Hillinger's evaluative voting — the paper, made runnable¶
Level: 301 · deep dive
Claude Hillinger's "Voting and the Cardinal Aggregation of Judgments" (Munich Discussion Paper 2004-9, DOI) argues that voting is the aggregation of judgments rather than of preferences, that judgments must be measured before they can be aggregated, and that the classic voting paradoxes are artifacts of ballots that restrict what a voter may say. His proposal is evaluative voting (EV): score every candidate on a uniform, unrestricted scale; the largest sum wins.
This folder runs his own worked example. The theory lives on the concept page → Cardinal utility.
The cases¶
| Case | What it shows | Source |
|---|---|---|
| Table 4 — three methods, three winners | his "mirror pathology" of IRV: the most popular candidate eliminated in round one | hillinger_t4_ev3.yaml |
| The same election, rescaled | what "cardinal" guarantees: totals move under u′ = 2u + 1, the winner does not |
hillinger_t4_affine.yaml |
| Table 3 — one approval result, two opposite Borda winners | his §10 answer to Saari–Van Newenhizen: a coarse score under-determines the ranking exactly as a ranking under-determines the score | hillinger_t3_arbitrariness.yaml |
Table 1 is already in the library — do not mint it again¶
Hillinger's Table 1 ("Most Disliked Candidate Wins Under PV") is 3 voters a > b > c, 2 b > c > a, 2 c > b > a: a wins the plurality count 3–2–2 while 4 of the 7 rank a last. That profile is identical to Felsenthal's Example 1, which this repo already runs and already has live on BetterVoting as BV2144 (mxfmhm) — same seven ballots, cast as Ana / Bo / Cal, where it demonstrates four plurality paradoxes at once (Condorcet winner ignored, Condorcet loser elected, absolute loser elected, and the spoiler/SCC effect).
Two authors, two decades apart, reached for the same seven ballots because it is the smallest profile that makes the point. Cite it as Hillinger Table 1 or Felsenthal Example 1 — but there is one election, and it already exists. The definitional question Hillinger raises alongside that table — what is a majority or minority candidate, given that almost nobody wins an arithmetic majority in a real field — is answered at Majority & minority candidates.
Table 4 — the mirror pathology¶
Hillinger's §12 point: plurality's famous defect is that an unpopular candidate may win; IRV's mirror defect is that the most popular candidate may be eliminated in the very first round. Thirty voters, three candidates:
| Voters | Order | EV-3 marks (Ana, Bruno, Chloe) |
|---|---|---|
| 9 | Ana > Bruno > Chloe | 2, 1, 0 |
| 10 | Bruno > Ana > Chloe | 1, 2, 0 |
| 11 | Chloe > Ana > Bruno | 1, 0, 2 |
Ana holds the fewest first choices (9) and is the one candidate nobody ranks last. She is the Condorcet winner — she beats Bruno 20–10 and Chloe 19–11 — and she wins the cardinal count with 39 to Bruno's 29 and Chloe's 22, reproducing Hillinger's table exactly.
What the engine adds to the paper. Hillinger reports only the STV failure. Run the same thirty ballots through everything and they split three ways:
| Method | Winner |
|---|---|
| Plurality | Chloe — most first choices (11), ranked last by 19 |
| RCV-IRV | Bruno — Ana eliminated first, her ballots transfer |
| STAR · Score · Ranked Robin | Ana — highest sum, and the Condorcet winner |
One electorate, one set of opinions, three different winners.
A note on the scale. The marks are Hillinger's EV-3 ballot
(−1, 0, +1)shifted onto this repo's 0–5 scale as(0, 1, 2). He states in §5 that the choice of origin does not affect the outcome; the rescaled twin checks that claim rather than taking his word for it.A typo, caught by running it. The paper gives the Ana–Bruno pairwise as
ab(20/11), which sums to 31 of 30 voters. It is 20–10 (theac(19/11)beside it is correct). Harmless to his argument — but it is what a runnable companion is for.
[Divergence from STAR]
STAR = Ana
Choose-One (Plurality) = Chloe (differs from STAR)
RCV-IRV = Bruno (differs from STAR)
Note: no ballots had tied scores, so RCV-IRV vs STAR here is a genuine
method difference, not a tie-breaking artifact.
Note: Ranked Robin (RCV-RR) agrees with STAR, so RCV-IRV is the lone
outlier — the classic center-squeeze signature.
Full round-by-round reports (generated for review):
RCV-IRV rounds: cases_tabulated/hillinger_t4_ev3_RCV-IRV_tabulated.txt
--- STAR Voting Method (single winner) ---
[STAR Voting]
Tabulating 30 ballots.
Count × Ana,Bruno,Chloe
11 × 1, 0, 2
10 × 1, 2, 0
9 × 2, 1, 0
[STAR Voting: Scoring Round]
The two highest-scoring candidates advance to the next round.
Ana -- 39 -- First place
Bruno -- 29 -- Second place
Chloe -- 22
Ana and Bruno advance.
[STAR Voting: Automatic Runoff Round]
The candidate preferred in the most head-to-head matchups wins.
Ana -- 20 -- First place
Bruno -- 10
Equal Support -- 0
Ana wins.
Runoff math:
30 ballots cast
− 0 Equal Support (no preference between the two finalists)
──
30 voters with a preference (majority = 16)
Ana 20 (67%) · Bruno 10 (33%)
[STAR Voting: Winner — STAR Voting Method (single winner)]
Ana
Related¶
- Cardinal utility — the concept page this folder supports: what makes a scale cardinal, Sen's measurability/comparability axes, the vNM trap, Harsanyi's theorem, and a claim-check of Hillinger
- Is Approval's outcome arbitrary? — the teaching page for §10 and Table 3: the Saari–Van Newenhizen criticism, why the Brams–Fishburn–Merrill defense fell short, and what the argument does to STAR
- Center squeeze — the same failure shape as a dedicated topic
- Condorcet efficiency — how often the head-to-head winner actually wins
- Range / Score voting — EV without a runoff, which is what Hillinger actually proposes
- Scores vs. ranks — the ballot-level distinction, without the theory