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.
\[
\mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]
= 3 \times \frac{35}{18}
= \boxed{\tfrac{35}{6}}.
\]
