====================================================================== SOURCE FILE: coombs_ex20_amalgamated.yaml TABULATED FILE: coombs_ex20_amalgamated_tabulated.txt ====================================================================== election_title: "Coombs Ex.20 — amalgamated: both districts chose B, their union chooses A" scenario_description: |- The amalgamation half of Felsenthal's Coombs reinforcement example. Source: Dan S. Felsenthal (2010), Appendix A7, Example 20. All 41 ballots from both districts in one election: District I's 34 (9×A>B>C, 9×B>C>A, 11×C>A>B, 5×C>B>A) plus District II's 7 (1×A>B>C, 6×B>A>C). Each district elected B on its own. Counted together the last-place tally changes hands: C is now last on 16 ballots — more than A's 14 or B's 11 — so Coombs deletes C instead of A, and the ballots C was holding lift A to a majority. A wins. Neither district wanted A; their union does. That is the reinforcement paradox, also called the inconsistency paradox, and it is the formal reason a method that cannot be summed district by district cannot be canvassed that way either. Labels are Felsenthal's own A/B/C so the case can be read side by side with the paper's table. Tabulated as RCV-IRV, the mirror-image count. IRV elects B here and in District II. In District I it does not have a determinate answer at all: A and B tie on nine first places each of 34, and that arbitrary first elimination decides the winner (this engine breaks it toward B, RCTab toward C in three of six declared candidate orders). So IRV is a clean control in two of the three files, not all three — a weaker contrast than it first looked, and worth stating rather than glossing. The Coombs reinforcement failure does not lean on it either way: Coombs' own eliminations are untied in all three files, which is why the paradox is still Coombs' alone. That contrast is why the three files are worth having separately. paradoxes: [multiple-districts] voting_method: RCV_IRV num_winners: 1 ballots: |- 9:A>B>C 9:B>C>A 11:C>A>B 5:C>B>A 1:A>B>C 6:B>A>C expected_winners: - B # file: coombs_ex20_amalgamated.yaml ====================================================================== TABULATION RESULTS ====================================================================== --- RCV / Instant-Runoff Voting (single winner) --- Coombs Ex.20 — amalgamated: both districts chose B, their union chooses A Tabulating 41 ballots (ranked ballots). ROUND 1 Candidate Votes Status ----------- ------- -------- C 16 Hopeful B 15 Hopeful A 10 Rejected FINAL RESULT Candidate Votes Status ----------- ------- -------- B 25 Elected C 16 Rejected A 0 Rejected Winner(s) — RCV / Instant-Runoff Voting (single winner) B --- 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 (3 of 3): A, B, C Outside (0): — More than one member ⇒ NO Condorcet winner: the top of the tournament is a cycle, so the strongest "candidate" is a set, not a person. Which member of the set should win is exactly what Minimax / Ranked Pairs / Schulze disagree about — see 05_Ranked_Robin/01_Learn/cycle_resolution.md. RCV-IRV winner B is INSIDE the Smith set. ✓ Not guaranteed — RCV-IRV is not Smith-efficient — but it holds here. More: 07_Concepts/topics/smith_set.md