Democratic symmetry (fwd)

[Blake Cretney]

On Fri, 10 Mar 2000, Rob Lanphier wrote:
> ---------- Forwarded message ----------
> Date: Fri, 10 Mar 2000 12:45:04 -0600 (CST)
> From: d_saari at borda2.math.nwu.edu
> To: Rob Lanphier <robla at eskimo.com>
> Cc: Donald Saari <dsaari at nwu.edu>
> Subject: Re: Democratic symmetry (fwd)
 
> REVERSAL SYMMETRY
> 
> If everyone thought Alice was Barb, and Barb was Alice,
> the correct election outcome can be obtained by interchanging
> names.  If everyone marked their ballots in the reversed order --
> the last place candidate was marked first, etc. (this actually
> happened in our department because of poor wording on the ballot),
> it seems that the correct outcome can be obtained by reversing
> the election outcome.  Not necessarily.
> 
> 3 ABC, 4 ACB, 4 BCA, 3 CBA leads to a plurality ranking of
> A B C with a 7 : 4 : 3 tally.  Now, investigate the reversed
> ranking (i.e., now give points to the bottom ranked candidate).
> The tally is precisely the same.  A B C with 7:4:3.  
> 
> The reason for this behavior is that for each ABC ranking, there is
> a reversed CBA ranking; for each ACB ranking, there is a reversed BCA
> ranking.  One might want this data "reflection symmetry" to cancel.
> If you accept that, fine.  If not, fine.  The main point is that *all*
> differences between positional procedures (i.e., where weights are
> assigned to a voter's first, second, and third ranked candidate) are
> caused by data symmetry of this type.  
--snip--
> To see some implications of this comment, since the pairwise vote forces
> a cancellation of reversal symmetry, it is the reversal symmetry terms which
> causes differences between, say, the plurality and pairwise outcome.

This is interesting because we have always taken reversal symmetry to mean that if candidate X wins, and all ballots are reversed, candidate X should not still win.  And, if the method gives a complete ordering of the candidates, a reversal of all ballots should give the reverse ordering.

Saari proposes a different criterion, that if candidate X wins, and both a ballot, and its reverse are added, that X should still win.

> Consequences:

Let me first clarify something.  A positional method is a method where each ballot gives each candidate a certain number of points based on the rank.

In Borda, the number of points is equal to the number of candidates ranked bellow.  In plurality, the first ranking gets 1, the others get 0.  You can come up with lots of different point systems, although your choices may seem arbitrary.

The methods we advocate are generally not positional methods.  Therefore many of Saari's comments do not apply to them.  For example, the following:

> 3.  If a positional procedure does not lead to a cancellation of the
> reversal symmetry, then it need not respect the pairwise winners (e.g.
> it could elect a person who would lose all pairwise elections -- the
> Condorcet loser).   

This does not apply to any Condorcet-type method, as these are not positional methods.

--snip--

> ROTATIONAL SYMMETRY

> The surprising fact is that all problems with a lack of transitive outcomes,
> all differences between methods using binary information -- the Copeland
> method, Kemeny method, Condorcet winners and losers, agendas, Dodgson's
> method, and on and on can be completely understood by how each procedure
> handles data with rotational symmetry.  

40 A B C
60 C A B

Here, A wins in Borda.  C wins in most other methods.  There are no votes which form a rotational symmetry.  I think Saari probably would deal with this problem something like this.  By adding ballots and their reverse (which we might expect to cancel) we get this.

40 A B C
60 C A B
40 C B A
40 A B C

And, by removing the newly created rotational symmetries we get

20 C A B
40 A B C

Now, A is the clear winner.  I'm not advocating this approach, just trying to clarify what Saari means when he says that all differences are caused by rotational symmetry.

> Let me strongly recommend that
> you experiment with this and your favorite method. It is arguable that this data
> symmetry should cancel -- the Borda Count does.  

Presumably, it is arguable, as Saari amply demonstrates.  The more important question is, is it true?

I have seen so many sensible-sounding criteria, often in conflict with each other, that I have become cautious about accepting them.  For instance, Arrow's IIA criterion sounds plausible, but it isn't even consistent with democracy.

