election_title: "Minimax Ex.32 — amalgamated: both districts elected D, together they cannot"
scenario_description: |-
  The amalgamation half of Felsenthal's Minimax reinforcement example. Source: Dan S. Felsenthal (2010), Appendix A10, Example 32.
  All 14 ballots from both districts in one election: District I is Example 29's eleven voters (2×D>A>C>B, 3×D>B>A>C, 3×C>B>A>D, 1×B>A>C>D, 2×A>C>B>D), where Minimax elects D; District II is minimax_ex32_district2.yaml (2×D>A>B>C, 1×B>A>C>D), where D is the outright majority winner. Each district on its own elects D.
  Counted together the matrix goes almost flat: four of the six pairs are DEAD HEATS at 7–7 (A–D, B–C, B–D, C–D), and the only decided pairs are B over A 8–6 and A over C 11–3. So B and D are both unbeaten and both suffer a worst opposition of 7 — they tie for the smallest, and Minimax has no further clause. D's two clean district wins dissolve into a coin flip with a candidate who won neither district. That is the reinforcement paradox, also called the inconsistency paradox.
  The tie is the whole point, so this case is deliberately NOT determinate under Minimax and could never be published as a BetterVoting election with a meaningful winner. Note the electorate is EVEN (14), which is what allows exact pairwise ties at all; Felsenthal's other Minimax examples use odd electorates, where no pair can draw and the winning-votes and margins conventions cannot come apart.
  Labels are Felsenthal's own A/B/C/D so the case can be read side by side with the paper's table.
  Tabulated here as Ranked Robin for the pairwise matrix Minimax reads; Ranked Robin has its own answer, B, reached by its own tiebreak.
paradoxes: [multiple-districts, condorcet-cycle]
voting_method: RankedRobin
num_winners: 1
ballots: |-
  2:D>A>C>B
  3:D>B>A>C
  3:C>B>A>D
  1:B>A>C>D
  2:A>C>B>D
  2:D>A>B>C
  1:B>A>C>D
expected_winners:
  - B

# file: minimax_ex32_amalgamated.yaml