Question

heather 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: heather rolls the number cube once. she wins $1 if a 1 is rolled, $2 if a 2 is rolled, and $3 if a 3 is rolled. she loses $1 if a 4, 5, or 6 is rolled. (a) find the expected value of playing the game. dollars (b) what can heather expect in the long run, after playing the game many times? heather can expect to gain money. she can expect to win dollars per roll. heather can expect to lose money. she can expect to lose dollars per roll. heather can expect to break even (neither gain nor lose money).

Explanation:

Part (a)

Step 1: Define probabilities and outcomes

The number cube has 6 sides, so the probability of each outcome is $\frac{1}{6}$. The outcomes and their values:

• Roll 1: Value = $1$, Probability = $\frac{1}{6}$
• Roll 2: Value = $2$, Probability = $\frac{1}{6}$
• Roll 3: Value = $3$, Probability = $\frac{1}{6}$
• Roll 4, 5, or 6: Value = $-1$ (loses $1$), Probability = $\frac{3}{6}=\frac{1}{2}$

Step 2: Calculate expected value

The formula for expected value $E(X)$ is $E(X)=\sum x_iP(x_i)$, where $x_i$ are the outcomes and $P(x_i)$ are their probabilities.

$$\begin{align*} E(X)&=(1\times\frac{1}{6})+(2\times\frac{1}{6})+(3\times\frac{1}{6})+(-1\times\frac{3}{6})\\ &=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}-\frac{3}{6}\\ &=\frac{1 + 2 + 3 - 3}{6}\\ &=\frac{3}{6}\\ &=\frac{1}{2} \end{align*}$$

Since the expected value from part (a) is $\frac{1}{2}$ (or 0.5) dollars, which is positive, Heather can expect to gain money. The expected value per roll is the value we calculated in part (a), which is $\frac{1}{2}$ dollars.

Answer:

$\frac{1}{2}$ (or 0.5)

Part (b)