One way I try to puzzle through criteria is to return to a probabilistic view of the election.  I don't assume that majorities are always right, as that leads to absurdity.  I just try to analyze people's votes as evidence for which candidate is the best choice.

Let's take an example where rotational symmetry comes into play:

2 A B C  \
2 B C A   |- rotational symmetry
2 C A B  /
5 C B A
4 B C A

C wins in Borda, because the rotational symmetry is removed.  B is the Condorcet winner, and also the winner in IRV/AV.

So, the important question is, do the rotated votes provide additional evidence that we might expect would change our conclusion on the most probable best candidate, or does the information they provide simply cancel out.

Here's how I look at it.  The rotated votes give evidence in favour of 3 conflicting propositions.

A is better than B (Since 2 people agreed and 1 disagreed)
B is better than C (Since 2 people agreed and 1 disagreed)
C is better than A (Since 2 people agreed and 1 disagreed)

Even though each of these propositions may be more probable when considered individually, together, it is difficult to say.

Here's a related, more concrete, example.  Consider the police do a blood test to determine if a hair at the scene belongs to a particular suspect.  The test comes back that it is 90% sure that it is.  The police test another suspect.  To their annoyance, it comes back as 90% sure that the hair belongs to this suspect as well.

Each piece of evidence when considered individually is fairly strong, although probably not enough to get a conviction.  However, when considered together, they provide much weaker evidence.  If the choice was already narrowed to the two suspects, the evidence cancels out.

Returning to my reading of the rotated votes, they provide evidence for three conflicting propositions, which together cancel out.  But imagine what would happen if for some reason you knew that A is the worst candidate.

Now, if the rotated ballots truly provided no evidence, this combined with the knowledge that A was worst, would give you no help in deciding between B and C.  However, if you actually review the three propositions in light of the new knowledge:

A is better than B (Since 2 people agreed and 1 disagreed)
B is better than C (Since 2 people agreed and 1 disagreed)
C is better than A (Since 2 people agreed and 1 disagreed)

You will note that the first proposition is now impossible.  We also now know the last proposition to be true.  The second proposition, however, still stands, and is no-longer contradicted by a propositions that we can hold with equal evidence.  Therefore, it is reasonable to conclude that B is better than C, based on a balance of probabilities.

Now, of course, there is no way we can simply know that a particular candidate is the worst.  However, we could be given additional evidence to this effect, as was given in the example I showed.

5 C B A
4 B C A

Although A was not proven worse, enough evidence against A was given to allow a re-evaluation of the information in the reverse-symmetric votes.

By the way, I give further criticism of Borda's method, particularly in comparison to Tideman's method, on my page

http://www.fortunecity.com/meltingpot/harrow/124/path/pos.html

Note that this is not part of my Election Method Resource site, so I make no attempt to be impartial.

> 	As a final comment, even a *trace* of rotational symmetry can cause
> difficulties.  For instance, take a ranking wheel with six candidates
> and the initial ranking ABCDEF.  Now, only rotate the wheel three times
> (rather than the full six) to create a profile
> 	ABCDEF, BCDEFA, CDEFAB
>
> It is easy to argue that C probably is the favorite of these voters, and
> F (who is bottom ranked so often) is the least favored. In any case, 
> it is clear that C, D, E are ranked above F.    Yet, using the
> myopic nature of the pairwise vote (which cannot catch this symmetry), 
> it is easy to design a procedure to elect F.  Namely, F can only beat
> A, A beats B, B beats C, C beats D, D beats E.  So, start with an 
> agenda pairing D, E, with the winner against C, with the winner against
> B, with the winner against A, with the winner against F.  F wins.  In
> fact, each of the tallies are either by a unanimous or 2/3 vote, so 
> it may appear that the outcome is the true belief of the voters. 

Saari may be under-selling his case against F here.  In fact, every voter prefers C to F.  The same is true of D to F and E to F.

This means that in any method meeting the Pareto unanimity criterion, and of the commonly mentioned methods, only Robert's Rules of Order fails Pareto, F cannot win.

I'm not moved by the argument that even though my particular favourite method performs well in this example, it belongs to a class of methods, some of which do not.  Similarly, one could lump Borda in with the class of positional methods.  

---
Blake Cretney