Prefer Accept Reject voting

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Prefer Accept Reject (PAR) voting works as follows:

  1. Voters can Prefer, Accept, or Reject each candidate. Default is "Reject" for voters who do not explicitly reject any candidates, and "Accept" otherwise.
  2. Candidates with at least 25% Prefer, and no more than 50% reject, are "viable". The most-preferred viable candidate (if any) is the leader.
  3. Each voter gives 1 point to each candidate they prefer; and, if they did not prefer the leader, 1 point to each viable candidate they accept. Most points wins.

Relationship to NOTA

If all the candidates in the first round got a majority of reject, then the voters have sent a message that none of the candidates are good, akin to a result of "none of the above" (NOTA). PAR still gives a winner, but it might be good to have a rule that such a winner could only serve one term, or perhaps a softer rule that if they run for the same office again, the information of what percent of voters rejected should be next to their name on the ballot.

Criteria compliance

PAR voting passes the majority criterion, the mutual majority 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.

There are a few criteria for which it does not pass as such, but where it passes related but weaker criteria. These include:

  • It fails the Condorcet criterion, but for any set of voters such that an honest majority Condorcet winner exists, there always exists a strong equilibrium set of strictly semi-honest ballots that elects that CW. (Note that this is not true for any strictly-ranked Condorcet system!)
  • It fails O(N) summability, but can get that summability with two-pass tallying (first determine who's disqualified, then retally).
  • 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 majority loser criterion, the Strategy-free criterion, and the later-no-harm and later-no-help criteria.

Favorite betrayal?

PAR voting fails the favorite betrayal criterion (FBC). For instance, consider the following "non-disqualifying center-squeeze" scenario:

  • 35: AX>B
  • 10: B>A
  • 10: B>AC
  • 5: B>C
  • 40: C>B

None are disqualified, so C wins with 40 points (against 35, 25, 35 for A, B, and X). However, if 6 of the first group of voters strategically betrayed their true favorite A, the situation would be as follows:

  • 29: AX>B
  • 6: X>B
  • 10: B>A
  • 10: B>AC
  • 5: B>C
  • 40: C>B

Now, A is disqualified with 51% rejection; so B (the CW) wins.

However, there are several ways to "rescue" FBC-like behavior for this system.

For one, we could add a "compromise" option to the ballot, as described in FBPPAR.

For another, we could restrict the domain to voting scenarios which meet the following restrictions:

  1. Each candidate either comes from one of no more than 3 "ideological categories", or is "nonviable".
  2. No "nonviable" candidate is preferred by more than 25%.
  3. Each voter rejects at least one of the 3 "ideological categories" (that is, rejects all candidates in that category) and does not reject at least one of them (rejects none of the candidates in that category).
  4. There are no honest Condorcet cycles.

If the above restrictions hold, then PAR voting would meet FBC. It is arguably likely that real-world voting scenarios will meet the above restrictions, except for a negligible fraction of "ideologically atypical" voters. For instance, in the first scenario above, the categories appear to be {XA}, {B}, and {C}, so the B>AC voters would probably actually vote either B>A or B>C.

And finally, note that in any scenario where it fails that for some small group, there is a rational strategy for some superset of that group which does not involve betrayal. For instance, in first scenario above, if 11 of the AX>B voters switch to >AXB, then A is disqualified without any betrayal.

An example

Tennessee's four cities are spread throughout the state

Imagine that Tennessee is having an election on the location of its capital. The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities, and that everyone wants to live as near the capital as possible.

The candidates for the capital are:

  • Memphis on Wikipedia, the state's largest city, with 42% of the voters, but located far from the other cities
  • Nashville on Wikipedia, with 26% of the voters, near the center of Tennessee
  • Knoxville on Wikipedia, with 17% of the voters
  • Chattanooga on Wikipedia, with 15% of the voters

The preferences of the voters would be divided like this:

42% of voters
(close to Memphis)
26% of voters
(close to Nashville)
15% of voters
(close to Chattanooga)
17% of voters
(close to Knoxville)
  1. Memphis
  2. Nashville
  3. Chattanooga
  4. Knoxville
  1. Nashville
  2. Chattanooga
  3. Knoxville
  4. Memphis
  1. Chattanooga
  2. Knoxville
  3. Nashville
  4. Memphis
  1. Knoxville
  2. Chattanooga
  3. Nashville
  4. Memphis

Assume voters in each city preferred their own city; rejected any city that is over 200 miles away or is the farthest city; and accepted the rest.

City P A R tally
Memphis 42 0 58 NA
Nashville 26 74 0 100
Chattanooga 15 43 42 NA
Knoxville 17 41 42 NA

Memphis is rejected by a majority, and is disqualified. Chattanooga and Knoxville both get less than 25% preference, so they are also disqualified. Nashville wins with a tally of 100%. This is a strong equilibrium; no rational strategy from any faction or combination thereof would change the winner. Knoxville and/or Chattanooga could each prevent the other from being disqualified, but Nashville would still win with a tally of at least 68 (the ballots of Nashville and Memphis).

(If Memphis voters rejected Nashville, then Chattanooga or Knoxville could win by conspiring to reject Nashville and accept Memphis. However, Nashville could stop this by rejecting them. Thus this strategy would not work without extreme foolishness from both Memphis and Nashville voters, and extreme amounts of strategy from the others.)

Discussion

Logic for 25%-preferred threshold (step 2)

The 25%-preferred threshold in step 2 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.