1. **State the problem:** We have a game where an eight-sided die (values 1 to 8) is rolled. If the number rolled is a multiple of 3, the player wins double the amount rolled. Otherwise, the player loses the amount rolled. 2. **Define the random variable and expected value formula:** Let $X$ be the amount won or lost. The expected value $E(X)$ is given by: $$E(X) = \sum (\text{value} \times \text{probability})$$ 3. **Identify multiples of 3 on the die:** Multiples of 3 between 1 and 8 are 3 and 6. 4. **Calculate the value of $X$ for each outcome:** - For 3 and 6: win double the amount rolled, so $X = 2 \times 3 = 6$ and $X = 2 \times 6 = 12$ respectively. - For other numbers (1, 2, 4, 5, 7, 8): lose the amount rolled, so $X = -1, -2, -4, -5, -7, -8$ respectively. 5. **Calculate the expected value:** Each outcome has probability $\frac{1}{8}$. $$E(X) = \frac{1}{8}(6 + 12 - 1 - 2 - 4 - 5 - 7 - 8)$$ 6. **Simplify inside the parentheses:** $$6 + 12 = 18$$ $$-1 - 2 - 4 - 5 - 7 - 8 = -(1 + 2 + 4 + 5 + 7 + 8) = -27$$ So, $$E(X) = \frac{1}{8}(18 - 27) = \frac{1}{8}(-9)$$ 7. **Final expected value:** $$E(X) = -\frac{9}{8} = -1.125$$ **Interpretation:** On average, the player loses 1.125 units per game played.