Problem. Dennis rolls three fair six-sided dice, obtaining a, b, c ∈ {1,…,6}. Find \[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]. \]
Try the question first before scrolling down to read the solution.
Solution.
Step 1 — Linearity of expectation.
\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] \;=\; \mathbb{E}[|a-b|]+\mathbb{E}[|b-c|]+\mathbb{E}[|c-a|] \;=\; 3\,\mathbb{E}[|a-b|]. \]Step 2 — Expected absolute difference of two dice.
Let \(X = |a-b|\). Its distribution is
\[ \begin{array}{c|cccccc} d & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \Pr(X=d) & \tfrac{6}{36} & \tfrac{10}{36} & \tfrac{8}{36} & \tfrac{6}{36} & \tfrac{4}{36} & \tfrac{2}{36} \end{array} \] \[ \mathbb{E}[|a-b|] =\frac{1}{36}\bigl(0\cdot6 + 1\cdot10 + 2\cdot8 + 3\cdot6 + 4\cdot4 + 5\cdot2\bigr) =\frac{70}{36} =\frac{35}{18}. \]Step 3 — Final answer.
\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] = 3 \times \frac{35}{18} = \boxed{\tfrac{35}{6}}. \]
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