Democratic symmetry (fwd)

[Rob Lanphier]

---------- 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)

Dear Mr. Lanphier,

Thanks for posting my message.  When I glanced at your discussion
group yesterday, I was impressed by some of the comments. So, between
appointments today, I typed up the following which may help out
some of the participants to use these new tools before they 
have a chance to read my articles.  I would appreciate it if you could post
this one too.  I promise; I won't bother you again.

	Don Saari

-----------------------------


I have been impressed with how people in this discussion group 
have been using my reflection and rotational symmetries. 
While I recommend you read my Economic Theory articles, let 
me offer a couple of comments -- hints -- that may help you use
these tools before hand. 

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.  

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.
Let me illustrate with my favorite example.  

6 prefer Milk>Wine>Beer, 5 have Beer>Wine>Milk, 4 have Wine>Beer>Milk.

The plurality outcome is MBW.  The pairwise outcome is the exact opposite
of WBM.  Why?  Well, notice that the first two preferences reverse one
another.  So, as far as the pairwise vote is considered, the real data
set is that [5 MWB, 5 BWM] cancels leaving a tie, so we are left with
1 MWB (what is left over from the 6) and 4 WBM.  Here, the outcome is
the natural WBM.  So, the main difference between the pairwise and the
plurality vote in this example is that the pairwise vote treats [5 MWB, 5 BWM]
as a tie, while the plurality vote treats it as giving M = B (> W) each
five points.

Consequences:

1.  If you think reversal symmetry should lead to a cancellation,
then the Borda Count is the only positional procedure you can use.
(Just tally [ABC, CBA] with (1, s, 0) where s = 0 is the plurality
vote, s = 1 is where you vote for two, and s = 1/2 is the BC, to
see how different methods are affected by this symmetry in data.)

2.  If you think ABC, CBA should NOT lead to a cancellation, then you
must decide whether the outcome should be A=C > B (which means you
choose procedure with  s < 1/2), or B > A=C (which means you have to choose
a s > 1/2).   I don't care whether you accept cancellation or not; 
but all differences in methods is the choice as explained here.  The choice
will determine all subsequent properties that your procedure satisfies.  

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).  (Reason: the reversal symmetry has an effect which is
ignored by the pairwise vote.)  In fact, the likelihood of this happening
increases as s gets near 0 (the plurality vote) or 1 ("vote for two")
because these methods place an added emphasis on the reversal symmetry data.

4.  For examples where the plurality vote elects a majority candidate and
other procedures, the data  MUST involve a strong amount of reversal symmetry.
This leads to the somewhat surprising fact that maybe the outcome is not
the correct one.  Rather than argue, let me recommend you play around with
this and other examples.  I would expect that you will find some new properties
that have not been discussed before.

5.  What surprised me is that ANY property showing how one positional 
method differs from another is strictly due to how they handle reversal
symmetry.  In other words, while there are many, many properties out 
there, but all of them for positional methods are consequences of reversal 
symmetry.  In fact, it turns out that you can start with the BC tally
(BUT, using (1, 1/2, 0) rather than (2, 1, 0)) and then get the tally
of any other (1, s, 0) procedure strictly by how the procedure treats the 
reversal symmetry terms of the data.

ROTATIONAL SYMMETRY

For reversal symmetry, start with a disk.  If you have three candidates,
place the numbers 1, 2, 3 evenly spaced around the edge (so 120 degrees apart).
If you have more candidates, do the same with the number.  (So, if you have
10 candidates, place the numbers 1, 2, ..., 10 evenly spaced along the
edge.)  Now attach this disk at its center on a wall.  On the wall next
to each number list a candidate.  For the three candidate example, this
might be in the order B A C.  Now, rotate the disk so that the number 1 is
by the next candidate and read off the ranking -- A C B.  Keep this up 
until all candidates have been number 1 once.  In this example, we have
the three rankings BAC, ACB, CBA.   (So, for three candidates, there are
two sets of rotational symmetries -- the other one comes from starting
with B C A.  For four candidates, there are six different sets of examples,
for six candidates there are 120.) 

By construction, each candidate has been ranked first, second, third, 
(and, with more candidates, each of the other positions) exactly once.
It is difficult to argue that any one candidate is favored over another,
so this is a portion of the data which should be viewed as causing a tie.
However, the pairwise vote (and Arrow's binary independence condition, etc.)
cannot see the full symmetry.  Nevertheless, the symmetry still is there, and
it manifests itself in terms of a cycle B>A, A>C, C>B all by 2:1 tallies.

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.  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.  In fact, the surprising
(at least to me) result is that the ALL differences between a BC ranking
and the pairwise rankings are due to the rotational symmetry terms!  As
a further consequence, ALL differences between the BC and your favorite
method -- Kemeny's method, Dodgson's method, Copeland, etc., etc., all 
differences in properties (whatever they may be!) are strictly due to 
the fact that these other methods do NOT cancel rotational
symmetry.  So, all differences in properties manifest the different ways
these rotational symmetries are handled by different procedures.

To give you a hint about further properties of rotational symmetry, consider
the 11 voter example of five candidates created with the ranking wheel
in the following manner:

	One voter has the ranking ABCDE.

	Now, choose a ranking for the ranking wheel approach.  Say,
ACEBD.  Use the ranking wheel to  generate five rankings; assume there are
two voters with each ranking.  The final profile has

	1 ABCDE, 2 ACEBD, 2 CEBDA, 2 EBDAC, 2 BDACE, 2 DACEB

	If you compute the election outcome for all five candidates with
the Borda Count, plurality -- whatever weights, the ranking will be some
version of ABCDE.  (I.e., with the plurality vote, BCDE are tied, with
a method with all distinct weights, you get a strict ranking.)  This is
because all of these methods ignore the data from the ranking wheel.

	Now, compute the rankings of the five sets of four candidates.  You
will observe a cyclic pattern in the outcomes because the four candidate
elections cannot observe the full ranking wheel symmetry over five candidates.  
A more complicated pattern cyclic pattern emerges when all three candidate
elections are tallied.  For all pairs, it is the expected cycle from
the ranking wheel where A>C, C>E, E>B, B>D, D>A (with some more complicated
setting for the other pairs).  ALL behavior of this type is due to 
rotational symmetries.

	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. 


The analysis of voting procedures is an important concern.  I don't really
care which procedure you prefer or use.  But, since all major procedures can
be analyzed in terms of these (and, actually, others for more alternatives)
data symmetries, this should be viewed as a new tool to better understand
voting.  In other words, take the very practical problem that we cannot 
rank all candidates.  OK, adjust the symmetries accordingly, and different
answers emerge.  I.e., I am hopeful that these tools will allow us to 
start with pragmatic problems and then understand what happens and WHY.
(As I stated earlier, I favor a modified version of the BC -- modified
for realistic concerns, but based on the BC because I cannot accept not
canceling these data symmetries.)

Don Saari

===============================
Donald G. Saari, Pancoe Prof. of Mathematics
Northwestern University, Evanston, IL 60208-2730
Ph: 847 491-5580	Fax: 847 491-8906
dsaari at nwu.edu