# [EM] smith/schwartz/landau

Curt accounts at museworld.com
Tue Mar 27 22:43:06 PDT 2018

```Hi Juho, you question whether it is a “natural extension of the majority rule to require that the winner should always be picked from the Smith Set in majority oriented elections”, and it appears your argument depends on the concept of a “strong” cycle.

My view is that we cannot assume a 66:33 Smith Set is a “strong” cycle. And that the *only* way to see 66:33 as a “strong” cycle is to imbue it with utility or strength-of-preference data.

Because rankings are ordinal, whether the margin is 66:33, 98:1, or 50:49, we cannot infer anything about “strength of victory” in those margins. All we know is that the one with more votes is majority preferred over the other. We have zero conception of “how much” they are preferred by. Ordinal rankings cannot be used to simulate utility or strength of preference. 98:1 could be a whisker, and 50:49 could be a gulf.

Since ABC is preferred to d by 50:49, ABC being the winner makes sense to me because it is the winner by definition. It is axiomatic (when we are talking about Condorcet-style elections).

I respect the view that utility would be helpful data in an example such as the 66:33 example, but I also see it as intrinsically related to the view that utility data would be helpful in a two-candidate 50:49 election.

To illustrate how we cannot rely on 66:33 to indicate any information supporting d as a winner, here are two examples of the 66:33 scenario with utility data added in:

Here, any of A, B, or C are clearly preferred to d, and ABC are clearly tied in utility despite the 66:33 margins:

17: A:100, B:99, d:2, C:1
16: A:100, d:3, B:2, C:1
17: B:100, C:99, d:2, A:1
16: B:100, d:3, C:2, A:1
17: C:100, A:99, d:2, B:1
16: C:100, d:3, A:2, B:1

A: 1700 + 1600 + 17 + 16 + 1683 + 32 = 5048
B: 1684 + 32 + 1700 + 1600 + 17 + 16 = 5048
C: 17 + 16 + 1683 + 32 + 1700 + 1600 = 5048
d: 34 + 48 + 34 + 48 + 34 + 48 = 246

Here, d is clearly preferred to A, B, and C, but ABC are still tied:

17: A:4, B:3, d:2, C:1
16: A:100, d:99  B:2, C:1
17: B:4, C:3, d:2, A:1
16: B :100, d:99,  C:2,  A:1
17: C:4, A:3, d:2, B:1
16: C:100,  d:99,  A:2,  B:1

A: 1600 + 16 + 32 + 68 + 17 + 51 = 1784
B: 1600 + 16 + 32 + 51 + 68 + 17 = 1784
C: 1600 + 16 + 32 + 17 + 51 + 68 = 1784
d: 4752 + 34 + 34 + 34 = 4854

Curt

> On Mar 27, 2018, at 5:15 PM, Juho Laatu <juho.laatu at gmail.com> wrote:
>
>> On 28 Mar 2018, at 01:44, Curt <accounts at museworld.com <mailto:accounts at museworld.com>> wrote:
>>
>> 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.
>
> I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).
>
>> 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.
>
> The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.
>
>>
>> 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.
>
> I tried to avoid linking d to the utilities.
>
>>
>> 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;
>
> I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).
>
>> 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.
>
> There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.
>
>>
>> (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.)
>
> Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
>
> BR, Juho
>
>>
>> Curt
>>
>>
>>> On Mar 27, 2018, at 10:54 AM, Juho Laatu <juho.laatu at gmail.com <mailto: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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>
> ----
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