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MIKE OSSIPOFF | 1 Jul 2002 08:54
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Dave Ketchum reply, part 2


First, to clarify something I was saying before, it can be said that,
with Approval, sometimes a voter doesn't know how to vote, if
knowing how to vote means knowing a way of voting such that after the
election you won't say "I'd have gotten a better outcome if I'd voted
differently". That's what Dave was referring to. And I was replying
that the voter can still know how to _choose_ how to vote, and it
is that which constitutes participation in an Approval election. Everyone 
knows what to do, then, and, by some reasonable approximations,
Approval maximizes the number of people who consider the winner better
than what they'd expected.

In addition to that, though Approval doesn't meet all the strategy
criteria that Condorcet(wv) meets, it still does better than any
practically proposable method other than Condorcet(wv). Besides,
Approval meets a strategy criterion that Condorcet(wv) doesn't meet:
FBC.

Dave continued:

>You seem to be implying that because IRV lets you express your
>preferences, but sometimes ignores them, and Approval doesn't let
>you express all of them, that you believe that IRV is better than
>Approval.

I was not trying to say this - I was only saying that I see problems
with
each. Still, I lean toward IRV a bit, for new equipment that can do IRV
should be able to do Condorcet with little strain. Ballots are
identical,
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Tarr, Adam | 1 Jul 2002 16:48
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RE: Cyclic Ambiguities = misinformed voters? (was: The True Majority Ghost)

Alex wrote:

> If all candidates fit on a one-dimensional spectrum then 
> this  is certainly true.  The person whose preference 
> order is  Bush>Nader>Buchanan>Gore needs to start reading 
> some newspapers.  The person whose  preference order is
> Buchanan>Gore>Nader>Bush probably lives in Florida ;)
> But what if we have two dimensional candidates?  
> [example follows]

I'd like to generalize Alex's example by giving a simple geometric argument
why a sincere cyclic tie can occur in the electorate.  For simplicity, I
will assume there are only three candidates.  These candidates have opinions
on a wide range of issues, so they each have a point on the "issue space"
that corresponds to their stances.  Their three points in the issue space
define a plane, so we can effectively ignore any differences outside that
plane.  Voters will find the point on that plane that is nearest to them,
and prefer candidates who are closer to that point.

So without loss of generality, we have reduced the situation to a
two-dimensional issue space.  The candidates are represented by points in
this space.  Now draw three lines; each line is a perpendicular bisector of
the line connecting two of the candidates.  On one side of the line, all the
voters prefer A to B, and on the other side all the voters prefer B to A,
and so on.  The three lines (A/B, B/C, and C/A) will meet in one point and
divide the plane into six sectors.  This is guaranteed by plane geometry as
long as the three candidates are not perfectly collinear (which is very
unlikely).  These six sectors correspond to the six permutations of
preference in a three way election; ABC, ACB, BAC, BCA, CAB, and CBA.
(Continue reading)

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Alex Small | 1 Jul 2002 20:28
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RE: Cyclic Ambiguities = misinformed voters? (was: The True Majority Ghost)

Adam-

I like your geometric proof.  However, I'm rusty on geometry.  For a given
triangle, will the perpendicular bisectors of its 3 sides always meet at a
single point?

Also, when working in issue space, I assume you're measuring the distance
between two points (x1, y1) and (x2, y2) as |x1-x2| + |y1-y2|.  I don't
think it matters for this case, but it seems more logical than using the
pythagorean theorem.

On the subject of Donald's proposal of IRV-completed Condorcet, right
after sending the message I realized it would be non-monotonic.  However,
it looks like IRV may catch on in the US in the next few years.  We have
yet to reach consensus amongst ourselves on resolving cyclic ambiguities,
so if IRV catches on it might make sense to propose it as the "backup." 
Keeping a pre-existing method as backup satisfies the cautious, and using
a method that the public has already accepted avoids debates over the many
different proposals for resolving cycles.  I know people on this list tend
to dislike IRV (I'm no fan), but I always maintain that progress is about
improvement rather than perfection.

