Variance of the number of adjacent changes in a coin sequence

Question

A fair coin is flipped 10 times. Look at each of the 9 neighboring pairs of flips. Let X count how many of those pairs differ (one head and one tail, in either order). What is the variance of X?

Accepted Answer

Write X as a sum of 9 counters: Ik = 1 if flips k and k+1 differ. A specific neighboring pair differs with chance 	frac12 (the outcomes HT and TH out of four equally likely), so each counter has variance 	frac12 \cdot 	frac12 = 	frac14. Covariance between counters. Non-overlapping counters are independent, giving covariance 0. Now check an overlapping neighbor pair, positions 1-2 and 2-3: both differ means the sequence goes high-low-high or low-high-low, that is HTH or THT, which is 2 of the 8 equally likely three-flip patterns, chance 	frac14. Compare with the product of the separate ch…

Variance of the number of adjacent changes in a coin sequence

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