====================================================================== SOURCE FILE: minimax_ex32_district2.yaml TABULATED FILE: minimax_ex32_district2_tabulated.txt ====================================================================== election_title: "Minimax Ex.32 — District II: three voters, D wins outright" scenario_description: |- District II of Felsenthal's Minimax reinforcement example. Source: Dan S. Felsenthal (2010), Appendix A10, Example 32. Three voters, four candidates: 2×(D>A>B>C), 1×(B>A>C>D). D is ranked first by an absolute majority, so D is the Condorcet winner and every reasonable method — Minimax included — elects D. There is no paradox in this district on its own; it is one of the two halves that produce one when combined. District I is Example 29's eleven voters (bv2167_f3dxq9_star.yaml), where Minimax also elects D. Both districts elect D separately. Amalgamate them and Minimax no longer does — see minimax_ex32_amalgamated.yaml, where the worst losses of B and D tie and the winner falls to a lot. That is the reinforcement paradox: a method can be unanimous across districts and undecided over their union. Labels are Felsenthal's own A/B/C/D so the case can be read side by side with the paper's table, and they match Example 29's cast because District I IS Example 29. Tabulated here as Ranked Robin for the pairwise matrix Minimax reads; with a Condorcet winner present both methods agree on D. paradoxes: [multiple-districts] voting_method: RankedRobin num_winners: 1 ballots: |- 2:D>A>B>C 1:B>A>C>D expected_winners: - D # file: minimax_ex32_district2.yaml ====================================================================== TABULATION RESULTS ====================================================================== --- Ranked Robin (RCV-RR / Copeland) Method (single winner) --- Tabulating 3 ballots (ranked ballots). Ballots: 2 × D > A > B > C 1 × B > A > C > D Round-Robin — every pair, head-to-head (For – Against): D beats A 2 – 1 D beats B 2 – 1 D beats C 2 – 1 A beats B 2 – 1 A beats C 3 – 0 B beats C 3 – 0 --- Pairwise (Round-Robin) Matrix --- Head-to-head / pairwise comparison — the Ranked Robin tally Legend: For - Equal Support - Against (row vs column) | D | A | B | C | ---------------------------------------------------- D > | --- |2 - 0 - 1 |2 - 0 - 1 |2 - 0 - 1 | A > | 1 - 0 - 2 | --- |2 - 0 - 1 |3 - 0 - 0 | B > | 1 - 0 - 2 |1 - 0 - 2 | --- |3 - 0 - 0 | C > | 1 - 0 - 2 |0 - 0 - 3 |0 - 0 - 3 | --- | Win–loss record — Copeland score = wins + ½·ties (highest score wins; ties broken by total margin, then lot order): # Candidate W–L–T Copeland Margin Beats 1 D 3–0–0 3 +3 A, B, C 2 A 2–1–0 2 +3 B, C 3 B 1–2–0 1 +1 C 4 C 0–3–0 0 -7 — Winner — Ranked Robin (RCV-RR): D beats every opponent head-to-head — the Condorcet winner. --- Smith Set (the generalized Condorcet winner) --- The smallest group whose every member beats every candidate outside it — the honest answer to "who is even in contention?". Smith set (1 of 4): D Outside (3): A, B, C One member ⇒ D is the Condorcet winner, beating every rival head-to-head. Ranked Robin (RCV-RR) winner D is INSIDE the Smith set. ✓ Guaranteed: Ranked Robin (Copeland) is Smith-efficient — every member of the set outscores every outsider, so the top of the win–loss table is always inside the set, however the tie among them is then broken. More: 07_Concepts/topics/smith_set.md