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Steven Brams once contended it was very rare in practice for Approval and Condorcet winners to differ (when the latter existed). I disputed that, noting that computer simulations of random normal elections said different. Also Blake Cretney at condorcet.org had the idea they'd often differ and usually Approval in such cases was "wrong" and Condorcet "right." I now do some interesting analysis that tends to support Brams' view. Also, we note that D.Roderick Kiewiet in Polity 12,1 (1979) 170-181, on p180 claimed that Approval Voting was "far more likely to elect Condorcet Winners" (than plurality), a claim we and our computers fully agree with. For definitions of terminology, it may help to consult the glossary.
The simplest possible election example in which Condorcet and Approval disagree on the winner is:
|2||C > A > | > B|
|1||A > | > C > B|
Here A is the approval winner (unanimously approved by all 3) beating C (2 approvals) and B (0); C is the Condorcet winner (C>A by 2:1 ratio, C>B unanimously); C also is the plurality, Borda, and IRV winner; and we denote the approval threshold with "|".
However, knowing it was down to C versus A, the C>A>B approval-voters would strategically (and I wouldn't even call this "dishonest"; I'd call it "semi-honest" because it only shifts the "|" without altering any ">") disapprove A, in which case this election really would be
|2||C > | > A > B|
|1||A > | > C > B|
and C would win with every method, so actually this was not a realistic example of an Approval versus Condorcet "conflict."
So: what if we instead ask for an example of Condorcet versus Approval disagreement where all the approval votes are strategic about A versus C (i.e, always approve of exactly one of them).
AMAZING "NO-CONFLICT" THEOREM: There does not exist a 3-candidate tie-free election in which the approval voters are both honest about their candidate orderings and strategic about their threshold-placement, in which the Approval and Condorcet winners differ!
Proof: Maximize s subject to these inequalities (note I use 4-letter-long variable names):
CBTA+CTBA+ACTB+ATCB+CATB+CTAB ≥ s + BCTA+BTCA+ABTC+ATBC+BATC+BTAC,
CATB+CTAB+BCTA+BTCA+CBTA+CTBA ≥ s + ACTB+ATCB+BATC+BTAC+ABTC+ATBC,
ABTC+ACTB+BATC+CATB+ATBC+ATCB ≥ s + BCTA+BATC+CBTA+BTCA+BTAC+ABTC,
ABTC+ACTB+BATC+CATB+ATBC+ATCB ≥ s + CBTA+CATB+BCTA+CTBA+CTAB+ACTB,
ABTC ≥ 0, ACTB = 0, BCTA ≥ 0, BATC ≥ 0, CBTA ≥ 0, CATB = 0, ATBC ≥ 0, ATCB ≥ 0, BTCA = 0, BTAC = 0, CTBA ≥ 0, CTAB ≥ 0,
ABTC+ACTB+BCTA+BATC+CBTA+CATB+ATBC+ATCB+BTCA+BTAC+CTBA+CTAB = 1000 .
Remarks: For information about linear programs, standard form, and simplex method, see George B. Dantzig: Linear progamming and Extensions, Princeton University Press 1963. We can, by examining the optimum of the linear program, strengthen the theorem to state that the only possible way for Condorcet and Approval to disagree is if there is an exact tie for the approval winner and an exact tie for the Condorcet winner (which still isn't much of a "disagreement").
Warren D. Smith 15 August 2006
Actually, the above proof, while valid, is overkill. We can prove a better theorem with a simpler proof:
EVEN BETTER "NO-CONFLICT" THEOREM: Let N≥2. There does not exist an N-candidate tie-free election in which the Approval (A) and Condorcet (C) winners differ, provided that, if they were going to differ (A≠C) that all the approval voters would place their threshold strategically under the assumption the winner was going to be either A or C, i.e. would place it somewhere between them.
Proof: Assume for a contradiction that A≠C. Then the approval voters will strategically place their thresholds | between C and A. But that will cause C to be approved more times than A is approved (since C, being the Condorcet winner, is preferred over A by a majority). Hence the claim A was the Approval winner and A≠C, leads to a contradiction. Hence either A=C or there is no Condorcet winner C. Q.E.D.
In view of this fact, it seems that much of the so-called "conflict" between the Approval and Condorcet "philosophies," is illusory.
We now see that approval voting always will elect the Condorcet Winner C (if C exists) provided merely that the voters consider it likely that both C and whoever (A) the putative non-C approval winner might be, might be elected – and hence vote strategically about C versus A.
That sounds like a highly realistic assumption in practice.
So we see that in practice, the supposed "difference" between Approval and Condorcet actually will rarely exist – so for practical purposes, Approval is a Condorcet method!
Indeed – counterintuitively – it might actually be that Approval Voting is more likely to elect the Condorcet Winner in practice, than Condorcet Methods!
Why? Because in approval voting it is quite rare that strategically voting dishonestly, is wise. (And when it is wise, it is even rarer that people will actually realize it and do it.) In other words, with AV, people will tend to honestly order the candidates, and the only strategic decision they'll make is where to locate their approval | "threshold."
In contrast, with Condorcet methods with rank-order ballots, it seems a lot more common to see a way to be usefully strategic-dishonest in your vote. In other words, I suspect it will be comparatively common that voters will dishonestly misorder their preferences.
In that case, it might well be that Condorcet rank-order ballot systems will do something silly (like not electing whoever should have been the Condorcet Winner with honest votes) more commonly than AV will in practice fail to elect the Condorcet Winner.
In that case, for practical purposes AV will be a better Condorcet method than actual "official" Condorcet methods!
I really think this is actually quite likely, it is not just some unlikely theoretical speculation. And if so, it is quite a remarkable counterintuitive conclusion.
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