# [EM] smith/schwartz/landau

Juho Laatu juho.laatu at gmail.com
Mon Apr 2 14:19:25 PDT 2018

```> On 02 Apr 2018, at 15:13, Kristofer Munsterhjelm <km_elmet at t-online.de> wrote:
>
> On 03/28/2018 04:36 PM, Juho Laatu wrote:
>>> On 28 Mar 2018, at 14:05, Kristofer Munsterhjelm <km_elmet at t-online.de> wrote:
>
>>> But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.
>> Note that the problems between matrix and ballot information mainly
>> emerge from the clone criterion. One could say that pure clone
>> independence / existence of clone candidates can not be measured from
>> the matrix only (without doing the "overkill"). The non Smith Set
>> arguments are more neutral (e.g. minmax style arguments) with respect to
>> using the matrix only vs also the ballots (matrix is enough). The
>> "overkill" is the problem that forces d not to be elected also when
>> there are no technical clones. (Smith Set criterion is close to the
>> clone independence criterion.)
>
> It's not just clones. Since the example's collapsed ballots are of the form
>
> m: A>d
> n: d>A
>
> with m>n, majority implies that A should be elected. In the uncollapsed example, that means that the set {A, B, C} is first on a majority of the ballots, so any method that passes mutual majority must elect from this set.
>
> That's perhaps a stronger example of how a generalization of majority forces one of {A, B, C} to be elected, since the point of mutual majority (as I see it, at least) is that a majority can get the candidate they want to be elected, elected, without having to coordinate beforehand to rank the candidates in the same order.
>
> It's hard to be opposed to such a property. I imagine it's easier to say "okay, the pairwise matrix doesn't supply enough information" and require that the method use more than just the pairwise matrix to decide the winner.

In the case that we have mutual majority the mutual majority candidates could nominate only one of them, and that candidate would win (if voter preferences would stay the same). But if they all run, then there may be also defeats among them, and those defeats could be considered worse than the defeats of some candidates outside that set. The alternative strategy is to rank A, B and C in the same order. If all ABC supporters implement this strategy (100%), they could break the ABC loop that other voters maybe generated.

Having the (theoretical) possibility to implement a working strategy should not be seen as a right to win without implementing that strategy. Implementing the first strategy could be difficult since voters may not understand if some of the competing candidates would withdraw. In the second strategy many voters might not follow the recommended strategy. In both cases it could be difficult to know if the strategic opportunity exists, and all three candidates to agree that all of them will benefit of this strategy. If d seems to have only minority in any case, then why let one of the other mutual majority candidates win for free (assuming that these three are from competing parties).

You already mentioned the problem that If one wants mutual majority without doing an overkill, then one must use the ballots instead of the matrix. In practice one would however probably use the matrix and a Smith Set compatible method (i.e. exactly measured mutual majority methods may not be very practical).

We know that some sincere winner or strategy related criteria will be violated in any case when preferences are circular. If one likes methods that may elect outside the Smith Set, then making a matrix based method compatible with mutual majority criterion probably violates the "outside the Smith Set target". If you want both and you can't, then you abandon one of them, or (maybe preferably) violate both of them but only lightly. (Just noting these facts.)

The mutual majority ballots could be as follows.

50: circular_mix_of_A_B_C > d > e > f
49: d > e > f > circular_mix_of_A_B_C

The idea is just to show that d could still be quite popular despite of the mutual majority. If one thinks that sometimes candidates outside the Smith Set could win, this might be one of those cases (despite of the mutual majority). Note that candidates d, e and f are not very far from having mutual majority, and among them it is clear that d is the best.

In this example the majority related key facts (of sincere opinions) are. 1) d loses marginally to A, B and C, 2) ABC is marginally better in mutual majority measurements than def, 3) A, B and C lose quite a lot, each to one of the others. Is the marginal mutual majority result more important than the strong (majority) defeats of A, B and C?

BR, Juho

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