Reinforcement — combined 9 voters, counted by Ranked Robin (Cara wins)¶

Generated from reinf_combined_c3_b9_rr.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: Cara

▶ Live on BetterVoting: vote · results ↗ (election t4by6x · test BV2254).

Official tie-break (lot) order: Ada > Ben > Cara — consulted only if every deterministic tiebreaker stays tied (how the ladder works).

Scenario¶

North (6 voters) + South (3 voters) merged, from Brandt, Dong & Peters, "Condorcet-Consistent Choice Among Three Candidates" (2024), Theorem 2 (P1 + P2). Ada was a winner in BOTH districts (outright in South, a co-winner in North's dead-heat cycle) — so reinforcement/consistency says Ada should win the merged election. Instead a NEW Condorcet winner appears:

Cara beats Ada 5–4 · Cara beats Ben 5–4 · Ada beats Ben 7–2

Cara beats everyone head-to-head, so every Condorcet method — Ranked Robin included — elects Cara. Ada, the only candidate who won both parts, loses. That is the reinforcement paradox, which the paper proves is unavoidable for EVERY Condorcet extension once there are ≥ 8 voters.

The same 9 ballots counted by STAR: reinf_combined_c3_b9_star.yaml (the scoring round leads Ada, but the runoff flips to Cara — STAR's runoff catches the same pairwise flip). Additive methods (Score/Approval/Plurality) instead keep Ada and show no paradox — see the folder README.

Ballots¶

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

2:Ada>Ben>Cara
2:Ben>Cara>Ada
3:Cara>Ada>Ben
2:Ada>Cara>Ben

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 9 ballots (ranked ballots).

Ballots:
     2 × Ada > Ben > Cara
     2 × Ben > Cara > Ada
     3 × Cara > Ada > Ben
     2 × Ada > Cara > Ben

Round-Robin — every pair, head-to-head (For – Against):
   Ada   beats Ben    7 – 2
   Cara  beats Ada    5 – 4
   Cara  beats Ben    5 – 4

--- Pairwise (Round-Robin) Matrix ---
Head-to-head / pairwise comparison — the Ranked Robin tally
Legend: For - Equal Support - Against   (row vs column)
         |    Ada    |   Ben    |  Cara    |
--------------------------------------------
   Ada > |    ---    |7 - 0 - 2 |4 - 0 - 5 |
   Ben > | 2 - 0 - 7 |   ---    |4 - 0 - 5 |
  Cara > | 5 - 0 - 4 |5 - 0 - 4 |   ---    |

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  Cara       2–0–0         2      +2  Ada, Ben
    2  Ada        1–1–0         1      +4  Ben
    3  Ben        0–2–0         0      -6  —

Winner — Ranked Robin (RCV-RR): Cara
   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 3): Cara
   Outside (2):        Ada, Ben
   One member ⇒ Cara is the Condorcet winner, beating every rival head-to-head.
   Ranked Robin (RCV-RR) winner Cara 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/reinforcement_paradox/cases/reinf_combined_c3_b9_rr.yaml

Reinforcement — combined 9 voters, counted by Ranked Robin (Cara wins)¶

Scenario¶

Ballots¶

What the engine says¶

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

See also¶