1. **Problem Statement:** You pay 1 to play a game where two fair dice are rolled. You win 3 if the sum is 6, 7, or 8; you win 5 if the sum is 2 or 12; otherwise, you lose your 1. 2. **Goal:** Calculate the expected value (EV) of this game to determine if it is favorable. 3. **Step 1: Possible sums and their probabilities.** The sums of two dice range from 2 to 12. The number of ways to get each sum out of 36 total outcomes is: - 2: 1 way - 3: 2 ways - 4: 3 ways - 5: 4 ways - 6: 5 ways - 7: 6 ways - 8: 5 ways - 9: 4 ways - 10: 3 ways - 11: 2 ways - 12: 1 way 4. **Step 2: Calculate probabilities:** - $P(2) = \frac{1}{36}$ - $P(6) = \frac{5}{36}$ - $P(7) = \frac{6}{36}$ - $P(8) = \frac{5}{36}$ - $P(12) = \frac{1}{36}$ - Other sums combined probability = $1 - (P(2)+P(6)+P(7)+P(8)+P(12)) = 1 - \frac{18}{36} = \frac{18}{36}$ 5. **Step 3: Calculate expected winnings:** - Win 3 for sums 6,7,8: $3 \times (P(6)+P(7)+P(8)) = 3 \times \frac{16}{36} = \frac{48}{36}$ - Win 5 for sums 2,12: $5 \times (P(2)+P(12)) = 5 \times \frac{2}{36} = \frac{10}{36}$ - Lose 1 for other sums: $-1 \times \frac{18}{36} = -\frac{18}{36}$ 6. **Step 4: Total expected value:** $$EV = \frac{48}{36} + \frac{10}{36} - \frac{18}{36} = \frac{40}{36} = \frac{10}{9} \approx 1.11$$ 7. **Step 5: Interpretation:** The expected value is about 1.11, meaning on average you win 1.11 per game. Since you pay 1 to play, your net expected gain is $1.11 - 1 = 0.11$, so the game is favorable. **Final answer:** The expected net gain per game is approximately 0.11, so it is a favorable game to play.