election_title: "Minimal tilted cycle — 5 voters, margins 3–1–1 (Ranked Robin)"

scenario_description: |-
  The Fig. 1 profile from Brandt, Dong & Peters, "Condorcet-Consistent Choice
  Among Three Candidates" (arXiv:2411.19857) — the SMALLEST electorate that can
  produce a *tilted* (asymmetric) Condorcet cycle:

      Ada beats Ben   4–1   (margin +3)
      Ben beats Cara  3–2   (margin +1)
      Cara beats Ada  3–2   (margin +1)

  Every voter here is perfectly transitive; the electorate is not. Copeland
  (Ranked Robin) ties all three at 1–1 — the "three-way dead rung" shape — and
  falls through to LH's margins tiebreak, which is ALIVE here (+2 / 0 / −2), so
  Ada wins deterministically without a lot.

  The tilt is what makes this profile useful: on the SYMMETRIC 6-voter cycle
  (reinforcement_paradox/cases/reinf_north_c3_b6_rr.yaml) every Condorcet method
  returns all three candidates and even the margins rung is dead. Tilt it by one
  voter and Copeland STILL ties all three, while the maximin family (Minimax /
  Ranked Pairs / Schulze / Kemeny — one and the same rule at three candidates)
  drops Ben and returns {Ada, Cara}. That is the minimal proof that Ranked
  Robin's Copeland count is NOT in the maximin family.

  Why 5 voters and why 3–1–1: see the minimality proof in the folder README —
  the three cyclic margins always sum to at most n and share n's parity, so
  n = 3 forces the symmetric (1,1,1) cycle, n = 4 admits NO cycle at all, and
  n = 5 leaves (3,1,1) as the only tilted shape in existence.

  LH-only by design: the Copeland three-way tie is exactly the case where
  BetterVoting breaks ties at RANDOM, so a BV result here could not be frozen.

  Companion: tilted_cycle_c3_b5_irv.yaml (same ballots, RCV-IRV → Cara).

paradoxes: [condorcet-cycle]

voting_method: RankedRobin
num_winners: 1

lot_numbers: [Ada, Ben, Cara]

ballots: |-
  2:Ada>Ben>Cara
  1:Ben>Cara>Ada
  2:Cara>Ada>Ben

expected_winners:
  - Ada