Jessie is playing a dice game and bets \( \$ 6 \) on her first roll. If a 10 , 7 , or 4 is rolled, she wins \( \$ 6 \). This happens with a probability of \( \frac{1}{3} \). If an 8 or 2 is rolled, she loses her \( \$ 6 \). This has a probability of \( \frac{1}{6} \). If any other number is rolled, she does not win or lose, and the game continues. Find the expected value for Jessie on her first roll. Round to the nearest cent if necessary. Do not round until the final calculation.

To find the expected value for Jessie on her first roll, we need to consider the outcomes and their probabilities. 1. Jessie wins \( \$6 \) with a probability of \( \frac{1}{3} \). 2. Jessie loses \( \$6 \) with a probability of \( \frac{1}{6} \). 3. In the remaining cases, her total remains the same (no win or loss). The total probability of not winning or losing is the complementary probability: \[ 1 - \left(\frac{1}{3} + \frac{1}{6}\right) = 1 - \frac{2}{6} - \frac{1}{6} = 1 - \frac{3}{6} = \frac{3}{6} = \frac{1}{2}. \] Now we'll calculate the expected value (EV) based on the winnings and losses: - Winning \( \$6 \) with probability \( \frac{1}{3} \): \[ EV_{\text{win}} = 6 \times \frac{1}{3} = 2. \] - Losing \( \$6 \) with probability \( \frac{1}{6} \): \[ EV_{\text{loss}} = -6 \times \frac{1}{6} = -1. \] - No win or loss (which can be considered \( \$0 \) gain) with probability \( \frac{1}{2} \): \[ EV_{\text{no change}} = 0 \times \frac{1}{2} = 0. \] Now, combine all the expected values: \[ EV = EV_{\text{win}} + EV_{\text{loss}} + EV_{\text{no change}} = 2 - 1 + 0 = 1. \] Therefore, the expected value for Jessie on her first roll is \( \boxed{1} \).