Expected Maximum and Minimum of Three Dice Rolls
You roll a fair 6-sided die three times. Let $M$ be the maximum of the three rolls and $N$ be the minimum.
- Compute $E[M]$ (the expected maximum).
- Compute $E[N]$ (the expected minimum).
- Derive general formulas for $E[\max]$ and $E[\min]$ of $n$ rolls of a $k$-sided die.
Hints
- For the maximum, start with the CDF: $P(M \leq k) = (k/6)^3$ because all three rolls must be at most $k$.
- Use the tail-sum formula $E[M] = \sum_{k=0}^{5} [1 - P(M \leq k)]$ to avoid computing the PMF directly.
- The symmetry $X_i \to 7 - X_i$ maps the max to $7 - \min$. Use this to verify: $E[M] + E[N] = 7$.
Worked Solution
How to Think About It: The maximum of several dice rolls is one of those problems where the CDF approach is far cleaner than trying to compute the PMF directly. The key identity: $E[M] = \sum_{j=1}^{6} P(M \geq j)$, which for non-negative integer-valued random variables avoids computing the PMF entirely. Similarly for the minimum: $E[N] = \sum_{j=1}^{6} P(N \geq j)$. Before computing, your gut should say: the max of 3 dice should be around 5 (biased high), and the min should be around 2 (biased low). They should sum to roughly 7 by a loose symmetry argument.
Quick Estimate: For a single die, $E = 3.5$. The max of 3 should be above 3.5 -- maybe around 5. The min should be symmetric to the max in some sense. Exact symmetry gives $E[M] + E[N] = 7$ (which we'll verify). So if $E[M] \approx 5$, then $E[N] \approx 2$.
Approach: Use the CDF method for the maximum and the survival function for the minimum.
Formal Solution:
Part 1: $E[M]$
The CDF of the max: $P(M \leq k) = P(\text{all three rolls} \leq k) = (k/6)^3$ for $k = 1, 2, \ldots, 6$.
Using the tail-sum formula for non-negative integer random variables: $E[M] = \sum_{k=0}^{5} [1 - P(M \leq k)] = \sum_{k=0}^{5} \left[1 - \left(\frac{k}{6}\right)^3\right]$
$= 6 - \frac{1}{216}\sum_{k=0}^{5} k^3 = 6 - \frac{0 + 1 + 8 + 27 + 64 + 125}{216} = 6 - \frac{225}{216} = 6 - \frac{25}{24} = \frac{119}{24}$
$E[M] = \frac{119}{24} \approx 4.958$
Part 2: $E[N]$
The survival function of the min: $P(N \geq k) = P(\text{all three rolls} \geq k) = \left(\frac{7-k}{6}\right)^3$ for $k = 1, 2, \ldots, 6$.
$E[N] = \sum_{k=1}^{6} P(N \geq k) = \sum_{k=1}^{6} \left(\frac{7-k}{6}\right)^3 = \frac{1}{216}\sum_{j=1}^{6} j^3 = \frac{1 + 8 + 27 + 64 + 125 + 216}{216} = \frac{441}{216} = \frac{49}{24}$
$E[N] = \frac{49}{24} \approx 2.042$
Sanity check: $E[M] + E[N] = 119/24 + 49/24 = 168/24 = 7$. This confirms the symmetry: for a die with faces $\{1, \ldots, 6\}$, the transformation $X_i \to 7 - X_i$ maps the max to $7 - \min$ and preserves the distribution, so $E[M] + E[N] = 7$.
Part 3: General formula
For $n$ rolls of a $k$-sided die (faces
$E[\max] = \sum_{j=0}^{k-1} \left[1 - \left(\frac{j}{k}\right)^n\right] = k - \frac{1}{k^n}\sum_{j=0}^{k-1} j^n$
$E[\min] = \sum_{j=1}^{k} \left(\frac{k+1-j}{k}\right)^n = \frac{1}{k^n}\sum_{j=1}^{k} j^n$
And the symmetry identity: $E[\max] + E[\min] = k + 1$.
Answer: $E[M] = 119/24 \approx 4.958$ and $E[N] = 49/24 \approx 2.042$. They sum to 7 by symmetry.
Intuition
The CDF/survival-function approach to computing expectations of order statistics is one of the most useful tricks in probability. Instead of first computing the PMF of the max (which requires inclusion-exclusion) and then summing $k \cdot P(M = k)$, you directly sum $P(M \geq k)$. For order statistics of i.i.d. random variables, the CDF factors into a product, making the computation clean.
The symmetry identity $E[\max] + E[\min] = k + 1$ is a consequence of the fact that reflecting a die ($j \to k + 1 - j$) swaps the max and min. This kind of symmetry check is invaluable in interviews -- if you compute the max and the min independently and they don't sum to $k + 1$, you know you've made an arithmetic error. In practice, the expected max of order statistics appears in auction theory (expected highest bid), in risk management (expected worst loss in a portfolio), and in algorithm analysis (expected depth of a search tree).