[EM] Saari's Basic Argument
fsimmons at pcc.edu
Thu Feb 20 17:08:59 PST 2003
Below I state that these symmetries preserve the Borda count. That's
because I use the numbers 1,0, and -1 for the three rank positions, so
that the symmetrical distributions all give a Borda count of zero to all
So when you add or subtract symmetrical sets of ballots, you don't
change this version of the Borda count.
On Thu, 20 Feb 2003, Forest Simmons wrote:
> On Tue, 18 Feb 2003, Steve Barney wrote:
> > Here is a simpler example to illustrate the difference that the order in which
> > cyclic and reversal terms are canceled does not matter when using the strictly
> > correct method - as opposed to the method used by Forest Simmons and Alex
> > Small, and in some of Saari's popular expositions where he is merely trying to
> > illustrate basic concepts to a more general audience.
> It's true that in my earlier examples I didn't take the decomposition all
> of the way to its logical conclusion. But if you look at my last few
> examples, you'll see that I finally hit on a simple method of adding and
> subtracting the two kinds of symmetries that reduce each ballot set to a
> canonical ballot set consisting of either one faction or two adjacent
> factions with positive multiplicity, and all other factions zero.
> [Two factions are adjacent if one of them follows the other in the cyclic
> order ABC->ACB->CAB->CBA->BCA->BAC->ABC.]
> It is easy to prove that this decomposition is unique, since it preserves
> the Borda count of each candidate, and the Borda counts determine the
> number in each faction when there are two or fewer adjacent factions. So
> the order of application of the symmetries doesn't matter.
> As my examples show, this can be done very simply without matrices.
> Presumably Saari uses matrices because he wants to develop tools that will
> generalize to more than three candidates.
> But worrying about the details of symmetry cancellations is to bark up the
> wrong tree.
> The fact is that in the ballot set 65*ABC+35*BCA the candidates A, B, and
> C have the same respective average ranks as they do in the simpler ballot
> set 30*ABC+35*BAC , so according to Borda they are equivalent ballot sets,
> and B should be the winner of the first election as decisively as in the
> second (according to Borda and Saari).
> This result may make sense in the context of dispassionate decision making
> such as in robotics when a robot is trying to decide what movement to make
> or whether a visual image represents the letter U or V.
> But in the context of public elections, this supposed equivalence is
> almost ludicrous.
> So the question is not, "Why is Borda such a great method for public
> The question is, "Why does the symmetry argument lead us down the wrong
> At least that is the question I was trying to answer (and did answer to
> my own satisfaction).
> In a nutshell the answer to this question is that the symmetries in the
> distribution of ballots are at odds (more often than not) with Saari's
> In other words, Saari's transformations do not preserve the natural axes
> of symmetry that may (and do) exist in the ballot distributions.
> In the above example, the first ballot set 65*ABC+35*BCA consists of two
> factions that determine an axis of symmetry. The second ballot set
> 35*BAC+30*ABC also consists of two factions along an axis of symmetry, but
> this axis is rotated thirty degrees relative to the original axis.
> So there is an essential change in the symmetry of the distribution that
> the Borda count doesn't detect, and Saari's symmetry transformations
> cannot preserve.
> [The center of gravity of the distribution is preserved, but the principal
> axes of rotation and the radii of gyration are changed.]
> For me this insight is sufficient to explain the fallacy of the symmetry
> Most non-mathematicians don't care one whit about what went wrong with the
> symmetry arguments; rather than watch the gory details of an autopsy, they
> prefer to move onward and upward.
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