Dice Payoff: Product of Two vs. Square of One
You can play one of two games:
- Game 1: Roll two fair six-sided dice. Your payoff is the product of the two outcomes.
- Game 2: Roll one fair six-sided die. Your payoff is the square of the outcome.
(a) Without any calculation, argue intuitively which game has the higher expected payoff.
(b) Verify your answer with explicit calculations.
Hints
- For Game 1, use independence to factor the expectation: $E[D_1 D_2] = E[D_1] E[D_2]$. For Game 2, think about the relationship between $E[D^2]$ and $(E[D])^2$.
- Invoke Jensen's inequality: for any convex function $f$ and random variable $X$, $E[f(X)] \geq f(E[X])$. The function $f(x) = x^2$ is convex.
- Compute $E[D] = 21/6 = 3.5$, then $E[D^2] = (1+4+9+16+25+36)/6 = 91/6$. Compare to $(E[D])^2 = 12.25$. The gap equals $\text{Var}(D)$.
Worked Solution
How to Think About It: Both games square a die roll in some sense, but Game 1 multiplies two independent rolls while Game 2 squares a single roll. The key question: is $E[D_1 \cdot D_2]$ greater or less than $E[D^2]$? Since $D_1$ and $D_2$ are independent and identically distributed, $E[D_1 D_2] = E[D_1] E[D_2] = (E[D])^2$. And $E[D^2] \geq (E[D])^2$ by Jensen's inequality (squaring is convex). So Game 2 wins.
Quick Estimate: $E[D] = 3.5$. Game 1 pays approximately $3.5^2 = 12.25$. Game 2 will pay something above that -- the difference is $\text{Var}(D)$, which we expect to be around 2-3.
(a) Intuitive argument:
Game 1 produces the product of two independent dice. By independence: $E[D_1 D_2] = E[D_1] \cdot E[D_2] = (E[D])^2$.
Game 2 squares a single die: $E[D^2]$. By Jensen's inequality, since $f(x) = x^2$ is convex: $E[D^2] \geq (E[D])^2$, with equality only if $D$ is constant. Since a die is not constant, $E[D^2] > (E[D])^2 = E[D_1 D_2]$.
Game 2 has the higher expected payoff.
(b) Calculation:
For a fair six-sided die, $D$ is uniform on $\{1, 2, 3, 4, 5, 6\}$.
$$E[D] = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$
$$E[D^2] = \frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} = \frac{1+4+9+16+25+36}{6} = \frac{91}{6} \approx 15.17$$
$$E[D_1 D_2] = (E[D])^2 = 3.5^2 = 12.25$$
$$\text{Difference} = E[D^2] - (E[D])^2 = \frac{91}{6} - \frac{49}{4} = \frac{182}{12} - \frac{147}{12} = \frac{35}{12} \approx 2.92$$
This difference is exactly $\text{Var}(D)$, confirming the Jensen argument.
Answer: Game 2 has the higher expected payoff: $91/6 \approx 15.17$ vs. $12.25$. The gap is exactly $\text{Var}(D) = 35/12 \approx 2.92$.
Intuition
The gap between $E[D^2]$ and $(E[D])^2$ is the variance of $D$ -- this is not a coincidence, it is the definition: $\text{Var}(D) = E[D^2] - (E[D])^2$. Jensen's inequality, applied to any convex function, gives you this kind of result automatically. The squaring function is convex, so it rewards dispersion: a random variable that swings between extremes will have a higher expected square than a constant variable at the mean.
This principle has real applications. In options pricing, convexity (the gamma of a call) means that an option on a volatile asset is worth more than an option on a flat asset, even if the expected future price is the same -- because the payoff function $\max(S_T - K, 0)$ is convex in $S_T$. The variance term that shows up here ($\frac{1}{2}\sigma^2 S^2 \Gamma$) is the continuous-time analog of the dice variance gap. Whenever you see a nonlinear payoff on a random variable, the first question to ask is: what does Jensen say about direction, and how big is the convexity correction?