You are offered two versions of a dice game and must decide which to play.
Game A: Roll a single fair six-sided die. You receive the face value in dollars.
Game B: Roll two fair six-sided dice. You receive half their sum in dollars.
Both games cost the same to enter.
Compute the expected payoff $E[A]$ and $E[B]$ for each game.
Compute the variance $\text{Var}(A)$ and $\text{Var}(B)$ for each game.
Which game should you choose, and does your answer depend on your risk preferences?
Hints
Think about whether averaging two dice changes the expected payoff before doing any calculation -- linearity of expectation gives you the answer immediately.
For the variance of Game B, recall that $\text{Var}(cX) = c^2 \text{Var}(X)$ and that independent dice add in variance: $\text{Var}(X_1 + X_2) = \text{Var}(X_1) + \text{Var}(X_2)$.
Compute $\text{Var}(A) = E[X^2] - E[X]^2$ using $E[X^2] = \frac{1^2 + 2^2 + \cdots + 6^2}{6} = \frac{91}{6}$, then apply the scaling rule to get $\text{Var}(B) = \frac{1}{4} \cdot 2 \cdot \text{Var}(A)$.
Worked Solution
How to Think About It: Before touching any formulas, ask yourself: does averaging dice change the expected payoff? No -- expectation is linear, so halving the sum of two dice gives you the same mean as one die. That tells you right away both games pay $\\$3.50$ on average. The interesting question is variance. Averaging two dice compresses the distribution -- extreme rolls like 1 or 6 are much less likely. So Game B is the lower-risk version of the same expected payoff. Your choice is entirely about risk preference.
Quick Estimate: Single die: mean is the midpoint of 1 through 6, so $\frac{1+6}{2} = 3.5$. The spread from the mean is at most 2.5 (rolling a 1 or 6), and the outcomes are symmetric, so variance should be on the order of
^2 / 3 \approx 1$-ish -- the exact answer is $\frac{35}{12} \approx 2.92$. For two dice averaged: the distribution is triangular and much tighter. The extreme outcomes (both dice show 1, or both show 6) each have probability $\frac{1}{36}$, versus $\frac{1}{6}$ for the extremes in Game A. We expect variance to roughly halve, and it does: $\frac{35}{24} \approx 1.46$.
Approach: Compute $E[A]$ directly. Use $\text{Var}(A) = E[A^2] - E[A]^2$. Then use linearity of expectation and independence for Game B.
Formal Solution:
*Game A -- single die:*
Let $X$ be the outcome of one fair die, $X \in \{1, 2, 3, 4, 5, 6\}$ each with probability $\frac{1}{6}$.
Both games have the same expected payoff of $\\$3.50$, but Game B has variance exactly half that of Game A. The choice reduces to risk preference:
Risk-neutral: indifferent -- expected payoffs are identical.
Risk-averse: prefer Game B -- same return for less variance.
Risk-seeking: prefer Game A -- more variance for the same return.
Answer: $E[A] = E[B] = 3.5$, $\text{Var}(A) = \frac{35}{12}$, $\text{Var}(B) = \frac{35}{24}$. Game B has half the variance for the same expected payoff. A risk-averse player prefers Game B; a risk-neutral player is indifferent.
Intuition
The core principle here is that averaging reduces variance without touching the mean -- and the reduction is proportional to the number of observations. Averaging $n$ independent identically distributed variables multiplies variance by $\frac{1}{n}$. This is the mechanics behind why diversification works: adding uncorrelated positions to a portfolio keeps expected return fixed while shrinking portfolio variance. Game B is, in a literal sense, a two-asset equally weighted portfolio of dice.
The practical lesson for quant interviews is to always separate the expected value question from the variance question. Interviewers love this problem because candidates often conflate the two -- assuming that the game with a wider range of outcomes must have a higher expected payoff. It does not. Once you establish that both expectations are equal, the entire comparison collapses to a one-dimensional risk preference question, and the vocabulary of risk-neutral vs. risk-averse gives you a clean framework to answer it. In real trading, the same logic drives decisions about position sizing, lot aggregation, and the value of reducing execution noise.