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Allais paradox
You will learn: Show why common-consequence cancellation constrains both lottery comparisons.
Start with: Expectation and variance
Consider two separate lottery choices. In the first, A gives one prize unit for certain. B gives five units with probability 10%, one unit with probability 89%, and nothing with probability 1%. In the second choice, C gives one unit with probability 11% and nothing otherwise; D gives five units with probability 10% and nothing otherwise.
The characteristic Allais preference pair chooses A over B but D over C. A offers certainty, while the second comparison may make the larger prize attractive. Can a sufficiently curved utility function represent both preferences? Under ordinary expected utility with the same utility for each prize across lotteries, it cannot.
Inspect the hundred tickets
Each lottery uses a uniformly chosen ticket from 1 through 100. Prizes are 0, 1, or 5 units. The same prize has the same utility in every lottery.
| Lottery | Tickets 1–10 | Ticket 11 | Tickets 12–100 |
|---|---|---|---|
| A: certain one | 1 | 1 | 1 |
| B: mixed lottery | 5 | 0 | 1 |
| C: 11% one | 1 | 1 | 0 |
| D: 10% five | 5 | 0 | 0 |
No common expected-utility assignment represents this preference pair under the stated prize definitions. This compatibility check is separate from the inspected utility below.
Normalize utility of zero to 0 and five to 1. Expected-utility difference A − B = C − D = 0.00450000. This utility prefers A and C.
| Lottery | Prize on ticket 11 | Expected utility |
|---|---|---|
| A: certain one | 1 | 0.95000000 |
| B: mixed lottery | 0 | 0.94550000 |
| C: 11% one | 1 | 0.10450000 |
| D: 10% five | 0 | 0.10000000 |
The ticket representation puts every lottery on one common sample space. Tickets one through ten give the first outcome, ticket eleven gives the second, and tickets twelve through one hundred give the third. Each ticket is equally likely before it is inspected.
A and B give the same one-unit outcome on the last eighty-nine tickets. C and D instead both give zero on those tickets. Nothing changes on the first eleven tickets when moving from the A/B comparison to the C/D comparison. This shared replacement is the central construction.
The inspected ticket exposes these consequences side by side. Once a ticket is known, comparing its prizes is no longer the same decision as choosing a lottery before the random draw. Its purpose is to reveal the coupling, not to suggest that a chooser knows the winning ticket in advance.
Normalize the utilities
Write the utilities of prizes zero, one, and five as u₀, u₁, and u₅. Expected utilities are obtained by multiplying each utility by its known probability and adding. Subtracting B from A yields 0.11u₁ − 0.10u₅ − 0.01u₀. Subtracting D from C yields exactly the same expression.
A positive value therefore favors A and C; a negative value favors B and D; zero makes both comparisons indifferent. The signs cannot disagree. This algebra does not assume utility is linear in prize size or specify a particular amount of risk aversion.
Assuming a strictly better utility for five than for zero, we can normalize u₀ to zero and u₅ to one without changing preferences. The slider then specifies u₁ between zero and one. Both differences reduce to 0.11u₁ − 0.10. Indifference occurs at u₁ = 10/11, approximately 0.909091.
With linear utility in prize units, u₁ = 0.2. The differences are −0.078, favoring B and D. With u₁ = 0.95, the differences are +0.0045, favoring A and C. The second setting makes the first unit very valuable relative to the additional four, but still cannot produce A and D together.
What the independence requirement asks
The relevant requirement is that replacing a consequence shared by both alternatives on the same event should not reverse the preference between them. Here the shared event is the final eighty-nine tickets. Those outcomes cancel when expected utilities are subtracted, regardless of whether their common prize is one or zero.
The experiment checks whether some common utility assignment could represent the selected preference pair. That question differs from whether the currently inspected slider value represents it. A and C are jointly representable, for example, but a slider below 10/11 does not represent that pair.
There are substantive responses to the paradox: defend expected utility and revise the choices, use a decision theory that permits sensitivity to more than expected utility, or enrich the outcome descriptions to include features such as regret or certainty. The displayed incompatibility is conditional on the original outcome definitions. It does not by itself settle which normative account of choice is best.
Make a prediction
Would a different increasing utility function, used consistently in all four lotteries, allow strict preferences for A and D?
Explore the answer
No. Both utility differences are the same algebraic expression for every common assignment of the three prize utilities. To represent opposite signs, something beyond the curvature of that one utility function must change.
Allais’s 1953 analysis challenges expected-utility descriptions of choice under risk. Our hundred-ticket construction makes the common consequence directly inspectable. Compare the Ellsberg paradox, where the composition of an urn is undisclosed, and expected utility, where prize utility helps explain a different puzzle.