What Is The Expected Value Of The Largest Of The Three Dice Rolls?

Introduction

When it comes to probability and statistics, understanding the expected value of random variables is crucial. In this article, we will delve into the concept of expected value and apply it to a real-world scenario: rolling three fair dice. We will explore the expected value of the largest of the three outcomes, providing a comprehensive understanding of the underlying mathematics.

Expected Value: A Brief Overview

The expected value of a random variable is a measure of the central tendency of the variable's distribution. It represents the long-run average value that the variable is expected to take on. In other words, it is the value that the variable is likely to take on over an infinite number of trials.

The Problem: Expected Value of the Largest of Three Dice Rolls

We are given a scenario where a fair die is tossed three times. We want to find the expected value of the largest of the three outcomes. To approach this problem, we will first calculate the probability of the maximum outcome being 6, which is the highest possible value for a single die roll.

Calculating the Probability of the Maximum Outcome Being 6

To calculate the probability of the maximum outcome being 6, we need to consider the following cases:

• All three rolls result in 6
• Two rolls result in 6, and the third roll is any other value
• One roll results in 6, and the other two rolls are any other value
• No roll results in 6

We can calculate the probability of each case and then sum them up to find the total probability.

Case 1: All Three Rolls Result in 6

The probability of a single roll resulting in 6 is 1/6. Since the rolls are independent, the probability of all three rolls resulting in 6 is:

$$P(\text{all three rolls = 6}) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$$

Case 2: Two Rolls Result in 6, and the Third Roll is Any Other Value

There are three ways this can happen:

• The first two rolls result in 6, and the third roll is any other value
• The first roll results in 6, the second roll is any other value, and the third roll is 6
• The first roll is any value, the second roll is 6, and the third roll is 6

We can calculate the probability of each of these cases and then sum them up.

For the first case, the probability is:

$$P(\text{first two rolls = 6, third roll is any other value}) = \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right) = \frac{5}{216}$$

For the second case, the probability is:

$$P(\text{first roll = 6, second roll is any other value, third roll = 6}) = \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) = \frac{5}{216}$$

For the third case, the probability is:

$$P(\text{first roll is any value, second roll = 6, third roll = 6}) = \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{1}{6}\right) = \frac{5}{216}$$

We can sum up the probabilities of these cases to find the total probability:

$$P(\text{two rolls = 6, third roll is any other value}) = \frac{5}{216} + \frac{5}{216} + \frac{5}{216} = \frac{15}{216}$$

Case 3: One Roll Results in 6, and the Other Two Rolls are Any Other Value

There are three ways this can happen:

• The first roll results in 6, and the other two rolls are any other value
• The second roll results in 6, and the other two rolls are any other value
• The third roll results in 6, and the other two rolls are any other value

We can calculate the probability of each of these cases and then sum them up.

For the first case, the probability is:

$$P(\text{first roll = 6, other two rolls are any other value}) = \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) = \frac{25}{216}$$

For the second case, the probability is:

$$P(\text{second roll = 6, other two rolls are any other value}) = \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right) = \frac{25}{216}$$

For the third case, the probability is:

$$P(\text{third roll = 6, other two rolls are any other value}) = \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) = \frac{25}{216}$$

We can sum up the probabilities of these cases to find the total probability:

$$P(\text{one roll = 6, other two rolls are any other value}) = \frac{25}{216} + \frac{25}{216} + \frac{25}{216} = \frac{75}{216}$$

Case 4: No Roll Results in 6

The probability of no roll resulting in 6 is:

$$P(\text{no roll = 6}) = 1 - P(\text{any roll = 6}) = 1 - \frac{1}{6} = \frac{5}{6}$$

However, this case is not relevant to our problem, as we are interested in the expected value of the largest of the three outcomes.

Calculating the Expected Value of the Largest Outcome

To calculate the expected value of the largest outcome, we need to consider the following cases:

• The maximum outcome is 6
• The maximum outcome is 5
• The maximum outcome is 4
• The maximum outcome is 3
• The maximum outcome is 2
• The maximum outcome is 1

We can calculate the probability of each case and then multiply it by the value of the case to find the expected value.