The Condorcet first step does cure one flaw of IRV:  The
Hitler-Stalin-Washington case.  The whole idea of HSW is that there's a
unifying figure who commands respect (albeit not first-place support) from
other camps, and IRV fails to elect him.  However, in the presence of a
cycle it's impossible for any figure to say "I'm a uniter, not a divider"
since the electorate can always point to another candidate and say "Oh,
yeah?  We'll take this other guy over you."  Hence the HSW case doesn't
really apply.
(Continue reading)

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Alex Small | 1 Jul 2002 20:41
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Condorcet Criterion vs. Condorcet Efficiency

Donald has claimed that using the Condorcet criterion is inherently
prejudicial, since only methods in the Condorcet family can comply.

However, there's also a concept called "Condorcet Efficiency":  Quantify
the likelihood that a given non-Condorcet method will elect the Condorcet
winner.  We can debate the best way to measure this (Monte-Carlo seems
reasonable off the top of my head), but it is used in the literature. 
Brams and Fishburn, two men who are most definitely NOT Condorcet
partisans (why else would Brams devote a considerable portion of his
academic work to studying, promoting, and defending Approval Voting?) use
this concept in their book.

I'm curious if any work has been done comparing the Condorcet efficiencies
of Approval and IRV.  It's been a few months since I looked at Brams and
Fishburn, and I don't have a copy handy, so I don't know if they compared
the two.  When my copy arrives (ordered it used) I'll check it out.

Alex

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Bart Ingles | 1 Jul 2002 22:15
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Re: Condorcet Criterion vs. Condorcet Efficiency


Alex Small wrote:
> ...
> I'm curious if any work has been done comparing the Condorcet efficiencies
> of Approval and IRV.  It's been a few months since I looked at Brams and
> Fishburn, and I don't have a copy handy, so I don't know if they compared
> the two.  When my copy arrives (ordered it used) I'll check it out.

Check out "Making Multicandidate Elections More Democratic"
Samuel Merrill, III
Princeton University Press, 1988

You might be able to find a used copy, or any good university library
should have it.  Any public library should be able to obtain it through
inter-library loan -- I think there was a one-dollar charge when used
this.

Two chapters compare several methods using both Condorcet Efficiency and
Social Utility Efficiency, using random society and spatial models.  One
thing I noticed was that all models assume no strategic information, so
they appear inherently stacked against Approval (Merrill addresses this
in his comments, stating that in a strategic situation Approval should
do better and Condorcet worse than shown).

Still, Approval holds its own, and tends to give moderately high but
consistent results across the board w/ Condorcet efficiency, and very
good results w/ Social Utility.  Plurality, Runoff, and IRV all do
poorly when there are many candidates, or when the candidates are more
centrist than the voters.
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MIKE OSSIPOFF | 2 Jul 2002 03:45
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IRV completed pairwise-count example


First of all, I call it pairwise-count, because "Condorcet" properly
applies to Condorcet's own proposals for solving circular ties.

IRV-completed pairwise-count has been proposed many times. Here's an
example:

3 candidates: A, B, & C.

Sincere rankings:

40: ABC
25: BAC
35: CBA

Voted rankings:

40: A
25: BA
35: CB

The A voters, the ones who are in a position to make a strategic
circular tie, are also the ones who win that tie.

What does it take to keep A from winning? Truncation by the B voters
won't do it. The C voters could rank B equal to C, something that they
wouldn't have to do in the Condorcet(wv) versions. That's a general
pairwise-count defensive strategy that works against offensive truncation, 
but not against offensive order-reversal.
(Continue reading)

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Tarr, Adam | 2 Jul 2002 16:23
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RE: Cyclic Ambiguities = misinformed voters? (was: The True Majority Ghost)

> Adam-
> 
> I like your geometric proof.  However, I'm rusty on 
> geometry.  For a given triangle, will the perpendicular 
> bisectors of its 3 sides always meet at a single point?

Yes.

> Also, when working in issue space, I assume you're 
> measuring the distance between two points (x1, y1) and 
> (x2, y2) as |x1-x2| +  |y1-y2|.  I don't think it 
> matters for this case, but it seems more logical than 
> using the pythagorean theorem.

I hadn't really thought about it; you could probably make an argument for
either euclidean or taxicab geometry.  But as you say, it makes no
difference in this case.  The bisecting line remains the same.

