The Same Three and a Half Million, with a Thousandth of the Spread

Both games pay 3.5 million dollars on average, so "they match" is true and the inference that it is a wash is not. Shrinking the ticket by a million divides the spread by a million while adding a million independent rolls multiplies it back by only a thousand, so the ratio of standard deviations is exactly root of a million: 1,707,825 against 1,707.83. The whole argument rests on independence, and at a correlation of 1 the diversified game becomes the single roll exactly.

Two offers on a single fair die. Game 1: roll it once, and you are paid a million dollars per pip. Game 2: roll it a million times, and you are paid one dollar per pip. You dislike risk. Which do you take?

The expectations are identical, to the cent. Game 2 is nonetheless the one to take, and it is not close: it carries exactly a thousand times less spread than game 1, for the same average.

The averages really do match

A fair die has expectation

\mathbb{E}[X] = \frac{1+2+3+4+5+6}{6} = \frac72

(1)

Game 1 pays $10^6 X$ , so its mean is $10^6 \times 3.5 = \$3{,}500{,}000$ . Game 2 pays a sum of $10^6$ independent copies of $X$ , so by linearity its mean is $10^6 \times 3.5$ as well. Nothing here is approximate.

So "they match on average" is a true statement, and anyone who says it has done the arithmetic correctly. What is wrong is the next step, the one that concludes it is therefore a coin flip which you choose. Two offers agreeing in their first moment can differ in everything else.

Two effects, and they do not cancel

The die's variance is

\operatorname{Var}(X) = \frac{1}{6}\sum_{m=1}^{6}\left(m - \tfrac72\right)^2 = \frac{35}{12}, \qquad \sigma = \sqrt{\tfrac{35}{12}} = 1.70783

(2)

Going from game 1 to game 2 does two things to the spread. Shrinking the payout per pip from $10^6$ to 1 divides the standard deviation by $10^6$ , because scaling a random variable scales its standard deviation by the same factor. Adding $10^6$ independent rolls multiplies it back by only $\sqrt{10^6}$ , because variances add and standard deviations therefore grow as the square root of the count.

\frac{\sigma_1}{\sigma_2} = \frac{10^6\,\sigma}{\sqrt{10^6}\,\sigma} = \sqrt{10^6} = 1000

(3)

The $\sigma$ cancels, so equation (3) is exact and carries no rounding at all. The asymmetry between the two effects is the whole content of the problem: one of them is linear in the count and the other is a square root.

Variance adds, spread does not

For independent $X_1, \dots, X_N$ with common variance $\sigma^2$ , the sum has variance $N\sigma^2$ and standard deviation $\sigma\sqrt N$ . Only independence and finite variance are needed. Nothing about the shape of the distribution enters, so the factor of 1000 is not a Gaussian fact.

The two spreads, in dollars

\begin{aligned} \text{game 1} &: \; \sigma_1 = 10^6\sqrt{\tfrac{35}{12}} = \$1{,}707{,}825.13 \\ \text{game 2} &: \; \sigma_2 = \sqrt{10^6}\,\sqrt{\tfrac{35}{12}} = \$1{,}707.83 \end{aligned}

(4)

Game 1's spread is 48.795 percent of its own mean. Game 2's is 0.048795 percent. Since the two means are equal, the factor of 1000 applies to the relative spread just as it does to the absolute one, which is not automatic and is worth checking rather than assuming.

Fig. 1 — The two payoff distributions on one axis. Game 1 cannot land within five thousand dollars of its own mean at all, because its outcomes are a million dollars apart.

How tight a million rolls actually is

Five thousand dollars is $5000/1707.83 = 2.9277$ standard deviations, so the probability that game 2 lands within five thousand dollars of its mean is

\operatorname{erf}\!\left(\frac{2.9277}{\sqrt2}\right) = 0.996585

(5)

Better than 996 times in a thousand, on a payout of three and a half million. Equation (5) uses a normal approximation, which is the only asymptotic step anywhere in this problem, and sampling the exact distribution of the sum instead gives 0.996700. Close enough that the approximation is not doing any load-bearing work.

If you would rather avoid the normal approximation entirely, Chebyshev's inequality gives a crude bound with no distributional assumption at all:

\mathbb{P}\left(|S - 3.5\text{M}| > 5000\right) \le \frac{\sigma_2^2}{5000^2} = 0.1167

(6)

Weak, and enough to make the point: even without knowing the shape, the total is pinned near its mean. Meanwhile the corresponding probability for game 1 is exactly zero, since its six possible payouts are a million dollars apart and none of them is within five thousand of 3.5 million. Over 20,000 simulated plays, game 2's entire observed range spanned $13,172 and game 1 never once landed inside that window.

The general statement

Hold the total expectation fixed at $M$ and split it into $N$ independent pieces, each paying $c X$ with $Nc\,\mathbb{E}[X] = M$ . Then

\sigma_{\text{total}} = c\,\sigma\sqrt N = \frac{M}{N\,\mathbb{E}[X]}\,\sigma\sqrt N = \frac{M\sigma}{\mathbb{E}[X]}\cdot\frac{1}{\sqrt N}

(7)

so at fixed expectation the spread falls like $1/\sqrt N$ . Cutting the risk in half needs four times as many independent pieces, and cutting it by a factor of a thousand needs a million, which is exactly the arithmetic of the problem.

Fig. 2 — Equation (7) at four sample sizes. Each factor of a hundred in the count buys one factor of ten in the spread, which is a poor exchange rate and still the only one available.

What this does not say

Game 2 is not worth more than game 1. Their expectations are identical, and anyone indifferent to risk is genuinely indifferent between them. The question stipulated a taker who dislikes risk, and that stipulation is doing real work rather than decorating the problem.

Someone who actively wants risk should take game 1, and there are honest reasons to. A single roll is the only one of the two that can pay six million dollars. If your goal is a threshold rather than a total, the concentrated bet is the wrong instrument, and no amount of variance arithmetic changes that.

The assumption that carries everything

Equation (3) needs the million rolls to be independent, and the failure mode is dramatic rather than gradual. With a common correlation $\rho$ between rolls,

\operatorname{Var}\!\left(\sum_{i=1}^{N} X_i\right) = N\sigma^2\left[1 + (N-1)\rho\right]

(8)

At $\rho = 1$ , meaning one roll paid out a million times, equation (8) gives $N^2\sigma^2$ and game 2 becomes game 1 exactly. And because $N$ is a million, even a correlation of $10^{-6}$ roughly doubles the variance. A structure that looks diversified and shares a hidden common factor is not diversified at all, and this is the algebra that says so.

Finite variance is the other requirement, and it is easy to overlook because dice have it trivially. For a distribution with infinite variance the averaging in equation (7) does not concentrate anything: the average of $N$ independent Cauchy variables has the same distribution as one of them, no matter how large $N$ is. The thousand-fold reduction is a consequence of finite variance, and it fails outright without it.

Sources and further reading

The two moments in equations (1) and (2) — Discrete uniform distribution
The additivity used in equation (3) — Variance
The relative spread — Coefficient of variation
The bound in equation (6) — Chebyshev's inequality

Every constant above was checked in exact fraction arithmetic, and the concentration figure was checked twice: once against equation (5) and once by drawing the six face counts from their exact joint distribution over 20,000 plays, which is the true law of the million-roll total rather than an approximation to it. The two agree to three decimals.