Minimax Ex.32 — District II: three voters, D wins outright¶

Generated from minimax_ex32_district2.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: D

Scenario¶

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.

Ballots¶

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

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 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.

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 (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

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_district2.yaml

See also¶