accounts at museworld.com
Tue Mar 27 15:44:21 PDT 2018
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that *if* the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference. Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.
It’s true that these might be entirely sane assumptions given the numbers. 66:33 *probably* does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set; it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)
> On Mar 27, 2018, at 10:54 AM, Juho Laatu <juho.laatu at gmail.com> wrote:
>> If we knew the utilities, then we could determine whether this is a
>> 51: A: 100, B: 1, C: 0
>> 48: B: 100, C: 1, A: 0
>> 2: C: 100, B: 1, A: 0
>> election (A should win if we're counting utilities), or a
>> 51: A: 100, B: 99, C: 0
>> 48: B: 100, C: 99, A: 0
>> 2: C: 100, B: 99, A: 0
>> election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, […]
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