Arrow's Theorem

This scene stress-tests rules that merge many rankings into one. Switch profiles and rules and watch what breaks.

Majority rule can spin in a circle. A beats B, and B beats C. Then C turns around and beats A. The Borda count wobbles for another reason. Delete a hopeless alternative and the ranking flips. No rule keeps all four fairness conditions.

For example, three 🧑 friends run for class president. The class splits into three groups with clashing favorites. Head to head, A beats B and B beats C. Yet C beats A. Majority rule alone cannot pick a president.

no rule satisfies U + P + IIA + non-dictatorship\text{no rule satisfies U + P + IIA + non-dictatorship}

Part of these learning paths

Examples & Exercises

Counterexample

For the profile (1: A>B>CA > B > C, 2: B>C>AB > C > A, 3: C>A>BC > A > B), show by direct computation that pairwise majority cycles.

A>B,B>C,C>AA > B,\quad B > C,\quad C > A
Verdict

The claim is false — pairwise majority cycles, A>B>C>AA > B > C > A, so it yields no transitive social ranking.

Condorcet paradox — no transitive social ranking.
Show proof

Counterexample and verification

  1. AA vs BB: voters 1 and 3 support A>BA > B, so A>BA > B (2–1).

  2. BB vs CC: voters 1 and 2 have B>CB > C, so B>CB > C (2–1). CC vs AA: voters 2 and 3 have C>AC > A, so C>AC > A (2–1). Hence A>B>C>AA > B > C > A cycles — pairwise majority yields no transitive ranking.

Counterexample

For the profile (1, 2: B>A>CB > A > C, 3: A>C>BA > C > B), compute Borda scores and show how the AA–BB order changes when CC is removed (IIA violation).

full: A=B=4,without C: B=2>1=A\text{full: } A = B = 4,\quad \text{without } C\text{: } B = 2 > 1 = A
Verdict

The claim is false — the Borda rule violates IIA: removing the irrelevant alternative CC flips the AA and BB order.

The AA–BB order flips when CC is removed.
Show proof

Counterexample and verification

  1. Full Borda scores (2/1/0): A=1+1+2=4A = 1+1+2 = 4, B=2+2+0=4B = 2+2+0 = 4, C=0+0+1=1C = 0+0+1 = 1 — AA and BB tie.

  2. Without CC (1/0): B=1+1+0=2B = 1+1+0 = 2, A=0+0+1=1A = 0+0+1 = 1 — BB wins. Removing the irrelevant alternative CC changed the AA–BB ranking, violating IIA.

17. Social Choice and Welfare