1
Preference is a relation on options
A ≻ B says the chooser ranks A strictly above B. The options may be cakes, careers or lotteries; the relation does not care, and that is why one condition on it can apply to all of them.
Lesson · Decision and Rationality · Preference
Preference, Transitivity and the Money Pump
Preference is a relation on options, and rational choice needs it to be complete and transitive. Build a cyclic preference on the relation grid, read the pair that breaks transitivity, and compute what a trader earns by walking an agent round the cycle.
A ≻ B A is preferred to B
complete: of any two, one is preferred
transitive: A ≻ B, B ≻ C give A ≻ C
cycle: A ≻ B, B ≻ C, C ≻ A
pay ε a swap: lose 3ε every lap
Transitive, or a trader eats you
Core ideas
What this lesson establishes
A theory of choice has to start from what the chooser prefers. Before any number appears, there is a structure the preferences must have.
2
Two conditions make a ranking
Complete: for any two options one is preferred or they are tied. Transitive: if A ≻ B and B ≻ C, then A ≻ C. Together they let the options be lined up, with ties sharing a rung.
3
A cycle has no best option
If A ≻ B, B ≻ C and C ≻ A, every option is beaten by another, so choosing “the best” has no answer. The money pump shows the same defect from the outside, as a loss.
The material
Preference, and the loop that cannot be ranked
Two conditions, then the argument that the second one is not optional.
Choice theory begins by asking what the chooser would pick from a pair. Write A ≻ B when A is strictly preferred to B, and A ≽ B when A is at least as good: preferred or tied. Everything the course computes later, expected values and rules for ignorance included, assumes that this relation can be turned into a ranking.
Definition · Completeness
A preference is complete when, for any two options A and B, at least one of A ≽ B and B ≽ A holds. The chooser is never left without a view on a pair.
Definition · Transitivity
A preference is transitive when A ≽ B and B ≽ C together give A ≽ C. A comparison that passes through a third option does not change the answer.
The relation grid below makes both conditions things you can see. Read a 1 in row a, column b as “option a is preferred to option b”. A complete strict preference has a 1 on one side of the diagonal or the other for every pair, with no gap. A transitive one has no two-step path a to b to c without the direct step a to c; the property table names the first offending pair when there is one.
The preset is the relation a < b on five options, and as a strict preference it is both complete and transitive: option 1 beats all, option 5 beats none. Now click the cell in row 5, column 1. That adds the pair 5 ≻ 1 and closes a loop, because the preset already ranks 1 ≻ 5 through the cell in row 1, column 5. The table reports transitivity failing at the pair (1, 5) and (5, 1) without (1, 1): a cycle forces each option to be preferred to itself, which no one means by preference.
A ≻ B
B ≻ C
C ≻ A
Nothing in those three lines is false as a report of taste. The trouble is what a cyclic chooser will do. Suppose you hold A, and you prefer C to A, B to C and A to B. A trader offers C for A plus a small fee ε, and you accept because you prefer C. The trader then offers B for C plus ε, and you accept. Then A for B plus ε, and you accept again.
Example · One lap
You began with A and you end with A. You are 3ε poorer, and each trade was one you wanted at the moment you made it. Another lap costs another 3ε, and the trader can run as many laps as you will go round.
That is the money pump. It is an argument that transitivity is a condition on what you prefer, not a feature of what you like. The options can be anything and the fee as small as you please; the loss is 3ε a lap in every case, so a cycle among three kinds of tea is exposed in the same way as a cycle among three jobs.
The argument has a strongest objection, and it deserves a hearing. A person may rank three flats by different standards, price against size, size against light, light against price, and each pairwise verdict can be reasonable. The reply is that this is a cycle in the pairwise verdicts and not in the ranking of the flats as wholes, so one of the standards has to give way or a weighting has to be chosen before the person can act without being walked in circles. Accepting the cycle costs the chooser the loss above; rejecting a pairwise verdict costs a reason that looked sound.
The limit of the argument is worth stating. It assumes the chooser really will pay for each swap and has no memory of the lap. An agent who sees the trader coming and refuses the third offer has changed her preferences, which is exactly the repair that transitivity asks for. The pump does not prove cycles unthinkable; it prices them.
Interactive
Relation properties
Decided by checking, not by name
Method
Testing a preference, and pricing it
Four moves, in this order. The first three decide whether the ranking exists and the fourth puts a price on its failing.
1
List the strict preferences as pairs
Write each X ≻ Y the chooser holds, one per pair of options. If some pair has no entry and no tie, the preference is not complete, and a different question is being asked.
2
Follow every two-step chain
For each A ≻ B and B ≻ C, look for A ≻ C. A missing pair is the witness that transitivity fails, and it is the only thing the check has to find.
3
Look for a way round
If a chain of strict preferences returns to where it began, the preference has a cycle, and no option in the cycle is best.
4
Price one lap
Start at any option in the cycle and take each preferred swap in turn, paying the fee each time. A lap of k swaps costs k·ε, and the agent holds exactly what it started with.
Cycles and what they cost
Score 0 / 0Choose the answer you can justify.
A cycle of three, walked once
The chooser prefers C to A, B to C and A to B, and pays 1 cent for each swap to something preferred. She starts holding A.
holds A cash 0
A -> C pays 1 cent, prefers C to A cash -1
C -> B pays 1 cent, prefers B to C cash -2
B -> A pays 1 cent, prefers A to B cash -3
holds A cash -3 cents after one lap
after 100 laps cash -300 cents, still holds A
Each line, taken alone, is a choice the chooser endorses. It is the sequence that has no defence: she ends with what she began with and three cents fewer, and the pump may run for as long as she keeps saying yes.
Common mistakes
1
Treating transitivity as a matter of tasteTaste decides whether you prefer tea to coffee. Transitivity constrains how your preferences fit together, and it is violated by a cycle among any three options whatever they are. The cycle costs 3ε a lap whether the options are drinks, jobs or flats, which is a fact about the structure and not about the contents.
2
Reading a cycle as indifferenceIf A, B and C are tied, nothing is wrong: ties are transitive, and no one pays to swap between things they value equally. A cycle is the opposite case, a strict preference all the way round. Each swap in the pump is paid for because the agent prefers what it gets.
3
Expecting a single choice to show the defectPick from the pair A and B and nothing looks wrong, and the same holds for the other two pairs. The cost appears only in a sequence, and it is the sequence the argument concerns.
Check your understanding
✓
Finish when you can build a cycle and put a price on it.
Given a set of options and the pairs a chooser prefers, you should be able to say whether the preference is complete, name the pair that breaks transitivity if there is one, and compute what a trader collects over a given number of laps at a given fee.
Note: The lab here is the relation grid from the Discrete Mathematics path, used for a different job. Nothing from that path is assumed: a preference is a relation on the options, transitivity is the property defined above, and what this lesson adds is the reason to want it. “The Decision Matrix and Dominance” starts from preferences that are in order and asks how to choose among acts.