Questions: Dice Game A person pays 2 to play a certain game by rolling a single die once. If a 1 or a 2 comes up, the person wins nothing. If, however, the payer rolls a 3,4,5, or 6, the person wins the difference between the number rolled and 2. Find the expectation for this game. Is the game fair? Dice roll: 1 2 3 4 5 6 Gain X 0-2 0-2 (3-2)-2 (4-2)-2 (5-2)-2 (6-2)-2 Gain X -2 -2 -1 0 1 2 P(X) X * P(X) Mean or expected value: [ mu=mathrmE(mathrmX)=sum[X * P(X)]= ] Is the game fair?

Solution Steps

To find the expected value $\mu$ of the game, we compute:

$$\mu = E(X) = \sum (X \cdot P(X)) = -2 \cdot \frac{1}{6} + -2 \cdot \frac{1}{6} + -1 \cdot \frac{1}{6} + 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6}$$

Calculating this gives:

$$\mu = -\frac{2}{6} - \frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} = -\frac{6}{6} = -0.333$$

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

$$\sigma^2 = \sum (X - \mu)^2 \cdot P(X)$$

Substituting the values:

$$\sigma^2 = (-2 - (-0.333))^2 \cdot \frac{1}{6} + (-2 - (-0.333))^2 \cdot \frac{1}{6} + (-1 - (-0.333))^2 \cdot \frac{1}{6} + (0 - (-0.333))^2 \cdot \frac{1}{6} + (1 - (-0.333))^2 \cdot \frac{1}{6} + (2 - (-0.333))^2 \cdot \frac{1}{6}$$

Calculating this gives:

$$\sigma^2 = (1.667)^2 \cdot \frac{1}{6} + (1.667)^2 \cdot \frac{1}{6} + (0.667)^2 \cdot \frac{1}{6} + (0.333)^2 \cdot \frac{1}{6} + (1.333)^2 \cdot \frac{1}{6} + (2.333)^2 \cdot \frac{1}{6}$$

This results in:

$$\sigma^2 = 2.222$$

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

$$\sigma = \sqrt{\sigma^2} = \sqrt{2.222} \approx 1.491$$

A game is considered fair if the expected value $\mu$ is equal to 0. Since:

$$\mu = -0.333 \neq 0$$

The game is not fair.

Final Answer