Expected Minimum of Three 100-Sided Dice
You roll three fair 100-sided dice (each showing values 1 through 100 uniformly). What is the expected value of the minimum of the three rolls?
Hints
- Use the tail-sum formula: $E[X] = \sum_{k=1}^{100} P(X \ge k)$. For the minimum to be at least $k$, what must be true of each die?
- By independence, $P(\min \ge k) = \left(\frac{101 - k}{100}\right)^3$. Substitute $j = 101 - k$ to get a sum of cubes.
- Apply the identity $\sum_{j=1}^{n} j^3 = \left(\frac{n(n+1)}{2}\right)^2$ with $n = 100$.
Worked Solution
How to Think About It: The minimum of multiple dice is a classic order statistics problem. The cleanest approach is the tail-sum formula for expectation: $E[X] = \sum_{k=1}^{100} P(X \ge k)$. For the minimum to be at least $k$, ALL three dice must show at least $k$. That factorizes beautifully by independence.
Quick Estimate: With 3 dice on $\{1, \ldots, 100\}$, the minimum is pulled toward the low end. A rough rule: the expected minimum of $m$ uniform draws from $[0, 1]$ is
Approach: Use the tail-sum formula and the sum-of-cubes identity.
Formal Solution:
Let $X = \min(D_1, D_2, D_3)$ where each $D_i$ is uniform on $\{1, 2, \ldots, 100\}$.
By the tail-sum formula for non-negative integer-valued random variables:
$E[X] = \sum_{k=1}^{100} P(X \ge k)$
The event $\{X \ge k\}$ means all three dice show $\ge k$. By independence:
$P(X \ge k) = \left(\frac{101 - k}{100}\right)^3$
Substituting $j = 101 - k$ (so $j$ runs from 100 down to 1):
$E[X] = \sum_{j=1}^{100} \frac{j^3}{100^3} = \frac{1}{100^3} \sum_{j=1}^{100} j^3$
Using the identity $\sum_{j=1}^{n} j^3 = \left(\frac{n(n+1)}{2}\right)^2$:
$\sum_{j=1}^{100} j^3 = \left(\frac{100 \cdot 101}{2}\right)^2 = 5050^2 = 25{,}502{,}500$
Therefore:
$E[X] = \frac{25{,}502{,}500}{1{,}000{,}000} = 25.5025$
Verification: Monte Carlo simulation with
```python import random
results = [min(random.randint(1, 100) for _ in range(3)) for _ in range(1000000)] print(sum(results) / len(results)) # approximately 25.50 ```
Answer: $E[\min(D_1, D_2, D_3)] = \frac{5050^2}{100^3} = 25.5025$.
Intuition
The tail-sum formula $E[X] = \sum_{k \ge 1} P(X \ge k)$ is one of the most useful tricks in discrete probability. It converts an expectation (which requires knowing the full PMF) into a sum of survival probabilities (which are often easier to compute, especially for order statistics where independence gives clean factorizations).
The sum-of-cubes identity $\sum j^3 = (\sum j)^2$ is a happy coincidence that gives a closed form for 3 dice specifically. For $m$ dice, you would need $\sum j^m$, which has formulas (Faulhaber's formulas) but gets messier. The continuous approximation $E[\min] \approx n/(m+1)$ for $m$ uniform draws from $\{1, \ldots, n\}$ is a quick sanity check worth remembering -- it comes from the fact that the $k$-th order statistic of $m$ uniform $[0,1]$ draws has mean $k/(m+1)$.