[EM] Modified Overall Preferences
Juho Laatu
juho.laatu at gmail.com
Wed Jun 26 00:55:39 PDT 2019
I studied the Modified Overall Preferences approach a bit more. Here's one new simple pairwise preference modification function. I'll explain the MOP approach again below, and describe the new simple modification function too.
The modification process starts from (1) some initial set of preferences. I'll use margins again, but also e.g. winning votes would be possible, although maybe not as natural as margins since the modification process already does quite similar things to the results that also the winning votes approach is supposed to do.
The next step (2) is to modify the strengths of the initial pairwise preferences. The simple modification function is (1-mod)*k*max + initial, where "initial" is the initial pairwise comparison result, "max" is the largest possible defeat size (= number of voters in margins), "k" refers to the easiness of influencing the preferences (e.g. k=4 means that if 1/4 of the voters (25%) think that A and B are "clones", then defeats between them are always weaker than any defeat between two candidates that nobody claims to be "clones"), "mod" refers to the proportion of voters that think that the compared candidates should be considered "clones" (protected from strong defeats to each others), and that the pairwise defeats should be modified / weakened (0 = nobody says so, 1 = all voters say so).
The third and final phase (3) is to find the winner based on the modified pairwise preferences. In this mail I'll use Minmax. Other approaches work too.
I'll describe the philosophy of the modification function briefly. Vote A>B>>C will be read as a normal ranked preference, but in addition to that the voter wants to say that A and B should be seen as "clones" (and defeats between them should be weakened). The presence of ">>" is interpreted so that the ">" preferences will be read as indicating "clones" / protection. In this modification function these opinions are symmetric (i.e. also (possible) B>A defeat should be weakened). If k=4 and we have 100 voters, that means that the modified preference strength values of defeats (that were initially margins in range 1..100) may range from 1 to 400. If mod=0, the strength range of defeats is from 301 to 400. If mod=1/4=1/k, the range is from 201 to 300. If mod=1, the range is from 1 to 100. This modification function can be said to use a "spread spectrum" technique since it expands the range of pairwise preference strengths (k times wider than the range of the initial strengths) to create space also for the weakened preferences.
Note that only the strengths of the defeats are modified. The (cyclic and transitive) preferences will stay just as they are in the initial preferences.
Rankings and protection ("clone indications") are in principle two separate opinions that could be given separately, but in practice protection is probably derived from enhanced rankings that may contain e.g. cutoffs, numeric gaps, positional gaps or strong and weak preference relations.
Next, few examples (with numeric results, in case you want to check the operation of the method). The first one describes a burying strategy example.
45: A > B >> C
15: B > A >> C
40: C > B > A
A and B clearly form a uniform group that has 60% support. B would win because it is a Condorcet winner, but then all the 45 A supporters decide to bury B and vote A>C>B. That would make A the winner. If k=4 and 31 C supporters (out of the 40) decide to protect B by voting C>B>>A, then B wins despite of the deceptive A supporters (A:-350, B:-346, C:-420). The method can thus defend against this kind of burying strategy in a quite natural way. Also the 15 B supporters could stop supporting A (as a "clone" of B) and vote B>A>C instead, in which case already 16 C supporters (voting C>B>>A) would make A supporters' strategy void (A:-410, B:-406, C:-420).
40: A (no "clones")
35: B>>C>A (A and C indicated as "clones")
25: C (no "clones")
This is the first challenge of Forest Simmons. A wins if k=4, as it should (A:-280, B:-405, C:-410). A wins also with k=1, k=2 and k=3.
40: A
35: B>C>>A
25: C
This is the second challenge. C wins if k=4, as it should (A:-420, B:-405, C:-270). C wins also with k=1, k=2 and k=3.
40: A
35: B>>C>A
25: C>>B>A
This is the third challenge. B wins if k=4, as it should (A:-320, B:320, C:-410). C wins also with k=0 (no modification), k=1, k=2 and k=3. Any number of C supporters could change their vote to C>B>A, and/or B supporters to B>C>A, and B will still win. That is because B is a Condorcet winner, and the modification function will keep initial defeats (and ties) as defeats (and ties), just with different strengths.
If you have examples where this kind of pairwise overall defeat modifying / weakening approach would not work well, please tell. I'm eager to evaluate MOP methods against such challenges.
> On 18 Jun 2019, at 09:01, Juho Laatu <juho.laatu at gmail.com> wrote:
>
> Chris Benham and I discussed (in private mail) on how well the example method below complies with the challenge that Forest Simmons proposed.
