Comparing Three Dice Expectations
You roll fair six-sided dice in three separate experiments:
- Roll one die. Let $X$ be the result. Compute $E[X^2]$.
- Roll two independent dice $X$ and $Y$. Compute $E[XY]$.
- Roll five independent dice and let $M$ be their median (the 3rd-smallest value). Compute $E[M^2]$.
Rank these three expectations from smallest to largest and justify the ordering.
Hints
- All three quantities share the same baseline $(E[X])^2 = 12.25$. Think about what each one adds on top of that.
- Use the identity $E[Z^2] = \text{Var}(Z) + (E[Z])^2$. For the product of independent variables, recall $E[XY] = E[X]E[Y]$. For the median, think about how order statistics concentrate the distribution.
- The median of 5 symmetric dice has $E[M] = 3.5$ by symmetry. Compute $\text{Var}(M)$ -- it must be smaller than $\text{Var}(X) = 35/12$ -- and use it to get $E[M^2]$.
Worked Solution
How to Think About It: All three quantities have the form "expected value of a product of dice-related things." The trick is that they look superficially similar but differ because of how squaring, multiplying independent copies, and taking order statistics interact with expectation. Before computing anything, think about it this way: $E[X^2]$ captures the spread of a single die (it equals $\text{Var}(X) + (E[X])^2$). The product $E[XY]$ of two independent dice is just $(E[X])^2$ since independence kills the covariance term. And $E[M^2]$ for the median sits in between -- the median has the same mean as a single die (by symmetry), but less variance because averaging over order statistics concentrates the distribution. So you should expect $E[XY] < E[M^2] < E[X^2]$.
Quick Estimate: All three revolve around the baseline $E[X] = 3.5$, so $(E[X])^2 = 12.25$. For $E[X^2]$, the variance of a single die is $35/12 \approx 2.92$, giving $E[X^2] \approx 12.25 + 2.92 = 15.17$. For $E[XY]$, independence gives exactly
Approach: Compute each expectation exactly. The first two are quick; the third requires the PMF of the median of 5 discrete uniform dice.
Formal Solution:
(1) $E[X^2]$ where $X \sim \text{Uniform}\{1,\ldots,6\}$:
$E[X^2] = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = \frac{91}{6} \approx 15.167$
(2) $E[XY]$ where $X, Y$ are independent $\text{Uniform}\{1,\ldots,6\}$:
By independence, $E[XY] = E[X] \cdot E[Y] = 3.5 \times 3.5 = \frac{49}{4} = 12.25$.
(3) $E[M^2]$ where $M$ is the median (3rd order statistic) of 5 i.i.d. $\text{Uniform}\{1,\ldots,6\}$ rolls:
The PMF of $M$ can be computed by direct enumeration over all $6^5 = 7776$ outcomes. By symmetry of the uniform distribution, $P(M = k) = P(M = 7-k)$, and we get:
| $k$ | $P(M=k)$ | |---|---| | 1 |
By the symmetry $P(M=k) = P(M=7-k)$, we immediately get $E[M] = 7/2 = 3.5$. Then:
$E[M^2] = \sum_{k=1}^{6} k^2 \, P(M=k) = \frac{2207}{162} \approx 13.623$
As a check: $\text{Var}(M) = E[M^2] - (E[M])^2 = \frac{2207}{162} - \frac{49}{4} = \frac{445}{324} \approx 1.373$, which is indeed less than $\text{Var}(X) = 35/12 \approx 2.917$.
Answer:
$E[XY] = \frac{49}{4} = 12.25 \;\;<\;\; E[M^2] = \frac{2207}{162} \approx 13.62 \;\;<\;\; E[X^2] = \frac{91}{6} \approx 15.17$
The ordering follows from the identity $E[Z^2] = \text{Var}(Z) + (E[Z])^2$. All three quantities share the same $(E[Z])^2 = 12.25$ baseline (since $E[X] = E[M] = 3.5$ and independence gives $E[XY] = (E[X])^2$). The differences come entirely from variance: $E[XY]$ adds zero variance, $E[M^2]$ adds the median's reduced variance, and $E[X^2]$ adds the full single-die variance.
Intuition
The unifying idea here is the bias-variance decomposition $E[Z^2] = (E[Z])^2 + \text{Var}(Z)$. All three experiments have the same "center" of 3.5 (or equivalently 12.25 when squared), so the ordering is entirely determined by how much variance each quantity carries. The product of two independent dice has no variance contribution at all -- independence strips it out. The median of five dice has some variance, but less than a single die because order statistics concentrate the distribution toward the center. And a single die squared carries the full variance.
This pattern shows up constantly in quantitative work. Whenever you are comparing estimators, pricing rules, or risk measures, the question is often: same expected level, but how much dispersion? The median-vs-mean comparison is especially practical -- it is the reason robust estimators (which behave like order statistics) give you tighter confidence intervals at the cost of some efficiency. In trading terms, the median is like a position with lower vol but the same expected P&L.