calculate expected value of max of 2 n-sided dice rolls

To calculate the expected value of the maximum of two \( n \)-sided dice rolls, we can proceed step by step.

### Step 1: Probability of Maximum Being a Specific Value

Let \( X \) and \( Y \) be the outcomes of the two independent dice rolls, both of which are uniformly distributed between 1 and \( n \). We want to calculate \( \mathbb{E}[\max(X, Y)] \), the expected value of the maximum of the two dice.

First, consider the probability that the maximum of the two dice rolls is a specific value \( k \), where \( k \in \{1, 2, \dots, n\} \). The maximum is \( k \) if at least one of the dice shows \( k \), and all other outcomes are less than or equal to \( k \). The probability of this can be calculated as follows:

- The probability that both dice show values less than \( k \) is \( \left(\frac{k-1}{n}\right)^2 \), since both dice must show one of the \( k-1 \) numbers from \( \{1, 2, \dots, k-1\} \).
- Therefore, the probability that the maximum is \( k \) is the complement of this, or:
  \[
  P(\max(X, Y) = k) = \left(\frac{k}{n}\right)^2 - \left(\frac{k-1}{n}\right)^2
  \]

### Step 2: Expected Value Calculation

Now, we calculate the expected value using the formula for the expectation:
\[
\mathbb{E}[\max(X, Y)] = \sum_{k=1}^{n} k \cdot P(\max(X, Y) = k)
\]

Substituting the expression for \( P(\max(X, Y) = k) \), we get:
\[
\mathbb{E}[\max(X, Y)] = \sum_{k=1}^{n} k \left[\left(\frac{k}{n}\right)^2 - \left(\frac{k-1}{n}\right)^2\right]
\]

### Step 3: Simplification

Simplifying the expression, the difference of squares gives:
\[
\mathbb{E}[\max(X, Y)] = \sum_{k=1}^{n} k \left[\frac{2k-1}{n^2}\right]
\]

This can now be computed as a sum of two separate terms:
\[
\mathbb{E}[\max(X, Y)] = \frac{1}{n^2} \sum_{k=1}^{n} k(2k - 1)
\]
\[
= \frac{1}{n^2} \sum_{k=1}^{n} (2k^2 - k)
\]
Now split the summation:
\[
\mathbb{E}[\max(X, Y)] = \frac{1}{n^2} \left[ 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k \right]
\]

Using the known formulas for sums of integers and squares:
\[
\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]

Substitute these into the expression:
\[
\mathbb{E}[\max(X, Y)] = \frac{1}{n^2} \left[ 2 \cdot \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \right]
\]
Simplifying:
\[
\mathbb{E}[\max(X, Y)] = \frac{n(n+1)}{6n^2} \left[ 2(2n+1) - 3 \right]
\]
\[
\mathbb{E}[\max(X, Y)] = \frac{n(n+1)}{6n^2} \cdot (4n - 1)
\]
\[
\mathbb{E}[\max(X, Y)] = \frac{(n+1)(4n-1)}{6n}
\]

### Final Answer:
\[
\mathbb{E}[\max(X, Y)] = \frac{(n+1)(4n-1)}{6n}
\]

This is the expected value of the maximum of two \( n \)-sided dice rolls.

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### 5 Related Questions:
1. What is the expected value of the sum of two \( n \)-sided dice?
2. How does the expected value of the maximum change if we use three \( n \)-sided dice instead of two?
3. What is the variance of the maximum of two \( n \)-sided dice rolls?
4. How does the expected value of the maximum behave as \( n \to \infty \)?
5. Can you calculate the expected value of the minimum of two \( n \)-sided dice?

### Tip:
For dice problems, knowing the basic formulas for the sum of integers and squares can significantly simplify your calculations.

Learn how to calculate the expected value of the maximum of two n-sided dice rolls using probability theory and summation formulas. This detailed solution includes key concepts such as expected value and probability, suitable for high school students in Grades 11-12.

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