>
>> Forest Simmons fsimmons at pcc.edu
>> Thu May 30
>>
>>> In the example profiles below 100 = P+Q+R, and?? 50>P>Q>R>0.??
>>>
>>> I am interested in simple methods that always ...
>>>
>>> (1) elect candidate A given the following profile:
>>> P: A
>>> Q: B>>C
>>> R: C,
>>>
>>> and
>>> (2) elect candidate C given
>>> P: A
>>> Q: B>C>>
>>> R: C,
>>>
>>> and
>>> (3) elect candidate B given
>>> P: A
>>> Q: B>>C?? (or B>C)
>>> R: C>>B. (or C>B)
>
>
> I copy my answers to those questions below.
>
> Note that I'm struggling a bit with presentation of votes since Forest Simmons' challenge seems to talk about strengthening some preferences, while my mail talked about weakening some preferences (= making the defeats friendlier). When interpreting Forest's presentation of votes I assumed weakened preferences to be present only in places where the stronger (">>") preferences make the presence of weaker preferences obvious. I thus assumed that e.g. vote B>C (= B>C>A) in Case (3) to not contain any weakened preferences. Same e.g. with vote A in all of the cases. Only votes like B>>C (= B>>C>A) were assumed to contain weakened preferences (between C and A).
>
>> I tried to see how it relates to Forest's requests.
>>
>> Case (2)
>> Since all candidates lose to one of the others, and the only defeat to be weakened is the B>C defeat, and there are always more than 25% of the voters demanding that (Q>25%), then the "4*" version always lowers the strength of that defeat to 0. Therefore C always wins with that method.
>>
>> Case (1)
>> The example method didn't have any special "dislike" cutoff (only a "near clones this far" cutoff). I.e. only votes like A>B>C>>D>>E were allowed. The easiest way to introduce richer use of preference strengths would be to allow any preference (in the ranked vote) to be either ">" or ">>". In this case that would lead to votes Q: B>>C>A. That would weaken A's defeat to C. Again, the number of Q voters is higher than 25%, and with the "4*" factor the strength of A's defeat to C would become 0, and A would win.
>>
>> Case (3)
>> Also here I must assume that votes Q: B>>C>A and R: C>>B>A are allowed. Since (Q+R)>25%, strength of A's defeat would be 0. Candidate B has however no defeats nor ties, so B will win in any case. Having votes of form B>C>A or C>B>A would make A's defeat more severe (at least once we get below the 25% "clone treatment recommendation" limit). Candidate B (Condorcet winner) wins thus also in the original example method.
>>
>> In summary, the method seems to meet Forest's requirements if we allow also weakening of the lower preferences.
>
> I note that the method doesn't even have a name yet. Maybe MMM-MOP will do. It refers to the base method (Minmax(margins)) and the modification to it (Modified Overall Preferences). That leaves some space to using different parameters and approaches in modifying the pairwise preferences in the matrix.
>
> P.S. A bit more on weakened preferences / indicated clones (vs strengthened preferences). I'll use square brackets to indicate clones/weakening here. One can vote [A>B]>C (corresponds to A>B>>C), which means that B's possible defeat to A should be just a friendly one. One could vote also [A=B]>C if ties are allowed. This means that also ties can be weak. Or maybe one should rather say that also tied candidates can be considered clones/friendly (maybe even as default). One option would be to allow also truncated candidates (equal last) be indicated as being friendly. As noted in the beginning of my original mail, the friendliness indicators / weak preferences can be seen as a separate opinion that doesn't have much to do with the original ranking of the candidates (those rankings are as strong as ever in determining the (possibly looped) order, even though the final strength of preferences will be weakened, once the (possibly looped) order has been determined). In typical Condorcet methods weakening of overall pairwise preferences has no meaning except when there are loops, and strengths of different pairwise defeats need to be compared. In the example method of the original mail I stuck with one approval like cutoff only, but in order to respond to the challenge of Forest Simmons we need a bit richer preference structure. If we would go for three preference strengths (>, >>, >>>), maybe that would lead to a "clones among clones" approach in the weakened votes terminology.
>
> P.P.S. In the original mail, ignore words "which means".
