of the number of adjacent head-head pairs

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