On Mon, 3 Mar 2003, Steve Barney wrote:
Forest:
In message # 10970, why did you say wisely, as follows?:
Kemeny (wisely) doesn't believe in cyclic symmetry removal
[...]
Do you mean to imply that the KR tie between ACB and BAC is more
reasonable than the ABC outcome yielded by both
:BCA
5:BAC
that the order of those operations matters in some cases. Please show me what
you mean.
SB
--- In [EMAIL PROTECTED], Forest Simmons [EMAIL PROTECTED]
wrote:
On Thu, 27 Feb 2003, Steve Barney wrote:
Forest:
Apparently, as I thought, your method of decomposition
On Wed, 26 Feb 2003, Gervase Lam wrote:
Nevertheless, could MCA be tweaked a very tiny amount to get closer to
the better fairness that Cardinal Ratings can give? May be this could be
done by having a different Preferred cut-off point. Using an example, I
suggested 2/3 of the votes instead
On Thu, 27 Feb 2003, Steve Barney wrote:
Forest:
Apparently, as I thought, your method of decomposition is to simply to remove
cycles first, and then reversals. My point remains, then, that your
decomposition method does NOT NECESSARILY yield the same outcome as Saari's
matrix decomposition
On Fri, 21 Feb 2003, Steve Barney wrote:
Forest:
How do you decompose my example (from my last email, #10873), and what do you
get?:
3:ABC
5:ACB
0:CAB
5:CBA
0:BCA
5:BAC
Subtract out five copies of the cycle ACB+CBA+BAC.
That leaves 3*ABC.
Forest
For more information about this
On Thu, 20 Feb 2003, Alex Small wrote:
snip
However, people electing politicians are clearly not machines. We have
our idiosyncracies and legitimate differences of opinion, and we debate
matters that don't have obvious, objectively correct answers. Because we
don't behave or think like
On Wed, 19 Feb 2003, Rob LeGrand wrote:
Obviously, I couldn't agree more. :-) Thanks so much to all of you who
have joined CAV. Those who haven't, please check us out at
http://www.approvalvoting.org/ .
Rob, I just checked out the CAV website. Very Good!
For more information
In the main version of MCA, the fifty percent plus of voters is only
needed for electing a candidate on the basis of favorite status. In other
words, if no candidate has favorite status on more than fifty percent of
the ballots, then the candidate with the most approval is elected, even if
no
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
Alex, you're right!
That's what happens when you try to simplify part of an argument while
permuting letters so that the default is in alphabetical order :-)
I'll forward a copy of the untampered original in a minute.
On Tue, 18 Feb 2003, Alex Small wrote:
snip
Seems like the best option is
In what follows you will find (most of) an argument showing that the
Favorite Betrayal Criterion (FBC) is incompatible with neutrality and
other weak assumptions when fully ranked ballots are employed.
By neutrality the ballot set XYZ+ZYX cannot yield either X or Z as winner,
so there are two
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
Suppose for the sake of argument that Blake is right in his (as I
understand it) main reason for preferring margins over winning votes:
Margins are better estimators than sheer numbers for deciding which
candidate is likely to do the best job.
Let's assume this is true if the margins are
It seems to me that any neutral method that gives a three way tie to a
reverse order pair (like the following ballot pair) cannot satisfy both
Pareto and the strong FBC:
1 ABC
1 CBA .
Here's my reasoning. Suppose that there are only two voters and one has
already voted ABC. Suppose further
It seems reasonable that if S is a ballot set with a definite winner X,
and T is any other ballot set, then sufficiently many copies of S added to
T should result in a ballot set supporting X.
As far as I can tell, all seriously considered deterministic methods
(including IRV and Borda) satisfy
On Thu, 6 Feb 2003, Alex Small wrote:
Forest Simmons said:
It seems reasonable that if S is a ballot set with a definite winner X,
and T is any other ballot set, then sufficiently many copies of S added
to T should result in a ballot set supporting X.
Let me see if I can understand what
On Wed, 5 Feb 2003, Narins, Josh wrote:
A bias over time for small states?
I think they are entirely incorrect.
In actuality if the small states had exact proportional representation,
they would have less voting power per citizen than the large states.
See EM archives messages 8541 and 9054
On Tue, 4 Feb 2003, Narins, Josh wrote:
The only issue is the overall standard deviation between district
sizes can sometimes be helped by _REDUCING_ the number of seats. FOr
instance, at the last Apportionment (2000). Although 435 seats were given
out, if only 432 had been, the
On Wed, 29 Jan 2003, Alex Small wrote:
I think the Partial Decisiveness condition removes the possibility of
fractal boundaries, since I specified that the ties occur on a set of
4 dimensions (or N!-2 dimensions for N candidate races). I don't know
much about fractal curves in a
On Thu, 30 Jan 2003, Alex Small wrote:
Well, what's the weakest condition we could impose to guarantee that the
boundaries have normals? You've said that fractal boundaries don't
necessarily have normals. Obviously boundaries specified by linear
equations would have normals, except
On Wed, 29 Jan 2003, Steve Barney wrote:
...