> So, overall, I'd say IRV-completed Condorcet is better 
> than IRV, even if other methods might be preferable.

Well I certainly agree with that.  IRV-completed Condorcet has some
strategic problems that better Condorcet completion methods lack, but it's
able to avoid some of the most eggregious flaws of IRV.

That said, this does not mean we should be promoting IRV or accepting it
where we find it.  While IRV promoters have the strongest organization of
any USA electoral reform group, it's still very small in the grand scheme of
things.  I think we should mostly concentrate on promoting election methods
we like, such as Approval and Condorcet.  If you happen to run into an IRV
(Continue reading)

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Forest Simmons | 3 Jul 2002 02:43
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Re: Dave Ketchum reply, part 1


On Sat, 29 Jun 2002, MIKE OSSIPOFF wrote:

> Dave continued:
>
> - after you tell me I can select more than one, I have to decipher
> how many is a reasonable selection in each case.
>
> I reply:
>
> I remind you that you also have to do that now, with Plurality.
> With Approval, a good strategy is to vote for the candidate whom you'd
> vote for in Plurality, and also for everyone whom you like better.
> Voting in Approval is _not_ more complicated than voting in Plurality.
> Approval is an obvious undeniable improvement over Plurality. That
> can't be said for IRV.

As Craig Layton has affirmed more than once, the elections in Australia
that make use of ranked ballots have resulted in most voters either voting
above the line (for a party's preference order) or in voters following
candidate voting cards rather than figuring out a preference order for
themselves.

Following party or candidate recommendations would be even easier with
Approval.

And if you disagree with the party or candidate recommendations, then you
could easily modify your own ballot accordingly.

Forest
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MIKE OSSIPOFF | 3 Jul 2002 03:14
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Dave Ketchum reply, part 3


I'd said:

>Doesn't every county have an at-large board of supervisors? In
>that case, each county has provision for accepting & counting votes
>for several candidates per ballot.

Dave replied:

WRONG, TWICE:

[...]

Yes, though I knew that boards of supervisors are usually elected
the same way, I for some reason forgot which way that is. Of course
they're usually or always elected in single-member districts, so as
to give local representation in the relatively large county
jurisdictions.

But lots of cities elect their councilmembers at-large, and so they're
set up to receive and count, on each ballot, votes for several
candidates, recording and incrementing vote totals for all the
candidates.

True, many cities elect their councils in districts, but there are
so many at-large cities that it surely wouldn't be difficult for a
county govt or a district city to obtain what it takes to count
on each ballot votes for several candidates, recording all their
vote totals, as the at-large cities already do.
(Continue reading)

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Alex Small | 6 Jul 2002 00:19
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Strong FBC and the SVM

A few months ago we discussed the SVM and the Strong Favorite Betrayal
Criterion.  I haven't solved the problem of Strong FBC, but I have some
thoughts on the matter that might help us answer the question.

(As a reminder, the SVM is a machine that takes each voter's preference
order as input, assigns to each voter an optimal strategy given everybody
else's preferences, to achieve a Nash equilibrium, where no voter has an
incentive to change his strategy.  The question is whether one ever has an
incentive to rank somebody equal to his favorite candidate when giving
input to the machine.  If so then it violates the Strong Favorite Betrayal
Criterion.  The weak criterion is that you should have no incentive to
rank another candidate above your favorite.)

First, must the machine be deterministic?  Yes.  We already have a
non-deterministic election method that satisfies strong FBC:  Random
ballot.  If we’re going to admit non-deterministic election methods to
satisfy strong FBC then there’s really no point in building the machine. 
However, I don’t know anybody (except maybe Alan Natapoff, who likes the
Electoral College because it increases your odds of casting the deciding
vote) who wants a method where you gamble on having the deciding vote.

Second, in general, a game can have more than one Nash equilibrium.  If
for a given voting method and electorate there are two or more different
outcomes that correspond to Nash equilibria the machine has a decision to
make.  There are two ways to resolve this:

The first is to first figure out what all possible Nash equilibria are,
given an electorate and election method.  If more than one outcome is
possible then we need to choose among those equilibria.  We can’t use the
strategies assigned to the voters because the machine HASN’T assigned any
(Continue reading)

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