Solution
Counting Adjacent Heads
Show the problem again
A fair coin is flipped 10 times. Let N be the number of positions i (1 ≤ i ≤ 9) such that flips i and i + 1 are both heads. Find Var(N) as a reduced fraction.
Worked solution
Write N = Σ Iᵢ with Iᵢ = 1 when flips i and i + 1 are both heads. Each Iᵢ has P = 1/4, hence variance 3/16, and the nine of them contribute 27/16. Indicators Iᵢ and Iⱼ are independent when |i − j| ≥ 2, since they use disjoint flips. Neighbours Iᵢ and Iᵢ₊₁ overlap in one flip: P(both) = P(three heads in a row) = 1/8, so Cov = 1/8 − 1/16 = 1/16. There are 8 neighbouring pairs, each counted twice in the variance of the sum, adding 2 · 8 · 1/16 = 1. Total: 27/16 + 1 = 43/16. The answer is 43/16.
Source: Original problem written for AxiomIQ; variance of a sum of dependent indicators.