Minimax Ex.32 — amalgamated: both districts elected D, together they cannot¶

Generated from minimax_ex32_amalgamated.yaml — do not edit by hand. Regenerate: python STARVote_LH_tabulation_engine/tools_adam/scripts/build_yaml_pages.py.

Method: Ranked Robin (RCV-RR / Copeland) · 1 seat · Expected winner: B

Scenario¶

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.

Ballots¶

Each row is one voter's ranking, most-preferred first (N: prefix = N identical 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

What the engine says¶

The count, step by step — the rounds and how the winner is reached:

--- Ranked Robin (RCV-RR / Copeland) Method (single winner) ---
 Tabulating 14 ballots (ranked ballots).

Ballots:
     2 × D > A > C > B
     3 × D > B > A > C
     3 × C > B > A > D
     2 × B > A > C > D
     2 × A > C > B > D
     2 × D > A > B > C

Round-Robin — every pair, head-to-head (For – Against):
   D  ties  A    7 –  7
   D  ties  C    7 –  7
   D  ties  B    7 –  7
   A  beats C   11 –  3
   B  beats A    8 –  6
   C  ties  B    7 –  7

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
      |      D       |     A       |     C       |     B       |
----------------------------------------------------------------
  D > |     ---      | 7 -  0 -  7 | 7 -  0 -  7 | 7 -  0 -  7 |
  A > |  7 -  0 -  7 |    ---      |11 -  0 -  3 | 6 -  0 -  8 |
  C > |  7 -  0 -  7 | 3 -  0 - 11 |    ---      | 7 -  0 -  7 |
  B > |  7 -  0 -  7 | 8 -  0 -  6 | 7 -  0 -  7 |    ---      |

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  B          1–0–2         2      +2  A
    2  A          1–1–1       1.5      +6  C
    3  D          0–0–3       1.5      +0  —
    4  C          0–1–2         1      -8  —

Winner — Ranked Robin (RCV-RR): B
   unbeaten, but draws D, C — a *weak* Condorcet winner, not a strict one (highest Copeland score, 2).

Full audit — preference matrix, Condorcet, and score distribution¶

--- 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 (4 of 4): B, D, A, C
   Outside (0):        —
   More than one member ⇒ NO Condorcet winner: the top of the tournament is a
   group held open by draws, so the strongest "candidate" is a set, not a
   person. Some members DO beat others, but no member beats them all — a draw
   blocks the sweep. No loop closes either, so there is no cycle for Minimax /
   Ranked Pairs / Schulze to resolve: which member wins is left to the
   tiebreak, not to a cycle rule. See
   05_Ranked_Robin/01_Learn/rr_tiebreak_lh_vs_bv.md.
   Note: the Copeland leaders (B) are only part of the set — the
   win–loss table's top block understates how wide the contention is.
   Ranked Robin (RCV-RR) winner B 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.
   Fine print: this set contains a pairwise DRAW, and a draw is enough to keep a
   candidate in the Smith set but not in the tighter Schwartz set — so Schwartz
   may be smaller here.
   More: 07_Concepts/topics/smith_set.md

Everything in one file: the _tabulated mirror (regenerated on every run; every analysis forced on).

Run it yourself:

python STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/felsenthal_paradoxes/cases/minimax_ex32_amalgamated.yaml

Minimax Ex.32 — amalgamated: both districts elected D, together they cannot¶

Scenario¶

Ballots¶

What the engine says¶

Full audit — preference matrix, Condorcet, and score distribution¶

See also¶