1. **State the problem:** You pay 10 to play a game where two fair dice are rolled, and you receive a payout equal to the sum of the dice. The net gain is the payout minus 10. We want to find the expected value of the net gain and determine if the game is fair. 2. **Formula and rules:** The expected value $E(X)$ of a discrete random variable $X$ is given by $$E(X) = \sum x_i P(x_i)$$ where $x_i$ are possible outcomes and $P(x_i)$ their probabilities. 3. **Possible sums and probabilities:** The sums of two dice range from 2 to 12. The number of ways to get each sum and their probabilities are: - 2: 1 way, $P=\frac{1}{36}$ - 3: 2 ways, $P=\frac{2}{36}$ - 4: 3 ways, $P=\frac{3}{36}$ - 5: 4 ways, $P=\frac{4}{36}$ - 6: 5 ways, $P=\frac{5}{36}$ - 7: 6 ways, $P=\frac{6}{36}$ - 8: 5 ways, $P=\frac{5}{36}$ - 9: 4 ways, $P=\frac{4}{36}$ - 10: 3 ways, $P=\frac{3}{36}$ - 11: 2 ways, $P=\frac{2}{36}$ - 12: 1 way, $P=\frac{1}{36}$ 4. **Calculate expected payout:** $$E(\text{payout}) = \sum_{k=2}^{12} k \times P(k) = \frac{1}{36}(2\times1 + 3\times2 + 4\times3 + 5\times4 + 6\times5 + 7\times6 + 8\times5 + 9\times4 + 10\times3 + 11\times2 + 12\times1)$$ Calculate the sum inside: $$2\times1=2, 3\times2=6, 4\times3=12, 5\times4=20, 6\times5=30, 7\times6=42, 8\times5=40, 9\times4=36, 10\times3=30, 11\times2=22, 12\times1=12$$ Sum these: $$2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 = 252$$ So, $$E(\text{payout}) = \frac{252}{36} = 7$$ 5. **Calculate expected net gain:** $$E(\text{net gain}) = E(\text{payout}) - 10 = 7 - 10 = -3$$ 6. **Interpretation:** The expected net gain is -3, meaning on average you lose 3 per game. Therefore, the game is not fair; it favors the house. **Final answer:** The expected payoff of the game is $-3$.