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.