Difference between revisions of "Talk:Definite Majority Choice"
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== more criteria... ==
== more criteria... ==
Have we established whether DMC passes monotonicity or clone independence? There are probably more criteria that can be listed on this page. --[[User:
Have we established whether DMC passes monotonicity or clone independence? There are probably more criteria that can be listed on this page. --[[User:|]] 18:47, 22 Nov 2005 (PST)
Latest revision as of 19:47, 22 November 2005
Please let us avoid the term "majority" when there need not be any majority involved! Look at this:
1 A>>B>C 1 B>>C>A 1 C>>A>B 3 A=B=C
Here no "majority agrees" that any candidate should be eliminated! [Heitzig-j]
- So ties have to be discussed
- No, I didn't talk about ties but about majorities! In the above example, there are defeats but no majorities in the usual sense of more than half of the voters. [Heitzig-j]
- I think I sent a suggestion in private email, but here it is again.
- The initial page I put up was intended as a public elections proposal. So I wasn't thinking about ties.
- In DMC, we eliminate candidates that lose pairwise matches to higher-approved candidates. Call the set of remaining candidates P.
- If there is a tie, or if in a public election there is a near-tie (difference of, say, 0.01%), what about forming the superset P*, the union of all P's resulting from all possible reversed close races.
- Then choose the winner by Random Ballot.
If a faction sees a Condorcet paradox looming, can it gain an advantage by insincerely approving of the contender that it expects to defeat pair-wise?
Your polling data: Approval runs narrowly A>B>C, but pairings run A>C>B>A, so 'B' stands to win under DMC. You're the tactician for faction 'A'. Can your supporters insincerely increase approval of 'C' in order to eliminate 'B'?
If so, then it doesn't even matter if you over-do it: As long as you expect to win head to head, you may hand out approval points with impunity and perhaps steal a victory. Of course, once this leaks out, 'C' will boost 'B', and 'B' will boost you... and the faction controlling the most first place votes will bury its nemesis and prevail.
So, in practice, will DMC degenerate into basic Condorcet with whatever limitations that has? Jrfisher 19:12, 29 Aug 2005 (PDT)
Note: There's little risk in extending approval to inferiors even when one does NOT know if a paradox looming. Therefore, approval extension will probably become automatic voter behavior, rendering DMC's approval results misleading, and that will lead to voter confusion and dissatisfaction as apparently highly popular candidates are regularly declared to be losers. Perhaps this effect on all elections, rather than the effect during paradoxes, is what will render DMC completely unworkable. Jrfisher 10:46, 14 Sep 2005 (PDT)
First of all, your example is not complete. There are 15 possible ballots that could be submitted in a 3 candidate race.
A's approval comes from (1-5,9,12); B's approval comes from (2,6-10,14); C's approval comes from (4,7,11-15). A>C votes come from (1-5,9,10); etc. If you first specify some range of conditions, such as all approval ratings > 50%, or all < 50%, then it is possible to generate ballot combinations from the space of possible solutions. Could you give an example of a set of ballots that demonstrate your case?
Secondly, polling is imprecise, usually accurate only to within 3%. So "narrow" difference of approval makes strategizing risky, if there is a possibility of the strategy backfiring.
Third, consider how a 3 candidate cycle might arise. With lower barriers to entry, the field is likely to be more crowded. So the 3 candidate cycle might not be known in advance.
Fourth, even if there are only 3 candidates: if the weakest defeat is B>A, then A and B are more closely aligned and are each other's compromise candidates over C. You describe a situation in which A's camp attempts to wrest the victory from their allies. This is sometimes known as the prisoner's dilemma betrayal strategy. If A's camp tries that, B's camp is likely to withhold approval from A. If the approval ratings are as narrow as you say, A will no longer be the highest approved candidate, and either B or C might win. --Araucaria 10:44, 15 Sep 2005 (PDT)
Araucaria responds again:
Here are some other thoughts on Jrfisher's example. I'm assuming he is imagining a scenario similar to
37: A>>C 33: B>>A 30: C>>B
that has statistically significant approval differences between the candidates, but no faction is willing to compromise approval of any other.
In the case of a divided cyclic electorate, any strong Condorcet voting scheme---for example either DMC or Schulze(wv)---provides an incentive for a fourth compromise candidate to run, if only as a write-in candidate. In other words, there is the opportunity for candidate D, either as
37: A>>D>C 33: B>>D>A 30: C>>D>B
37: A>D>>C 33: B>D>>C 30: C>D>>B
So this example is somewhat artificial. More worrisome and possibly more common is whether a faction would deliberately induce a cycle in order to take advantage of a Condorcet completion scheme's elimination of either the weakest defeat in the cycle or the weakest candidate. In this case, DMC has a major advantage over Schulze(wv), because the strategic unpredictability of the approval cutoff makes such maneuvering riskier.
But even if such a situation arises, what could happen? A might have an incentive to inflate C's approval. B becomes lowest approved and is eliminated. If C's faction wants to prevent an A victory, they can elevate B's approval and A is eliminated, electing C. B's faction might object to both A and C's insincerity, but by avoiding insincere promotion of A, they effectively create a poison pill against A's tactic.
So the question in any campaign tactician's mind has to be, can I profit through a Mexican standoff? It is very risky. The best bet in that case is to avoid shooting and attempt to forge an ally. --Araucaria 13:19, 21 Sep 2005 (PDT)
Have we established whether DMC passes monotonicity or clone independence? There are probably more criteria that can be listed on this page. --James Green-Armytage 18:47, 22 Nov 2005 (PST)