Hillinger Table 3 — one approval result, two opposite Borda winners¶

Generated from hillinger_t3_arbitrariness.yaml — do not edit by hand. Regenerate: python STARVote_LH_tabulation_engine/tools_adam/scripts/build_yaml_pages.py.

Method: Approval Voting · 1 seat · Expected winner: Ada

Scenario¶

Claude Hillinger, "Voting and the Cardinal Aggregation of Judgments" (Munich Discussion Paper 2004-9), section 10, Table 3 — his answer to the Saari-van Newenhizen (1988) criticism of Approval Voting.

SVN's argument: a voter with the strict preference a > b > c cannot express it on an approval ballot. They must arbitrarily pick (1,0,0) or (1,1,0), and one can construct profiles where that arbitrary choice decides the election. So Approval - and, SVN generalize, cardinal voting at large - is indeterminate.

Hillinger's inversion: the argument only bites if the STRICT ORDERINGS are the true preferences. Assume instead that the approval marks are what the voters actually mean, and it is the rankings that become arbitrary - because a coarse score under-determines the ranking in exactly the same way.

These are the seven approval ballots of his Table 3. Ada 5, Ben 2, Cora 4; Ada wins. Now complete each ballot to a strict ranking. TWO completions are consistent with these very marks:

reading 1 3x Ada>Ben>Cora 2x Ben>Cora>Ada 2x Ada>Cora>Ben Borda: Ada 10, Ben 7, Cora 4 -> ADA wins reading 2 3x Ada>Cora>Ben 2x Ben>Cora>Ada 2x Cora>Ada>Ben Borda: Ada 8, Ben 2, Cora 11 -> CORA wins

Same ballots, same voters, opposite Borda winners. The ordinal formalism is as under-determined by the marks as the marks are by the formalism - which is Hillinger's point, and it is a genuine standoff rather than a refutation.

Teaching page: 04_Approval/01_Learn/approval_indeterminacy.md Concept page: 07_Concepts/topics/cardinal_utility.md

Ballots¶

Row 1 = candidate names; each later row is one voter's approvals (1 = approve, 0/blank = not approved).

Ada,Ben,Cora
1,0,0    # approves Ada only
1,0,0    # approves Ada only
1,0,0    # approves Ada only
0,1,1    # approves Ben and Cora
0,1,1    # approves Ben and Cora
1,0,1    # approves Ada and Cora
1,0,1    # approves Ada and Cora

What the engine says¶

Full report from the _tabulated mirror (regenerated on every run; every analysis forced on):

--- Approval Voting (single winner) ---
 Tabulating 7 ballots (any non-zero score = approval).

Ballots:
   columns = Ada, Ben, Cora      (1 = approve; 0 = not approved)
     3 × 1,0,0
     2 × 0,1,1
     2 × 1,0,1

   Ada  -- 5 (71%) -- Elected
   Cora -- 4 (57%)
   Ben  -- 2 (29%)

[Approval Distribution] (how many candidates each ballot approved)
   11 approvals across 7 ballots — average 1.6 of 3 (range 1–2).
     approved 1: 3 ballots
     approved 2: 4 ballots

[Co-Approval Matrix]
 Of the voters who approved the ROW candidate, the % who ALSO approved the COLUMN candidate.
         |  Ada   |  Cora  |  Ben   |
   ----------------------------------
   Ada   |   --   |  40%   |   0%   |
   Cora  |  50%   |   --   |  50%   |
   Ben   |   0%   |  100%  |   --   |

Winner — Approval Voting (single winner)
  Ada

Run it yourself:

python STARVote_LH_tabulation_engine/starvote_larry_hastings.py method_comparisons/hillinger_evaluative_voting/cases/hillinger_t3_arbitrariness.yaml

Hillinger Table 3 — one approval result, two opposite Borda winners¶

Scenario¶

Ballots¶

What the engine says¶

See also¶