BTW, one reason given in a news article for dropping Nanson's Method and
reverting back to the plurality with a runoff was that they preferred voting
twice, and felt that they could be more informed voters the second time
around. What to do about
Good work, Alex. I think the argument can be simplified so that
it will generalize easier, but nobody else has faced up to it like you
have.
BTW it seems like every N+3 candidate election has a three candidate
election embedded within it as far as each faction is concerned, since
each faction
Remember that Joe W. once suggested approving as far down your preference
order as you can without exceeding 50 percent probability of the winner
coming from your approved set.
Whether or not you should approve the next candidate (the one that would
tip the scales to more than fifty percent)
You might be interested in Demorep's ideas along these lines in message
8228 (Oct 20,2001) of the EM archives. The subject line is Low Tech
Proxy P.R. Method. Further messages under that heading are 8233 and
8241.
See also message 8249 of Oct 26, 2001 for a related idea of mine under the
On Sat, 18 Jan 2003, Steve Barney wrote:
If you don't like Condorcet's example, how about this one, which I have looked
at before:
5 ABC
3 BCA
Can you give me the decomposition profile, T(p), for this example?
This example is essentially the same as the
66% ABC
34% BCA
example.
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
In your example,
66 ABC
34 BCA
If you give second preferences any more than 16/33 of the weight which you
give to the first prefs, the winner is B; since:
No need of giving weights to see all the mischief that could come from
giving the
the radii of gyration.
On Fri, 17 Jan 2003, Alex Small wrote:
Forest Simmons said:
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
In your example,
66 ABC
34 BCA
No need of giving weights to see all the mischief that could come from
giving the win to B.
Moreover
On Thu, 16 Jan 2003, Alex Small wrote:
Forest Simmons said:
Could we consider this a canonical reduction of some type?
Do you mean canonical transformation? You mentioned variational
principles in another post, which are related to canonical transformations
in Lagrangian and Hamiltonian
On Thu, 16 Jan 2003, Tom Ruen wrote:
I judge that Bucklin has pretty much the same strengths and weaknesses as
MCA. Ranking versus Rating is the difference.
The option of allowing equal rankings is nice, although not clear that it is
in the best interest of anyone to do so.
Bucklin might
On Fri, 17 Jan 2003, Forest Simmons wrote:
...
In fact, suppose that your favorite candidate is A, and that there are two
front runners B and C, of which your preferred is B, so that your sincere
preferences (restricted to these three candidates) look like ABC.
What would be the worst
On Fri, 17 Jan 2003, Alex Small wrote:
Forest Simmons said:
66 ABC
34 BCA
No need of giving weights to see all the mischief that could come from
giving the win to B.
Moreover, if candidate C weren't there then we'd all agree that A trounced
B conclusively. Then we throw in C
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
Isn't that just another way of saying Kemeny's Rule does not respect cyclic
symmetry?
Or we could say that cyclic symmetry doesn't respect the minimal
distance criterion, since that is what Kemeny's rule is.
A more neutral statement is that
On Thu, 16 Jan 2003, Alex Small wrote:
Anyway, after exploring what happens when we require election methods to
respect symmetry, I'm forced to conclude that symmetry isn't a very useful
criterion.
When stabbing around in the dark you can use symmetry as a guide for where
to start
On Thu, 16 Jan 2003, Forest Simmons wrote:
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
Isn't that just another way of saying Kemeny's Rule does not respect cyclic
symmetry?
Or we could say that cyclic symmetry doesn't respect the minimal
distance criterion, since that is what
I cannot do better, but here is a consideration:
Suppose each voter is to mark one candidate and write down one number in
the space provided on the ballot, and that the winner of the election is
the candidate marked on the ballot that has the largest number in the
space provided.
Does this
I've already given an example in which Borda gives the wrong answer after
the symmetry is removed. Now you have given an example in which symmetry
removal shows the CW to be wrong. So that evens the score :-)
In other words, neither Borda nor Condorcet can claim to be superior on
the basis of
One could limit the space to so many ASCII symbols, so the person most
ingenious at describing large numbers would choose the winner.
Say there are ten symbols allowed, which would be larger
9!
or
9^9^9^9^9! ?
If there were enough symbols to write out the phrase
The sum of all the
Actually, Cantor proved that there are infinitely many distinct infinities
on the same day he proved that the cardinality of the reals is greater
than the cardinality of the rationals.
Here's the proof in modern notation:
Let X be any set (finite or infinite, it doesn't matter). Let P(X) be the
On Wed, 15 Jan 2003, Alex Small wrote:
I'm not convinced
that symmetry is a particularly compelling reason to pick an election
method,
especially not the symmetry of {ABC,BCA,CAB}, which has a rotational bias.
