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Prefer Accept Reject voting: Difference between revisions

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Prefer Accept Reject voting (view source)

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Category:Graded Bucklin systems → Category:Graded Bucklin methods
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# '''Tally 1 point for each "Prefer"''' for each candidate.
# Out of the candidates (if any) with no more than 50% "Reject", find the one with the most points. '''For every ballot which doesn't "Prefer" this frontrunner, add 1 point for each "Accept".'''
# If the frontrunner changed,still addhas 1the formost eachpoints, "Accept"they that'swin. notOtherwise, yet tallied (the oneswinner which "Prefer"is the originalcandidate frontrunner).with '''Mostfewest points"Reject" winsratings.'''
 
To express it in a single sentence: if the most-preferred non-majority-rejected candidate, X, has more non-reject votes than any other candidate has non-reject votes that aren't below X, then X wins; otherwise, the least-rejected candidate wins.
 
Note that the procedure above will always elect a candidate with no more than 50% "Reject", if any exist. This is because, if any exist, one of them will be the frontrunner, and they will thus score points equal to at least 50% of the voters.
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== Criteria compliance ==
 
PAR voting passes the [[majority criterion]], the [[mutual majority criterion]], the [[majority loser criterion]], [[Local independence of irrelevant alternatives]] (under the assumption of fixed "honest" ratings for each voter for each candidate), [[Independence of clone alternatives]], [[Monotonicity]], [[polytime]], [[resolvability]], and O(N²) [[summability]]. (It is also possible to run it in no more than 3 counting rounds, each of which is O(N) summable.)
 
There are a few criteria for which it does not pass as such, but where it passes related but weaker criteria. These include:
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* It fails [[Independence of irrelevant alternatives]], but passes [[Local independence of irrelevant alternatives]].
 
* It fails the [[Condorcet criterion]], but forif anythere setis ofa voters such that an honestvoted majority Condorcet winner existsX, therethen alwaysany existsfaction a(defined strongas equilibriumthe set of strictlyvoters semi-honestwho ballotsprefer thatX>Y,Z, electsor thatalternately CW.as (Notethe thatset thiswho isprefer not trueY>X>Z, for anysome strictlyY and Z) can ensure that X beats Z, using semi-rankedhonest Condorcetballots system!(X>>Z or YX>>Z, respectively).
 
* It fails the [[participation criterion]] but passes the [[semi-honest participation criterion]].
 
* It fails O(N)the [[summabilitylater-no-help criterion]], but canpasses getif thatthere summabilityis withat two-passleast tallyingone (firstcandidate determinerejected who's disqualified,by thenunder retally)50%.
 
* It may pass the majority Condorcet loser criterion (?).
 
* It fails the [[later-no-help criterion]], but passes if there is at least one candidate above the qualification thresholds (which is always true, for instance, if there are some three candidates who get 3 different ratings on every ballot).
 
It fails the [[consistency criterion]], the [[Condorcet loser criterion]], [[reversibility]], the [[Strategy-free criterion]], and the [[later-no-harm criterion|later-no-harm]] criterion.
 
=== Favorite betrayal? ===
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* 40: C>B
 
None are majority-rejected, and C is the frontrunner. Points are: A, 60; B, 55; C, 55; X, 35. A wins. However, if 611 of the last group of voters strategically betrayed their true favorite C, the situation would be as follows:
 
* 30: AX>B (That is, on 35 ballots, A and X are preferred, B is accepted, and C is rejected)
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* 15: B>A
* 10: B>AC
* 3429: C>B
* 611: B
 
Now, C is not viable with 51% rejection; so B is the leader. Since C is no longer the leader, B gets the 34 points from C voters, and wins. The strategy succeeded; the strategic voters are better off.
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=== Logic for 25%-preferred threshold (step 2) ===
 
The 25%-preferred threshold in step 21 is not purely arbitrary; it is exactly enough so that, in a 3-candidate election where all voters give all three grades, there will always be at least 1 candidate who passes the thresholds to not be disqualified. In other words: if a minority supports a rejected candidate, while a majority divides preferences between two candidates while accepting the other, then at least one of those two will not be disqualified. This does not hold for an election with 4 or more candidates, because the majority could split its preferences more than two ways; but even in those cases, it is usually reasonable to hope that the top 3 candidates combined will get enough preferences to ensure that at least one of them is above the 25% threshold.
 
[[Category:Graded Bucklin systemsmethods]]
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