Now that Chris has brought up this clone problem, maybe I'll hold off.
Some thoughts on other possible ways of de-cloning WMA.
1. Proceed as in my de-cloning of Copeland. (Use an "irreducible beat clone set hierarchy structure" to modify the weights used in WMA.)
2. Does the problem persist in the version of WMA that uses fractional approval at the level where half of the weight is used up?
3. Break down and use Cardinal Ratings style ballots along with weighted mean instead of weighted median.
With regard to (1): A four candidate election always gives rise to what I recently described under the moniker of "a proper beat clone set."
If the election has a CW then the other three candidates form a proper beat clone set, because they are all beaten by evey (i.e. the only) candidate outside the set.
A similar analysis takes care of the case of a Condorcet Loser.
If there is neither a Condorcet Loser nor a CW (and assuming no pairwise ties) then there will be two candidates C1 and C2 such that both of these beat a third candidate B, and both are beaten by the other candidate A. [To see this, try putting arrows on the edges of a tetrahedron in such a way that no vertex has an in-degree or out-degree of three. All such arrangements of arrows are isomorphic.]
The set {C1,C2} is a proper beat clone set, and this should be taken into account when assigning the weights for WMA, or better yet, just eliminate the member C that is beaten by the other C, i.e. keep the one that is in the Banks set.
[De-cloned Copeland would give a tie to A, B, and the remaining C.]
To summarize in slightly different terms, we can say that a four candidate set can never have more than three uncovered candidates.
So "top four" is just as easy to handle as "top three" and might have even less incentive for "burying" than "top three." [If I am not mistaken, the main advantage of "top three" over "top two" is that it has less incentive for burying.]
I wonder how common it is that a Ranked Pairs winner (for example) will not be the winner in all of its four candidate supersets. If this is extremely uncommon, then giving the win to the candidate that wins in all of its four candidate supersets (if there is such a candidate), else random ballot among the Banks set candidates, would be a nice method.
I'm not sure where these ideas will lead; I haven't sorted them all out, but here's some more related ideas.
The idea of WMA is to automate near optimal approval strategy based on a set of ordinal ballots in order to compensate for Approval's Achilles' heel (bad information can lead one to make unfortunate approval cutoffs).
From what I understand, the only positive probability winners underperfect information with sophisticated player strategies would be the members of the Banks set. So keeping in mind the purpose of WMA, why not restrict to the Banks set in the first place.
Here's a five candidate example that is pleasant to contemplate:
Start with the transitive closure of A beats B beats C beats D, before adding in E such that E beats A and C but is beaten by B and D.
There are no proper beat clone sets in this example, so there is a beat path from every candidate to every other candidate. But C and D are both covered by B, so they cannot be sophisticated strategy winners, and hence should be eliminated.
WMA could then be applied to the cycle A beats B beats C beats A, to give near optimal approval cutoffs, at least much better than the voters could determine from typical quality horse race information.
And remember, if the voters have a strong approval or disapproval feeling that they want to have respected by WMA, they can always use the equal ranking options at the top and bottom of their ordinal ballots.
[To respect this feature, we modify Chris' rule of completely throwing out the unworthy candidates; we'll keep them on the ballots that rank them at the top, by ranking them equal to the top worthy candidate. Also symmetric completion might has to be modified to completely respect these strong preferences; yes, use the weights from the symmetric completion, but apply those weights in the ballots in their form immediately before the symmetric completion.]
The other line of thought that brought me to this topic was this: random ballot to choose among Banks set members might be a reasonable improvement on plain old random ballot. The weights in plain WMA are proportional to the probabilities of winning under random ballot. So why not make the weights proportional to winning under the improved random ballot method, i.e. under random ballot Banks?
I believe that random ballot Banks is clone free. If so, it would make sense for the fractional weight version of Banks//WMA to be clone free as well.
I hope that these musings stimulate some more thought along the lines that Chris and Jobst have been advocating.
Forest
Chris Benham wrote ...
From: Chris Benham <[EMAIL PROTECTED]>
James G-A (and interested others),
As I mentioned in my last message (in the "Condorcet completed by IRV" thread), in September this year I described and recommended Schwartz // SC-WMA. (The letters stand for "Symmetrically Completed-Weighted Median Approval").