True, it favors no candidate, but it does favor its three orders over
rigidly into a
three dimensional coordinate system:
ABC -- [1,0,-1]
ACB -- [1,-1,0]
CAB -- [0,-1,1]
CBA -- [-1,0,1]
BCA -- [-1,0,1]
BAC -- [0,1,-1]
Forest
On Mon, 13 Jan 2003, Forest Simmons wrote:
A simple example illustrates the Achilles heel of the VMO (Voter Median
Order).
3 ABC
2 BCA
In EM archives message #8999 Alex recounts in his own words Saari's idea
of subtracting out the symmetrical part of a ballot pattern and then
deciding the winner on the basis of the residual ballots.
Let's do an example:
7 ABC, 5 ACB , 9 CAB, 3 CBA, 7 BCA, 8 BAC.
If we subtract three of each
greater than unity, but with
only a few of degrees of freedom to be shared between the numerator
and denominator, we cannot have much confidence in this conclusion.
Note that the top VMO candidate is also the IRV winner. That's not too
comforting.
Forest
On Fri, 10 Jan 2003, Forest Simmons wrote
the winning order is BAC, in this case the same as
as the Borda order, rather than the Ranked Pairs order of ABC.
Forest
On Wed, 8 Jan 2003, Forest Simmons wrote:
Suppose that (in a certain election) candidate X is preferred over any
other candidate Y on any issue Z by some
far, the VMO order agrees with Black.
Is there an example with three candidates in which Black does not yield
the VMO?
-- Forwarded message --
Date: Thu, 9 Jan 2003 19:23:41 -0800 (PST)
From: Rob LeGrand [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: Forest Simmons [EMAIL
the shape of the distribution of
voters in Voter Space, which presumably reflects the distribution of
voters in issue space.
It remains to be seen if there is a three candidate example in which Black
does not agree with the VMO.
Forest
On Fri, 10 Jan 2003, Forest Simmons wrote:
Rob LeGrand
is called the Footrule aggregation of the
voter rankings of the candidates. It is now easy to see why this Footrule
order is suboptimal; the old coordinate axes represent variables that are
correlated in varying degrees.
Forest
On Wed, 8 Jan 2003, Forest Simmons wrote:
Suppose
On Wed, 8 Jan 2003, Forest Simmons wrote:
Suppose that (in a certain election) candidate X is preferred over any
other candidate Y on any issue Z by some majority (depending on Y and Z).
Such a candidate would seem like a logical choice for winner of the
election, if there were such a candidate
Suppose that (in a certain election) candidate X is preferred over any
other candidate Y on any issue Z by some majority (depending on Y and Z).
Such a candidate would seem like a logical choice for winner of the
election, if there were such a candidate.
How could we locate such a candidate if
(if not differential game theory).
[Differential Game theory would be ideal for the continuous (joy stick)
version of the CRAB race method.]
Forest
On Thu, 2 Jan 2003, Dave Ketchum wrote:
On Tue, 31 Dec 2002 17:46:56 -0800 (PST) Forest Simmons wrote:
Linear algebra, graph theory, probability
://electionmethods.org/?
SB
--- In [EMAIL PROTECTED], Forest Simmons [EMAIL PROTECTED]
wrote:
Your example is correctly done.
Despite the intractability of the method for large numbers of candidates,
it seems like an ideal method for some situations.
One application could be in choosing between
Adam's response is precisely correct and well written.
I would add only this:
Determining the Kemeny order from ranked preference ballots suffers the
combinatorial explosion because there is no way of getting around one by
one testing of most of the N! permutations of candidates in order to see
Your example is correctly done.
Despite the intractability of the method for large numbers of candidates,
it seems like an ideal method for some situations.
One application could be in choosing between several orders that have been
found by other means.
[The main computational difficulty of
On Fri, 27 Dec 2002, Alex Small wrote:
snip
Another point is that Approval is a very BAD idea for the primary. Say
that we're narrowing it down to 4 candidates. The largest faction could
all approve their favorite and the 3 Stooges. The second stage would
likely include those 4
Somewhere up this thread Blake Cretney brought up the idea that some
Condorcet methods may satisfy a certain modified consistency criterion.
If the method gives a complete ranking as output, and if two subsets of
ballots produce the same ranking, then the output based on the union of
the two
This sounds like a good idea to me for situations in which it is not too
inconvenient to do an actual two stage process.
As Alex said in his filter posting, if we're going to go to all the
trouble of having primaries, why not do them right?
Using two stage cumulative approval we could say that
On Wed, 18 Dec 2002, Forest Simmons wrote:
By the way, it turns out that in the three candidate case, if the
preference ballots are generated in this way, regardless of the metric
used in step 2, a CW is assured; there can be no cycle.
