Minimax Ex.30 — before: all 19 vote, Minimax elects B¶

Generated from minimax_ex30_noshow_before.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: A

Scenario¶

The BEFORE half of Felsenthal's Minimax no-show pair. Source: Dan S. Felsenthal, "Review of Paradoxes Afflicting Various Voting Procedures Where One Out of m Candidates (m ≥ 2) Must Be Elected", University of Haifa / LSE, revised 26 May 2010; Appendix A10, Example 30 (credited to Hannu Nurmi, private communication 22.2.2010). 19 voters, four candidates: 5×(D>B>C>A), 4×(B>C>A>D), 3×(A>D>C>B), 3×(A>D>B>C), 4×(C>A>B>D). The social ordering cycles (C>A>D>B>C), so there is no Condorcet winner and Minimax must fall back on its second clause — elect whoever's worst pairwise loss is smallest. Worst losses: A 13, B 11, C 12, D 14, so Minimax elects B. Then compare minimax_ex30_noshow_after.yaml, where three of the four C>A>B>D voters stay home and A wins instead — an outcome those absent voters PREFER to B. That is the no-show paradox, and read in the other direction it is the twin paradox. Labels are Felsenthal's own A/B/C/D so the case can be read side by side with the paper's table; this is an academic reproduction, not a scenario with a cast. Minimax has no tabulator in the LH engine or on BetterVoting, so the file is tabulated here as Ranked Robin — which prints the full pairwise matrix, the exact object Minimax reads. Ranked Robin then breaks the Copeland tie by margin and elects A, not B: same matrix, different cycle-breaker. For the Minimax count itself run tools_adam/pref_voting_tabulation_engine/minimax_report.py, which is cross-checked against pref_voting.

Ballots¶

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

5:D>B>C>A
4:B>C>A>D
3:A>D>C>B
3:A>D>B>C
4:C>A>B>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 19 ballots (ranked ballots).

Ballots:
     5 × D > B > C > A
     4 × B > C > A > D
     3 × A > D > C > B
     3 × A > D > B > C
     4 × C > A > B > D

Round-Robin — every pair, head-to-head (For – Against):
   D  beats B   11 –  8
   D  beats C   11 –  8
   A  beats D   14 –  5
   B  beats C   12 –  7
   A  beats B   10 –  9
   C  beats A   13 –  6

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
      |      D       |     B       |     C       |     A       |
----------------------------------------------------------------
  D > |     ---      |11 -  0 -  8 |11 -  0 -  8 | 5 -  0 - 14 |
  B > |  8 -  0 - 11 |    ---      |12 -  0 -  7 | 9 -  0 - 10 |
  C > |  8 -  0 - 11 | 7 -  0 - 12 |    ---      |13 -  0 -  6 |
  A > | 14 -  0 -  5 |10 -  0 -  9 | 6 -  0 - 13 |    ---      |

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  A          2–1–0         2      +3  D, B
    2  D          2–1–0         2      -3  B, C
    3  B          1–2–0         1      +1  C
    4  C          1–2–0         1      -1  A

Winner — Ranked Robin (RCV-RR): A
   *** 2 candidates tie for the most wins (D, A) — tied on the tally, not a cycle (some of them beat others head-to-head, but no loop closes). Resolved by total margin, then lot order.

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): D, 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.
   Note: the Copeland leaders (D, A) are only part of the set — the
   win–loss table's top block understates how wide the contention is.
   Ranked Robin (RCV-RR) winner A 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_ex30_noshow_before.yaml

Minimax Ex.30 — before: all 19 vote, Minimax elects B¶

Scenario¶

Ballots¶

What the engine says¶

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

See also¶