# [EM] smith/schwartz/landau

Kristofer Munsterhjelm km_elmet at t-online.de
Tue Mar 27 08:11:20 PDT 2018

On 03/26/2018 11:18 PM, Juho Laatu wrote:

> In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).

Isn't this analogous to majority vs utility? Suppose you have a ranked
election like this:

51: A>B>C
48: B>C>A
2: C>B>A

A is the majority winner, but it only takes two voters changing from
A>B>C to B>A>C to make B the majority winner, analogous to how {ABC} is
the Smith set but {d} is close to being the CW. A voting method that
passes the majority criterion will always elect A, but a more
consensus-focused method like Borda elects B.

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, and so there seem to be only two ways to get
out of the problem.

The first way is to say that we have a model of how the electorate
behaves, which lets us determine which rated election is most likely
given the ranked data. (E.g. in a very polarized electorate, it's more
likely to be the former than the latter.)

The second way is to say "we have no idea of which may be true, but we
want to pass the majority criterion, so that settles the matter for us".
In that case, the hypothetical "we" would choose a method that elects A.

If the Smith situation is analogous to majority, then it seems that the
most consistent choice with respect to majority (as opposed to utility)
is to elect from the Smith set. A method that doesn't elect from the
Smith set may be superior in a utilitarian sense, but it may also not
be. There's no way to know (short of taking the first approach, or by
proving that e.g. for a very wide range of utility models, some
non-Smith method has better utilitarian performance than some Smith method).

I suppose what I'm saying is that if we choose to have a method that
passes majority on the grounds on the grounds that the majority is
right, then the argument can be adapted pretty easily to Condorcet and
then to Smith. Any counter of the sort that electing from Smith can get
the wrong winner can be similarly "ported back" to the majority
situation by showing that the majority criterion can also get it wrong.

Of course, if there's a more indirect argument behind wanting majority,
then that can break the symmetry. For instance, one could argue that
passing majority is a sort of DSV; in strategic Range, everybody votes
Approval style and so the majority winner wins anyway, so why not ease
the burden of the voter? Or "a large majority has the backing to get
what it wants, better let it have what it wants on the ballot than risk
a riot". It's not obvious how such arguments would support Smith,
whereas it is pretty clear that they would support the majority criterion.

But then that argument has to be made, first; in the absence of such an
argument, Smith seems a reasonable extrapolation from majority (by way
of Condorcet).