1. **State the problem:** You roll three fair 6-sided dice. The payouts are: - $20 if all three dice show the same number. - $10 if exactly two dice show the same number. - Lose $2 if all three dice show different numbers. We want to find the expected return per roll. 2. **Formula for expected value:** $$E(X) = \sum (\text{value} \times \text{probability})$$ 3. **Calculate probabilities:** - Total outcomes: $6^3 = 216$ - Probability all three dice are the same: There are 6 possible triples (111, 222, ..., 666), so $$P(\text{all same}) = \frac{6}{216} = \frac{1}{36}$$ - Probability exactly two dice are the same: Choose the number that appears twice: 6 choices. Choose the number that appears once: 5 choices (different from the pair). Choose which two dice show the pair: $\binom{3}{2} = 3$ ways. Total favorable outcomes: $$6 \times 5 \times 3 = 90$$ So, $$P(\text{exactly two same}) = \frac{90}{216} = \frac{5}{12}$$ - Probability all different: Total outcomes minus above cases: $$216 - 6 - 90 = 120$$ So, $$P(\text{all different}) = \frac{120}{216} = \frac{5}{9}$$ 4. **Calculate expected value:** $$E(X) = 20 \times \frac{1}{36} + 10 \times \frac{5}{12} - 2 \times \frac{5}{9}$$ Calculate each term: $$20 \times \frac{1}{36} = \frac{20}{36} = \frac{5}{9} \approx 0.5556$$ $$10 \times \frac{5}{12} = \frac{50}{12} = \frac{25}{6} \approx 4.1667$$ $$-2 \times \frac{5}{9} = -\frac{10}{9} \approx -1.1111$$ Sum: $$E(X) = \frac{5}{9} + \frac{25}{6} - \frac{10}{9} = \left(\frac{5}{9} - \frac{10}{9}\right) + \frac{25}{6} = -\frac{5}{9} + \frac{25}{6}$$ Find common denominator 18: $$-\frac{5}{9} = -\frac{10}{18}, \quad \frac{25}{6} = \frac{75}{18}$$ So, $$E(X) = -\frac{10}{18} + \frac{75}{18} = \frac{65}{18} \approx 3.6111$$ 5. **Interpretation:** On average, you expect to earn about 3.61 per roll.