Voters rank the candidates, truncation ok. Non-last equal preferences also ok (but a version that doesn't allow them is also good and might be a more practical proposition). Eliminate the non-members of the Schwartz set (and henceforth continue as though they had never stood). Symmetrically complete the ballots. Now apply the "Weighted Median Approval" method to pick the winner, thus: Each (remaining) candidate is assigned a "weight" which is equal to the number of first-preferences they get. The sum of the "weights" is equal to the total number of non-empty ballots. Each ballot approves the candidate they rank in first-place. If the weight of candidates so far approved by a ballot sums to less than half the total weight of all the candidates, then that ballot also approves the candidate they rank second. And so on until each ballot has approved at least half the candidates "by weight". The candidate with the highest total (thus derived) approval score wins.
If there are three candidates, all in the Schwartz set, then each ballot approves the two highest-ranked candidates and those ballots that only rank one candidate approve that candidate and half-approve the other two.
At the time I wrongly believed that this method is Clone Independent, but since then Douglas Woodall has showed that it fails Clone-Winner. This can only arise when the Schwartz set contains more than three members (maybe at least two more). Here is his demonstration.
Consider this example:
42: A B|C 28: B C|A 30: C A|B
Here the weights are A 42, B 28, C 30, total 100. The sum of any two candidates' weights is greater than 50, and so the cutoffs are as shown, giving A 72, B 70, C 58, and A wins. Now suppose we clone A:
14: A1 A2 A3 B|C 14: A2 A3 A1 B|C 14: A3 A1 A2 B|C 28: B C|A1 A2 A3 10: C A1 A2|A3 B 10: C A2 A3|A1 B 10: C A3 A1|A2 B
Here the weights are A1 14, A2 14, A3 14, B 28, C 30, total 100. The cutoffs are in the analogous positions to those in the previous election, except that since 30+14+14 = 58 > 50, the last 30 voters approve only 3 candidates each instead of the 4 that one might expect. Now the approval votes are A1 62, A2 62, A3 62, B 70, C 58, and B wins.
So WMA fails clone-winner.
CB: Clone Independence has been one of my "essential" criterion compliances, but in general, as a practical matter, I don't consider problems that can only arise when there are more than three (or four?) candidates in the Schwartz set to be very serious.
But I do have an idea for a bizarre "patch" to fix the clone problem!
It seems to me that lots of methods have clone (and/or sometimes other) problems that only arise when there are many or more than three candidates. It occurred to me that, as an interesting alternative to pairwise comparisons, triowise comparisons could be used as a clone-proof way to "trim" the field to three (or two) candidates.
Use some rule to determine the loser in each triowise comparison, and a way of scoring each loser (so that it can be compared with other losers). Repeatedly eliminate the worst loser in any of the triowise comparisons among remaining candidates, until three (or two) candidates remain. Three possible simple rules spring to mind. Based on the symmetrically completed ballots (I prefer, but alternatively not): (1) the number of top preferences (2) the number of top plus middle preferences (quasi-Bucklin) (3) the number of top minus bottom preferences (Borda)
A couple of pure and simple examples of complete methods that employ one of these "triowise trimmers". (1) "SC-TT(tp)//IRV" Based on the symmetrically completed ballots, repeatedly eliminate the candidate with the fewest top preferences in any of the triowise comparisons among the remaining candidates until two remain. Elect the winner of the pairwise comparison between these two.
This reduces to IRV when there are three candidates, but it fixes (at least the worst of) the horror IRV examples with many candidates (like Adam Tarr's "Lucky Right" example). A candidate could have no first preferences at all, and yet not be last in any triowise comparison, and so able to win.
(2) "SC-TT(Borda)//Borda-Elimination" Based on the symmetrically completed ballots,repeatedly eliminate the candidate with the lowest top-minus-bottom preferences score in any of the triowise comparisons among remaining candidates until two remain. Elect the winner of the pairwise comparison between these two.
This fixes the clone problem with Borda-Elimination.
My patch for Schwartz//SC-WMA ? Eliminate non-members of the Schwartz set. Symetrically complete the ballots, and then repeatedly eliminate the candidate with the lowest top-plus-middle preferences score in any of the triowise comparisons among remaining candidates, until three remain. Ignoring eliminated candidates, elect the WMA/Bucklin winner among these three.
(WMA and Bucklin are equivalent when there are three candidates). This does look a bit makeshift and inelegant. The method meets (at least 3-candidate) Minimal Defense. If the field is trimmed to two candidates like in the other two methods, then it doesn't.
Chris Benham
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