Roll a die, then flip that many coins

Question

You roll a fair six-sided die to get a number N, then flip N fair coins and let X be the number of heads. Use the law of total variance to find \operatorname{Var}(X).

Accepted Answer

Condition on the die roll N. Given N = n, the head count is 	ext{Binomial}(n, 	frac12), so \mathbb{E}[X \mid N] = 	frac12 N, \qquad \operatorname{Var}(X \mid N) = N \cdot 	frac12 \cdot 	frac12 = 	frac14 N. The law of total variance gives \operatorname{Var}(X) = \mathbb{E}[\operatorname{Var}(X \mid N)] + \operatorname{Var}(\mathbb{E}[X \mid N]) = 	frac14\mathbb{E}[N] + 	frac14\operatorname{Var}(N). For a fair die, \mathbb{E}[N] = 3.5 and \operatorname{Var}(N) = 	frac{35}{12} \approx 2.917 (from \mathbb{E}[N^2] = 	frac{91}{6} minus 3.5^2). So \operatorname{Var}(X) = 	frac14(3.5) + 	…

Roll a die, then flip that many coins

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