Two Seats, a Dollar Each, and Only One Worth Taking

Both seats in the marble game average a dollar a play, and that arithmetic stays true to the last line. Seat A carries variance 3/2 against seat B's 1, and seat A's law turns out to be seat B's law with one prize smeared outward, so every concave utility prefers B without variance ever being mentioned. Once both players stop flipping coins, seat B is ahead on the average too, at 1 against 3/4.

Two players each hold a red marble and a blue one, and each shows one, chosen by a fair coin flip. If both show red, seat A collects three dollars. If both show blue, seat A collects one dollar. If the colours differ, seat B collects two dollars. A third party at the table pays whoever collects, so nothing passes between the two seats. Which seat would you rather have?

The usual first answer is that it makes no difference, because both seats average a dollar a play. That arithmetic is correct and it stays correct all the way to the end of this article. It simply does not answer the question, because “would you rather” is a question about two whole distributions and an average is one number read off each.

Write down what each seat actually holds

The four colour pairs are equally likely, so each has probability $1/4$ . Two of them are mismatches. Collecting terms gives two payoff laws that look nothing alike:

A \sim \begin{cases} 0 & \tfrac12 \\ 1 & \tfrac14 \\ 3 & \tfrac14 \end{cases} \qquad\qquad B \sim \begin{cases} 0 & \tfrac12 \\ 2 & \tfrac12 \end{cases}

(1)

Seat A wins on agreement, with two winning amounts that differ by a factor of three. Seat B wins on disagreement and always for the same amount. Both sit idle half the time, which is what makes the comparison sharp: the seats fail equally often, so everything separating them is in the shape of the winnings.

Fig. 1 — Same mean, different shapes. Seat A replaces seat B's single prize with a pair of prizes straddling it.

A dollar each, and that part is true

Averaging over the four pairs, seat A collects $(3 + 1 + 0 + 0)/4 = 1$ and seat B collects $(0 + 0 + 2 + 2)/4 = 1$ . Nothing is being hidden here and there is no sleight of hand to expose. The equality is exact and it is the honest starting point of any answer.

One hypothesis carries it, and it is worth flagging early. Both players are flipping fair coins, which is a statement about behaviour rather than a property of the payoff table. The last section measures how far the equal averages move once either player thinks about it.

Second moment minus squared mean

With the means in hand the variances take one line each. For seat A the second moment is $(9 + 1 + 0 + 0)/4 = 5/2$ , and for seat B it is $(0 + 0 + 4 + 4)/4 = 2$ . Subtracting the squared mean, which is one in both cases:

\operatorname{Var}(A) = \frac{5}{2} - 1 = \frac{3}{2}, \qquad \operatorname{Var}(B) = 2 - 1 = 1.

(2)

Seat A carries half again as much variance for the same average return. In dollars rather than squared dollars the standard deviations are $1.2247$ and $1$ . A player who dislikes risk takes seat B, and that is the answer a room is looking for.

The argument that never mentions variance

Equal means plus a smaller variance is a weaker premise than it looks. It does not by itself imply that every risk-averse player prefers the tighter law, and the last section builds a counterexample. What rescues the conclusion here is that something stronger happens to be true, and it can be read straight off Fig. 1.

Take seat B's law. Its mass at zero is half, matching seat A exactly. Its remaining half sits on the single value two. Split that half symmetrically into a quarter at one and a quarter at three. The conditional average of the split is $(1 + 3)/2 = 2$ , so the operation moves no mass across the mean, and what it produces is precisely seat A. Seat A is seat B with one prize smeared outward.

Fig. 2 — The spread that turns B into A. Splitting the prize of two into one and three preserves the mean, so no concave player can gain from it.

Now let $u$ be any concave utility whatsoever, named or unnamed. Concavity applied to the midpoint of one and three says $\tfrac12 u(1) + \tfrac12 u(3) \le u(2)$ , and multiplying by the mass that was moved gives the comparison outright:

\mathbb{E}[u(A)] \;=\; \tfrac12 u(0) + \tfrac14 u(1) + \tfrac14 u(3) \;\le\; \tfrac12 u(0) + \tfrac12 u(2) \;=\; \mathbb{E}[u(B)],

(3)

with strict inequality as soon as $u$ is strictly concave. No variance was computed and no utility was chosen. This is what licenses the flat recommendation to take seat B, rather than a recommendation contingent on measuring risk by the second moment.

