Expected Value of the Maximum of Two Dice

Expectation · Easy · Free problem

You roll two fair six-sided dice and keep the one showing the higher value (if they are equal, you keep that value). What is the expected value of the die you keep?

Hints

  1. Think about the CDF: what is the probability that the maximum of two dice is at most $n$?
  2. If both dice must show $\le n$, there are $n^2$ outcomes out of 36. Use this to get $P(\max = n) = \frac{n^2 - (n-1)^2}{36}$.
  3. Compute $P(\max = n) = \frac{2n-1}{36}$ for $n = 1, \ldots, 6$, then evaluate $\sum n \cdot P(\max = n)$.

Worked Solution

How to Think About It: You are computing $E[\max(D_1, D_2)]$ where $D_1, D_2$ are independent uniform draws from $\{1, 2, 3, 4, 5, 6\}$. Before doing any math, think about the bounds. The expected value of a single die is $3.5$. Taking the max of two dice must be higher than $3.5$ -- you are always keeping the better outcome. But it cannot exceed $6$. A reasonable gut estimate: around $4.5$.

Quick Estimate: The max of two uniform draws on $[0,1]$ has expectation

/3$. Scaling to $[1,6]$: roughly
+ 5 \times (2/3) \approx 4.33$. The discrete case will be slightly different, but this puts us in the right ballpark.

Approach: Count $P(\max = n)$ for $n = 1, 2, \ldots, 6$ using the CDF method, then compute the expectation directly.

Formal Solution:

There are $36$ equally likely outcomes. The event $\{\max \le n\}$ means both dice show $\le n$, which happens in $n^2$ ways. So: $P(\max = n) = P(\max \le n) - P(\max \le n-1) = \frac{n^2 - (n-1)^2}{36} = \frac{2n - 1}{36}$

This gives:

| $n$ | $P(\max = n)$ | |-----|----------------| | 1 |

/36$ | | 2 | $3/36$ | | 3 | $5/36$ | | 4 | $7/36$ | | 5 | $9/36$ | | 6 |
1/36$ |

Sanity check:

+ 3 + 5 + 7 + 9 + 11 = 36$. Good.

Now compute the expectation: $E[\max] = \sum_{n=1}^{6} n \cdot \frac{2n-1}{36} = \frac{1}{36}\left(1 \cdot 1 + 2 \cdot 3 + 3 \cdot 5 + 4 \cdot 7 + 5 \cdot 9 + 6 \cdot 11\right)$ $= \frac{1 + 6 + 15 + 28 + 45 + 66}{36} = \frac{161}{36} \approx 4.472$

Answer: $E[\max(D_1, D_2)] = \dfrac{161}{36} \approx 4.472$.

Intuition

This is a fundamental order-statistics problem. The CDF trick -- $P(\max \le n) = P(D_1 \le n)P(D_2 \le n) = (n/6)^2$ -- is the cleanest way to handle max (and min) of independent random variables, discrete or continuous. The PMF falls out as the difference of consecutive CDF values, giving the neat formula $P(\max = n) = (2n-1)/36$. Notice the probabilities are consecutive odd numbers divided by 36 -- the max is heavily skewed toward higher values.

This pattern generalizes immediately. For $d$ dice, $P(\max \le n) = (n/6)^d$, and the expectation increases with $d$ but with diminishing returns. In quant interviews, this exact structure appears whenever you are selecting the best of several independent draws -- whether it is the best bid from multiple market makers, the maximum return in a portfolio, or the highest signal among multiple strategies. The key takeaway: taking the max always improves your expected outcome relative to a single draw, and the improvement grows with the number of draws but at a decreasing rate.

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