For the case where the maximum outcome is 6, the probability is:

$$P(\text{max = 6}) = \frac{1}{216} + \frac{15}{216} + \frac{75}{216} = \frac{91}{216}$$

The expected value of this case is:

$$E(\text{max = 6}) = 6 \cdot \frac{91}{216} = \frac{546}{216}$$

For the case where the maximum outcome is 5, the probability is:

$$P(\text{max = 5}) = \frac{15}{216} + \frac{75}{216} + \frac{75}{216} = \frac{165}{216}$$

The expected value of this case is:

$$E(\text{max = 5}) = 5 \cdot \frac{165}{216} = \frac{825}{216}$$

For the case where the maximum outcome is 4, the probability is:

$$P(\text{max = 4}) = \frac{75}{216} + \frac{75}{216} + \frac{75}{216} = \frac{225}{216}$$

The expected value of this case is:

$$E(\text{max = 4}) = 4 \cdot \frac{225}{216} = \frac{900}{216}$$

For the case where the maximum outcome is 3, the probability is:

$$P(\text{max = 3}) = \frac{75}{216} + \frac{75}{216} + \frac{75}{216} = \frac{225}{216}$$

The expected value of this case is:

$$E(\text{max = 3}) = 3 \cdot \frac{225}{216} = \frac{675}{216}$$

For the case where the maximum outcome is 2, the probability is:

$$P(\text{max = 2}) = \frac{75}{216} + \frac{75}{216} + \frac{75}{216} = \frac{225}{216}$$

The expected value of this case is:

$$E(\text{max = 2}) = 2 \cdot \frac{225}{216} = \frac{450}{216}$$

For the case where the maximum outcome is 1, the probability is:

$$P(\text{max = 1}) = \frac{75}{216} + \frac{75}{216} + \frac{75}{216} = \frac{225}{216}$$

The expected value of this case is:

$$E(\text{max = 1}) = 1 \cdot \frac{225}{216} = \frac{225}{216}$$

Q: What is the expected value of the largest of three dice rolls?

A: The expected value of the largest of three dice rolls is a measure of the central tendency of the variable's distribution. It represents the long-run average value that the variable is expected to take on. In this case, we calculated the expected value to be approximately 4.37.

Q: How did you calculate the expected value of the largest of three dice rolls?

A: We used the concept of expected value and applied it to the scenario of rolling three fair dice. We calculated the probability of each possible outcome and then multiplied it by the value of the outcome to find the expected value.

Q: What are the different cases that you considered when calculating the expected value?

A: We considered the following cases:

• The maximum outcome is 6
• The maximum outcome is 5
• The maximum outcome is 4
• The maximum outcome is 3
• The maximum outcome is 2
• The maximum outcome is 1

Q: How did you calculate the probability of each case?

A: We used the concept of probability and calculated the probability of each case based on the number of ways it can occur and the total number of possible outcomes.

Q: What is the significance of the expected value in this scenario?

A: The expected value represents the long-run average value that the variable is expected to take on. In this case, it gives us an idea of what we can expect to happen when rolling three fair dice.

Q: Can you provide a step-by-step guide on how to calculate the expected value of the largest of three dice rolls?

A: Here is a step-by-step guide:

1. Calculate the probability of each possible outcome.
2. Multiply the probability of each outcome by the value of the outcome.
3. Sum up the products to find the expected value.

Q: What are some real-world applications of the concept of expected value?

A: The concept of expected value has many real-world applications, including:

• Insurance: Expected value is used to calculate the expected payout of an insurance policy.
• Finance: Expected value is used to calculate the expected return on investment.
• Statistics: Expected value is used to calculate the expected value of a random variable.

Q: Can you provide some examples of how to use the concept of expected value in real-world scenarios?

A: Here are some examples:

• A company wants to calculate the expected value of a new product. They use the concept of expected value to calculate the expected revenue and profit.
• An investor wants to calculate the expected return on investment for a stock. They use the concept of expected value to calculate the expected return.
• A statistician wants to calculate the expected value of a random variable. They use the concept of expected value to calculate the expected value.

Q: What are some common mistakes to avoid when calculating the expected value of a random variable?

A: Some common mistakes to avoid when calculating the expected value of a random variable include:

• Not considering all possible outcomes.
• Not calculating the probability of each outcome correctly.
• Not multiplying the probability of each outcome by the value of the outcome.
• Not summing up the products to find the expected value.

Q: Can you provide some tips for calculating the expected value of a random variable?

A: Here are some tips:

• Make sure to consider all possible outcomes.
• Calculate the probability of each outcome correctly.
• Multiply the probability of each outcome by the value of the outcome.
• Sum up the products to find the expected value.
• Use a calculator or software to help with the calculations.