>
> Juho
>
>
>> On 16 Jun 2019, at 16:51, Juho Laatu <juho.laatu at gmail.com> wrote:
>>
>> The point of this mail is to promote the idea of separating the strength of different pairwise preferences from the ordinary rankings (of ranked or Condorcet methods). Instead of giving voters the ability to adjust the strength of their own vote or pairwise preferences (the more common approach), they can influence the strength of the final preferences of the whole electorate. In this approach the pairwise preferences of the electorate will not be changed from A>B to B>A. Only the strength of that preference (A>B or B>A) will be modified.
>>
>> You could make many complex tricks with this kind of tools, but I'll concentrate here on one example method that is very simple to the voters. It will consist of adding just one approval like threshold to a basic Condorcet method. I'll use Minmax(margins) as the base method to be modified. Instead of voting A >> B >> C >> D >> E (a normal ranked vote) voters can vote also A > B > C >> D >> E (i.e. one or more of the first preferences may be weakened/moderated). The philosophy of the latter vote is that A, B and C are promoted as "favoured candidates" or "near clones" or "protected candidates" (with the full strength of one vote).
>>
>> In the counting process, first count the normal pairwise preferences (i.e. the matrix). After that, some of the pairwise preferences are made weaker. And then the final results are counted, (almost) as in Minmax(margins).
>>
>> Our first approach is to count the number of votes that had preference A>B (moderated preference) (= Mab). Then the A over B preference count (of the whole electorate) will be weakened by multiplying it with 1-Mab/N (where N = number of votes). Moderated preferences are transitive in the sense that A > B > C increases also the Mac count. The idea is that if voters consider A and B to be "near clones", their defeats to each other should be seen and treated as "friendly defeats". Their strength can thus be weakened, although a large number of voters may have preferred one over another (in a friendly way).
>>
>> (I note that some alternatives to the presented example method could be to make it symmetrical by using factor 1-(Mab+Mba)/N, or one could use the number of votes that preferred either A to B or B to A instead of N.)
>>
>> I'll add one more trick to the example method. The "friendliness factor" can be stronger than presented above. Let's say that 50% of the voters (maybe one of two parties) think that A and B should be treated as "near clones". Half of those voters have voted A > B, and half B > A. This means that 25% of the voters have given (moderated) preference A > B. Maybe already this 25% is enough to convince us that A and B indeed are to be treated as full clones. If so, we can use factor max(1-4*Mab/N,0) instead of 1-Mab/N. Here reaching strength 0 means that those candidates will be declared as "clones" and treated as such.
>>
>> The counting process may have some problems with ties if multiple preferences will be equal to 0. I will not discuss this question much more in this mail. Let's just say that also those preferences (although they all seem to be equal to 0) can be seen to have different strength, e.g. based on the initial margins. And that if there are two parties with 50 votes, and there are some "clones" in one of the parties, they could be seen as one alternative when flipping the coin. (Winner among the clones to be decided separately if the clone party wins.)
>>
>> Here's one example set of votes.
>>
>> 45 A>B>>C --> A>>C>>B
>> 15 B>A>>C
>> 40 C>B>>A
>>
>> B is the sincere Condorcet winner. Supporters of A will however strategically bury B under C. In Minmax(margins) the worst defeats are A:-10, B:-70, C:-20. The strategic voters will get what they wanted. There were however 40 voters that said Mcb. This means that in the example method the strength of B's defeat to C will be 0 (using the "4*" in the factor). There are also 15 voters that said Mba. This helps A a bit (-10 --> -4), but not too much. If all C supporters (or 17 or more of them) had voted C>>B>>A, A would win. C supporters may thus vote C>B>>A sincerely or as a (still quite sincere) defensive tactic (after hearing about A supporters' plans).
>>
>> Another question where this ability to moderate the defeats of favourite near clones is interesting is whether to elect from the Smith set or outside of it.
>>
>> 17 A>>B>>C>>D
>> 17 B>>C>>A>>D
>> 17 C>>A>>B>>D
>> 16 D>>A>>B>>C
>> 16 D>>B>>C>>A
>> 16 D>>C>>A>>B
>>
>> A, B and C are not "clones" in the described sense (which means no weakening of their mutual defeats since not a single voter proposed that). D wins.
>>
>> 17 A>B>C>>D
>> 17 B>C>A>>D
>> 17 C>A>B>>D
>> 16 D>>A>>B>>C
>> 16 D>>B>>C>>A
>> 16 D>>C>>A>>B
>>
>> A, B and C are "clones" (weight of mutual defeats = 0, with the "4*" moderation). One of them wins.
>>
>>
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