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
For the profile (1: A>B>C, 2: B>C>A, 3: C>A>B), show by direct computation that pairwise majority cycles.
A>B,B>C,C>A
Verdict
The claim is false — pairwise majority cycles, A>B>C>A, so it yields no transitive social ranking.
Condorcet paradox — no transitive social ranking.Show proof
Counterexample and verification
A vs B: voters 1 and 3 support A>B, so A>B (2–1).
B vs C: voters 1 and 2 have B>C, so B>C (2–1). C vs A: voters 2 and 3 have C>A, so C>A (2–1). Hence A>B>C>A cycles — pairwise majority yields no transitive ranking.
Counterexample
For the profile (1, 2: B>A>C, 3: A>C>B), compute Borda scores and show how the A–B order changes when C is removed (IIA violation).
full: A=B=4,without C: B=2>1=A
Verdict
The claim is false — the Borda rule violates IIA: removing the irrelevant alternative C flips the A and B order.
The A–B order flips when C is removed.Show proof
Counterexample and verification
Full Borda scores (2/1/0): A=1+1+2=4, B=2+2+0=4, C=0+0+1=1 — A and B tie.
Without C (1/0): B=1+1+0=2, A=0+0+1=1 — B wins. Removing the irrelevant alternative C changed the A–B ranking, violating IIA.