Mean-preserving spread — second-order dominance

$Y$ is a mean-preserving spread of $X$ when $Y$ can be built from $X$ by replacing some of its mass with a distribution of the same conditional mean. Equivalently $\mathbb{E}[X] = \mathbb{E}[Y]$ and $\int_{-\infty}^{t} F_X \le \int_{-\infty}^{t} F_Y$ for every $t$ . Every concave utility then ranks $X$ at least as high as $Y$ , and $X$ is said to dominate $Y$ in the second order. Here the integrated cumulative laws are $0.75$ against $0.875$ at $t = 1.5$ , with equality at both ends.

What happens once both players stop flipping

Return to the fifty-fifty hypothesis. Suppose seat B keeps flipping and seat A stops. If seat A shows red every time, agreement happens half the time and it is always the expensive kind, so seat A collects $3 \times \tfrac12 = \tfrac32$ instead of one. The figure of a dollar each is a property of one pair of strategies, not of the game.

Solving the game properly, let $a$ and $b$ be the probabilities that seats A and B show red. The two expected payoffs are:

\pi_A(a,b) = 3ab + (1-a)(1-b), \qquad \pi_B(a,b) = 2\big[a(1-b) + (1-a)b\big].

(4)

Differentiating, $\partial \pi_A / \partial a = 4b - 1$ vanishes at $b = 1/4$ , and $\partial \pi_B / \partial b = 2(1 - 2a)$ vanishes at $a = 1/2$ . Each payoff is linear in its own variable, so those are the only points at which neither player has a direction to move:

(a^\star, b^\star) = \left(\tfrac12, \tfrac14\right) \quad \Longrightarrow \quad \pi_A = \tfrac34, \qquad \pi_B = 1.

(5)

Once both players optimise, seat B is ahead on the average as well as on the spread. Seat A cannot beat three quarters against that mix. Both statements hold at once because of the third party: the payments are never netted between the seats, so no accounting identity forces one seat's gain to be the other's loss.

Where variance stops being the right order

Variance is not a complete order on distributions with a common mean. It reads one number off the squared deviations, so it cannot see which side of the mean a deviation falls on, and a rare catastrophe enters only through its small probability.

Here is an exact counterexample. Let $X$ pay nothing with probability $1/100$ and $100/99 = 1.0101$ otherwise, and let $Y$ pay $1/2$ or $3/2$ with equal probability. Both average one dollar. Their variances are $\operatorname{Var}(X) = 1/99 = 0.0101$ and $\operatorname{Var}(Y) = 1/4$ , so $Y$ carries $99/4 = 24.75$ times the variance of $X$ . Now take the concave utility $u(x) = x - k\max(0,\ \tfrac12 - x)$ with $k > 0$ , which is linear except for a penalty on falling below half a dollar. Since $Y$ never falls below half a dollar, $\mathbb{E}[u(Y)] = 1$ , while $X$ hits zero one time in a hundred and $\mathbb{E}[u(X)] = 1 - k/200$ . For every positive $k$ this risk-averse player strictly prefers the prospect with almost twenty-five times the variance.

The reason the earlier argument does not rescue that pair is exactly the hypothesis it needs. The cumulative laws of $X$ and $Y$ cross, at $0.01$ against $0$ below half a dollar and at $0.01$ against $0.5$ above it, so neither is a mean-preserving contraction of the other and no second-order ranking exists. When distributions are ordered by spread, variance and every concave utility agree. When they are not, variance is a convention, and quoting it as though it settled the matter is the same error as quoting the mean.

Two smaller caveats belong with that one. A risk-seeking player should take seat A, which is a different preference rather than a mistake, so the verified claim is that a risk-averse player takes seat B. And the variances $3/2$ and $1$ are in squared dollars, so the pair of standard deviations is the honest thing to quote at a table.

Sources and further reading

Josef Hadar and William R. Russell, “Rules for Ordering Uncertain Prospects”, American Economic Review59 (1969), 25–34, where the second-order rule is stated in the form used above.
Michael Rothschild and Joseph E. Stiglitz, “Increasing Risk I: A Definition”, Journal of Economic Theory2 (1970), 225–243, on mean-preserving spreads and why variance is the wrong primitive.
John F. Nash, “Non-Cooperative Games”, Annals of Mathematics54 (1951), 286–295, for the equilibrium concept used in the last section.
Wikipedia: Variance, Risk aversion and Stochastic dominance.

The payoff space here has four points, so every expectation and variance above was computed in exact fractions rather than floating point, and the equilibrium was located by sweeping both strategies on a fine grid rather than by trusting the derivative alone.