1. **State the problem:** We have a game where a ten-sided die (faces 1 to 10) is rolled. If the result is less than 7 (i.e., 1 to 6), the player wins that amount in dollars. If the result is 7 or greater (7 to 10), the player pays 4 dollars. 2. **Formula for expected value:** The expected value $E$ is the sum of each outcome multiplied by its probability: $$E = \sum (\text{value of outcome} \times \text{probability of outcome})$$ 3. **Calculate probabilities:** Each face has probability $\frac{1}{10}$. 4. **Calculate expected winnings for outcomes less than 7:** $$\sum_{k=1}^6 k \times \frac{1}{10} = \frac{1}{10} (1+2+3+4+5+6) = \frac{1}{10} \times 21 = 2.1$$ 5. **Calculate expected loss for outcomes 7 or greater:** There are 4 such outcomes (7,8,9,10), each with probability $\frac{1}{10}$. The player pays 4 dollars, so the value is $-4$. $$4 \times \left(-4 \times \frac{1}{10}\right) = 4 \times \left(-\frac{4}{10}\right) = -\frac{16}{10} = -1.6$$ 6. **Sum the expected values:** $$E = 2.1 + (-1.6) = 0.5$$ 7. **Interpretation:** The expected value of the game is $0.5$, meaning on average the player wins 0.5 dollars per game. **Final answer:** $$\boxed{0.5}$$