Problem 2 - Algebra and Trigonometry

Question

Find the expected value (or expectation) of the games described.? $$ \begin{array}{l}{\text { Jane wins } \$ 10 \text { if a die roll shows a six, and she loses } \$ 1} \\ {\text { otherwise. }}\end{array} $$

Step-by-Step Solution

Answer

The expected value of the game is $0.84.

Step 1: Understand the Problem

Jane plays a game where she wins $10 if a six is rolled on a die and loses $1 if any other number is rolled. A die has six faces: numbers 1 through 6.

Step 2: Determine Possible Outcomes

There are two possible outcomes when Jane rolls the die: she might roll a six or she might roll any one of the numbers from 1 to 5.

Step 3: Calculate Probability of Winning

The probability of rolling a six is \( \frac{1}{6} \). Thus, the probability that Jane wins $10 is \( \frac{1}{6} \).

Step 4: Calculate Probability of Losing

The probability of rolling any number other than six (i.e., rolling a 1, 2, 3, 4, or 5) is \( \frac{5}{6} \). Thus, the probability that Jane loses $1 is \( \frac{5}{6} \).

Step 5: Compute Expected Value

Use the formula for expected value: \( E(X) = p_1 \cdot v_1 + p_2 \cdot v_2 \), where \( p_1 \) and \( p_2 \) are the probabilities, and \( v_1 \) and \( v_2 \) are the winnings/losses associated with these probabilities. Substitute the known values: \( E(X) = \frac{1}{6} \cdot 10 + \frac{5}{6} \cdot (-1) \).

Step 6: Simplify the Expression

Calculate \( \frac{1}{6} \cdot 10 = \frac{10}{6} = 1.67 \). Calculate \( \frac{5}{6} \cdot (-1) = -\frac{5}{6} = -0.83 \).

Step 7: Sum the Results

Add the results to find the expected value: \( 1.67 + (-0.83) = 0.84 \). Therefore, the expected value, or expected gain per game, is \$0.84.

Key Concepts

Probability

Probability is the measure of the likelihood of an event to occur. It is expressed as a number between 0 and 1. A probability of 0 means the event is impossible, while a probability of 1 indicates certainty. In Jane's game, the probability is essential to determining whether she wins or loses. We can calculate probabilities using a simple formula:

• Probability of an event = (Number of favorable outcomes) / (Total number of possible outcomes)

For instance, the probability that Jane rolls a six and wins $10 is calculated as:\[P(\text{rolling a six}) = \frac{1}{6}\]Meaning there is a one in six chance for Jane to win in each die roll.

Outcomes

In probability and decision-making, outcomes are the possible results of a random event. When Jane rolls the die, there are several outcomes to consider. In this scenario, the outcomes are either rolling a six, which is favorable, or rolling any other number, which is unfavorable.

• Favorable outcome: Rolling a six, winning $10.
• Unfavorable outcomes: Rolling 1, 2, 3, 4, or 5, losing $1 each time.

By analyzing these outcomes, we can better understand their impact on the expected value over numerous repetitions of the game.

Probability Theory

Probability theory is the branch of mathematics concerned with the analysis of random phenomena. It helps us quantify and predict the likelihood of various outcomes. In Jane's game, probability theory allows us to calculate the expected gain or loss per game using known probabilities of each outcome. One fundamental concept in probability theory is the expected value, which is key for Jane's scenario. By utilizing probability and outcomes, we calculate how much, on average, she can expect to win or lose per game. This theoretical framework helps guide decisions not just in games, but in situations involving risk and uncertainty in various fields.

Mathematical Expectation

Mathematical expectation, or expected value, is a crucial concept in probability and statistics. It represents the average outcome one can expect over a series of trials. To find the expected value, we multiply each possible outcome by its probability and then sum these products.For Jane, the formula for expected value is:\[E(X) = p_1 \cdot v_1 + p_2 \cdot v_2\]Where:

• \(p_1\) is the probability of rolling a six, \(\frac{1}{6}\), and \(v_1\) is the win, \(10.
• \(p_2\) is the probability of not rolling a six, \(\frac{5}{6}\), and \(v_2\) is the loss, -\)1.

Using these values, Jane's expected value calculation shows that, on average, she can expect to earn $0.84 per game. This concept is vital for assessing the fairness and potential profitability of a game or decision-making scenario.