Problem 5 - Algebra and Trigonometry

Question

Find the expected value (or expectation) of the games described.? $$ \begin{array}{l}{\text { Carol wins } \$ 3 \text { if a die roll shows a six, and she wins } \$ 0.50} \\ {\text { otherwise. }}\end{array} $$

Step-by-Step Solution

Answer

Carol's expected winnings per die roll are approximately $0.92.

Step 1: Understand the Problem

We need to find the expected value of Carol's winnings when rolling a die. This requires calculating the average winnings over all possible outcomes.

Step 2: Identify Possible Outcomes

When rolling a six-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Carol wins $3 if the die shows a 6, and she wins $0.50 for any other number.

Step 3: Assign Probabilities

Each outcome when rolling a fair die has a probability of \( \frac{1}{6} \). Therefore, the probability of rolling a 6 is \( \frac{1}{6} \), and the probability of rolling any other number (1-5) is also \( \frac{1}{6} \).

Step 4: Calculate Expected Value

The expected value is calculated as the sum of each outcome's value times its probability. Thus, the expected value \( E \) can be calculated by: \[ E = 3 \times \frac{1}{6} + 0.50 \times \frac{5}{6} \] \[ E = \frac{3}{6} + \frac{2.5}{6} \] \[ E = \frac{5.5}{6} \] \[ E \approx 0.9167 \]

Step 5: Interpret the Result

The expected value, approximately \( 0.917 \), represents Carol's average expected winnings per die roll.

Key Concepts

Understanding Probability

Probability is a way of quantifying the likelihood of an event occurring. It is a fundamental concept in understanding games of chance, like rolling a die. When we say the probability of an event, we're essentially talking about how often that event is expected to happen in a series of trials.

Consider a fair six-sided die. Each side has a distinct number from 1 to 6, and when the die is rolled, any of these numbers is equally likely to appear. Since there are 6 sides, the probability of rolling any specific number, say a 6, is calculated as:

• Probability of rolling a 6 = \( \frac{1}{6} \)

This means if you rolled the die many times, you would expect to get a 6 about one-sixth of the time.

Probability helps us determine not just the outcomes, but also the fairness and expected results in games like the one Carol is playing.

Exploring Die Roll Outcomes

Rolling a standard six-sided die is a perfect example of a random event with discrete outcomes. Each roll of the die results in one of six possible outcomes: 1, 2, 3, 4, 5, or 6.

These outcomes are considered to be mutually exclusive and collectively exhaustive, meaning that one, and only one, of these outcomes can occur in any single roll, and all possible outcomes are accounted for. This characteristic makes dice games simple yet excellent tools for understanding probability concepts.

In Carol's game, she wins $3 if the outcome is a 6 and $0.50 for any other outcome (1 through 5). To calculate her winnings, it's crucial to understand the tie between probability and these set outcomes. This is where expected value comes into play.

Calculating Average Winnings

"Expected Value" is a crucial concept when determining the average result of an uncertain event over time. It provides a way to calculate the mean or average winnings in games of chance.

In Carol's scenario, we calculate her expected winnings per roll by taking into account the different winnings based on the die outcomes and their probabilities.

The formula used to find the expected value (\( E \)) of her winnings can be represented as:

• Carol wins \(3 with a probability of \( \frac{1}{6} \)
• She wins \)0.50 with a probability of \( \frac{5}{6} \)

Hence, the calculation is:\[E = 3 \times \frac{1}{6} + 0.50 \times \frac{5}{6}\]This simplifies to:\[E = \frac{3}{6} + \frac{2.5}{6}\]Thus:\[E = \frac{5.5}{6} \approx 0.917\]

This means that, on average, Carol can expect to win about $0.917 per roll over a large number of rolls. This average takes into account all the possible outcomes and their probabilities.