Borda SCC Ex.15 — STAR: C wins (and stays won when a loser exits)¶

Generated from bv2161_q3h4fk_star.yaml — do not edit by hand. Regenerate: python STARVote_LH_tabulation_engine/tools_adam/scripts/build_yaml_pages.py.

Method: STAR (single winner) · 1 seat · Expected winner: C

▶ Live on BetterVoting: vote · results ↗ (election q3h4fk · test BV2161).

Scenario¶

Race 1 of 2 in the Borda-SCC election (BV2161, bvid q3h4fk; BV-confirmed). 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 A5, Example 15. 7 voters, three candidates: 2×(A>C>B), 2×(B>A>C), 3×(C>B>A). NOTE: the paper prints the third bloc as C>A>B, but its own Borda totals (6/7/8, summing to 21) are only consistent with C>B>A — this case uses the arithmetic-consistent profile. The Borda paradox lives on the case page: Borda elects C (8 points), but if B — a LOSING candidate — drops out, Borda on the remaining pair elects A 4–3: SCC violated. This STAR race (5/3/1 map) scores A 19, B 21, C 23; C beats B 5–2 in the runoff — STAR agrees with Borda's initial pick. Honest note: if B exits, A beats C head-to-head 4–3, so a two-candidate contest elects A under ANY method, STAR included — on a cyclic profile (B>A 5–2, A>C 4–3, C>B 5–2) every method's winner is exit-sensitive; Felsenthal's charge is that Borda's POINT COUNT did the flipping. The cycle is also why no Ranked Robin or IRV race exists here (random ties on BV). Live results: https://bettervoting.com/q3h4fk/results

Ballots¶

Row 1 = candidate names; each later row is one voter's 0–5 scores (a N × prefix = N identical ballots).

A,B,C
5,1,3
5,1,3
3,5,1
3,5,1
1,3,5
1,3,5
1,3,5

What the engine says¶

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

[Divergence from STAR]
  STAR     = C
  Approval = B   (differs from STAR)

--- STAR Voting Method (single winner) ---

[STAR Voting]
 Tabulating 7 ballots.
Count × A,B,C
    3 × 1,3,5
    2 × 5,1,3
    2 × 3,5,1

[STAR Voting: Scoring Round]
 The two highest-scoring candidates advance to the next round.
   C             -- 23 -- First place
   B             -- 21 -- Second place
   A             -- 19
 C and B advance.

[STAR Voting: Automatic Runoff Round]
 The candidate preferred in the most head-to-head matchups wins.
   C             -- 5 -- First place
   B             -- 2
   Equal Support -- 0
 C wins.
   Runoff math:
     7  ballots cast
   − 0  Equal Support (no preference between the two finalists)
     ─
     7  voters with a preference  (majority = 4)
           C 5 (71%)  ·  B 2 (29%)

[STAR Voting: Winner — STAR Voting Method (single winner)]
 C

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

--- Runoff (Preference) Matrix ---
Head-to-head / pairwise comparison
Legend: For - Equal Support - Against
        * indicates Top 2 Finalist
               |      A     |   * B     |   * C     |
-----------------------------------------------------
           A > |    ---     |2 - 0 - 5  |4 - 0 - 3  |
         * B > | 5 - 0 - 2  |   ---     |2 - 0 - 5  |
         * C > | 3 - 0 - 4  |5 - 0 - 2  |   ---     |

[Condorcet Winner]
  No Condorcet winner (majority cycle: A > C > B > A)

[Score Distribution] (how many ballots gave each star rating)
                Score
Candidate  5  4  3  2  1  0  | Total   Avg
A          2  0  2  0  3  0  |    19   2.7
B          2  0  3  0  2  0  |    21   3.0
C          3  0  2  0  2  0  |    23   3.3

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/bv2161_q3h4fk_star.yaml

See also¶

Methods disagree on this election — its entry in the divergence review ledger
Condorcet efficiency (topic hub)
Ties & tie-breaking (topic hub)
Runoff reversal (worked set)
Glossary · all cases by method