Chapter 3: Problem 38
Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either \((1 / 3.5)\) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]
Short Answer
Expert verified
Gamble, as the expected value of \(h(X)\) is greater.
Step by step solution
01
Understand the Problem
The problem asks to compare a guaranteed amount of \(\frac{1}{3.5}\) dollars against a gambling alternative based on a function \(h(X) = \frac{1}{X}\), where \(X\) is the outcome of a fair die roll. The decision depends on calculating the expected value of \(h(X)\) and comparing it with the guaranteed amount.
02
Calculate the Expected Value of a Die Roll
For a fair six-sided die, each outcome from 1 to 6 occurs with equal probability. Thus, the expected value \(E(X)\) of the die roll is given by:\[E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5\]
03
Calculate the Guaranteed Amount
The guaranteed amount offered is \(\frac{1}{3.5}\) dollars, which can be computed as follows:\[\frac{1}{3.5} = \frac{2}{7}\text{ dollars}\]
04
Compute the Expected Value of the Gamble
The expected value of \(h(X)\) is computed using the formula:\[E\left(\frac{1}{X}\right) = \sum_{i=1}^{6} \frac{1}{6} \cdot \frac{1}{i} = \frac{1}{6}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)\]Evaluate the expression inside the parentheses:\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \approx 2.45\]Then, the expectation \(E\left(\frac{1}{X}\right)\) is:\[\frac{1}{6} \times 2.45 \approx 0.4083\text{ dollars}\]
05
Compare the Values
Compare the guaranteed amount \(\frac{2}{7} \approx 0.2857\) dollars with the expected value of the gamble \(0.4083\) dollars:Since \(0.4083 > 0.2857\), the expected value of the gamble \(E\left(\frac{1}{X}\right)\) is greater than the guaranteed amount.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Fair Die
A fair die, commonly used in various games and probability exercises, is a six-sided cube with numbers from 1 to 6. Each side is equally likely to land face up when the die is thrown. This means that every face of the die has the same probability of appearing on top; hence the term "fair."
In probability theory, a fair die ensures each outcome occurs with equal likelihood. For a six-sided die, this translates to each number (1 through 6) having a probability of \(\frac{1}{6}\). This characteristic forms the basis for many calculations, including expected value, and fair probability models.
In probability theory, a fair die ensures each outcome occurs with equal likelihood. For a six-sided die, this translates to each number (1 through 6) having a probability of \(\frac{1}{6}\). This characteristic forms the basis for many calculations, including expected value, and fair probability models.
- Ensures unbiased results
- Equal probability for each side
- Used widely in probability and statistics
Exploring Probability
Probability measures the likelihood of an event occurring, and it ranges from 0 to 1. In the context of a fair die, the probability of any particular side showing up is \(\frac{1}{6}\). This uniform distribution is what makes probability calculations straightforward for a die.
The rules of probability help us calculate the outcomes of combined events. When rolling a die, for instance, if we want the probability of getting an even number, we count the possible outcomes (2, 4, 6) and find the probability, \(\frac{3}{6} = \frac{1}{2}\). These calculations are foundational in understanding broader statistical problems.
The rules of probability help us calculate the outcomes of combined events. When rolling a die, for instance, if we want the probability of getting an even number, we count the possible outcomes (2, 4, 6) and find the probability, \(\frac{3}{6} = \frac{1}{2}\). These calculations are foundational in understanding broader statistical problems.
- Probability of an event = number of favorable outcomes / total outcomes
- Ranges from 0 (impossible event) to 1 (certain event)
- Helps in predicting event outcomes
Delving into Expected Value Calculation
The expected value is a fundamental concept in probability that represents the average or mean value of a random variable over numerous trials. For a fair die, the expected value can be calculated using its outcomes and probabilities.
In this problem, rolling a six-sided fair die, each outcome (1 to 6) contributes equally to the expected value. Using the formula for expectation \(E(X) = \sum{(x_i \cdot p_i)}\), where \(x_i\) is the outcome and \(p_i = \frac{1}{6}\) for each outcome, the expected sum-value is computed as follows:
\[E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\]
This tells us the average number rolled on a die, highlighting how expected value helps make informed decisions around uncertainty.
In this problem, rolling a six-sided fair die, each outcome (1 to 6) contributes equally to the expected value. Using the formula for expectation \(E(X) = \sum{(x_i \cdot p_i)}\), where \(x_i\) is the outcome and \(p_i = \frac{1}{6}\) for each outcome, the expected sum-value is computed as follows:
\[E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\]
This tells us the average number rolled on a die, highlighting how expected value helps make informed decisions around uncertainty.
- Represents the mean outcome of a random variable
- Essential for decision-making in uncertain scenarios
- Used to evaluate gambles and predicts outcomes over time