Questions: Computing expected value in a game of chance Carmen is playing a game of chance in which she rolls a number cube with sides numbered from 1 to 6. The number cube is fair, so a side is rolled at random. This game is this: Carmen rolls the number cube once. She wins 1 if a 1 is rolled, 2 if a 2 is rolled, 3 if a 3 is rolled, and 4 if a 4 is rolled. She loses 5 if a 5 or 6 is rolled. (a) Find the expected value of playing the game. dollars (b) What can Carmen expect in the long run, after playing the game many times? Carmen can expect to gain money. She can expect to win dollars per roll. Carmen can expect to lose money. She can expect to lose dollars per roll. Carmen can expect to break even (neither gain nor lose money).

Solution Steps

To find the expected value \( E(X) \) of Carmen's game, we compute the weighted sum of the outcomes based on their probabilities:

\[ E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + (-5) \times \frac{1}{6} + (-5) \times \frac{1}{6} \]

Calculating this gives:

\[ E(X) = \frac{1 + 2 + 3 + 4 - 5 - 5}{6} = \frac{0}{6} = 0.0 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = E(X^2) - (E(X))^2 \]

First, we compute \( E(X^2) \):

\[ E(X^2) = (1^2) \times \frac{1}{6} + (2^2) \times \frac{1}{6} + (3^2) \times \frac{1}{6} + (4^2) \times \frac{1}{6} + ((-5)^2) \times \frac{1}{6} + ((-5)^2) \times \frac{1}{6} \]

Calculating this gives:

\[ E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 25}{6} = \frac{80}{6} \approx 13.333 \]

Now, substituting back into the variance formula:

\[ \sigma^2 = 13.333 - (0.0)^2 = 13.333 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{13.333} \approx 3.651 \]

Final Answer