So somewhere in this 5 step process the cyclical
Don might be happy to be expelled; then he can claim that his barbs were
so sharp that the EM list members couldn't cope with them, so they banned
him. It might give the EM list a reputation for closed mindedness.
Even undeserved reputations can be bad PR.
Forest
For more information about
A brief progress report:
Let the entry in the ith row and jth column of a matrix be a zero or a one
depending on whether or not the ith voter approves of the jth candidate.
Call this matrix A.
The rows of this matrix represent the approval ballots of the voters, so
we could call the row space
As I understand it, Kemeny's Rule amounts to minimizing a certain metric
on rankings, and that this minimization is an NP complete problem, making
it intractable for elections with more than four or five voters when there
are as few as twenty candidates.
Forest
On Wed, 18 Dec 2002, barnes99
I took it for granted that favorite would also be among the approved on
Majority Choice ballots, and that favorites would be determined from the
rankings or ratings in the case of CR or ranked ballots.
But I still think that CS as I proposed it suffers from a fault. If the
race is perceived as
Here's the version of Candidate Space (CS) that I like the best now:
The ballots must have some way of determining favorite, so must have at
least the expressivity of Majority Choice ballots.
[The favorite on the expressive side of the ballot must have maximal
positive instrumentality in the
On Fri, 13 Dec 2002, Alex Small wrote:
Forest Simmons said:
Each voter marks one candidate on the ballot. These candidates become
proxies (for the voters that marked their names) in an Election
Completion Convention.
If there are n seats to be filled, and there is a subset of n
Here's an heuristic argument showing why proxy approval is apt to pick the
CW when there is one, regardless of the dimension of the issue space.
When there is a CW, rational approval players with perfect information
will concede the win to the CW, because the CW is the one and only stable
I agree with Alex, if the at large method is not even semi-proportional,
districts would be an improvement over the current situation.
However a simpler and much better solution is to adopt some kind of PR or
semi-PR method that uses the same ballot style and same ballot machinery.
Among these
I've been thinking about Richard Moore's Majority Potential idea which is
just Copeland with fictitious candidates distributed uniformly in issue
space.
The method was never intended for public proposal, rather it was intended
as a standard to measure the Majority Potential of other methods.
a ranked ballot for each voter, and find the Condorcet Winner, if
there is one.
This makes the method very easy to test in simulations comparing it with
other methods.
We need a name for the method. Any ideas?
Forest
On Fri, 13 Dec 2002, Forest Simmons wrote:
I've been thinking about Richard
) of the Election Completion Procedure
votes].
Forest
On Wed, 11 Dec 2002, Forest Simmons wrote:
Candidate Proxy with 3 candidates, 1 seat to be filled, and a 1
dimensional issue space:
The following conditions taken together are sufficient to ensure that the
Candidate Proxy Winner and the Condorcet
Candidate Proxy with 3 candidates, 1 seat to be filled, and a 1
dimensional issue space:
The following conditions taken together are sufficient to ensure that the
Candidate Proxy Winner and the Condorcet Winner will be one and the same
in a three candidate, single winner election.
(1) The issue
Dear Steven,
thanks for your insightful and thoughtful response.
An explanation of my apparent dismissal of Approval in the context of
that message is given below, along with some other related thoughts.
Best Wishes,
Forest
On Sun, 8 Dec 2002, Steven J. Brams wrote:
Dear Forest,
On Mon, 9 Dec 2002, Gervase Lam wrote:
Each voter may give a candidate Yes, No, or nothing. A candidate's
No votes are subtracted from his Yes votes, and the result is his
score. The candidate with highest score wins.
That's equivalent to CR, with -1, 0, 1. Which is equivalent to
0, 1, 2
method?
Does it depend on the number of candidates? If so, perhaps that should be
factored into the formula.
Forest
On Wed, 4 Dec 2002, Alex Small wrote:
Forest Simmons said:
Let F(V,M) represent the set of voter ballots that are optimal for the
voters with utility set V under method M
Here are some preliminary considerations that come to mind in the course
of trying to quantify manipulability.
The kind of manipulability that is most crucial (in my opinion) has to do
with the sensitivity of (near optimal) strategy to variations in
information. To the degree a method is
I meant to make this information available to the whole EM list.
-- Forwarded message --
Date: Tue, 3 Dec 2002 14:40:05 -0800 (PST)
From: Forest Simmons [EMAIL PROTECTED]
To: Elisabeth Varin/Stephane Rouillon [EMAIL PROTECTED]
Subject: Re: [EM] The Copeland/Borda wv hybrid
Rob
as the Election Completion
Procedure.
On Mon, 2 Dec 2002, Forest Simmons wrote:
Here's a message that I forwarded to a friend of mine who is a prominent
and influentual member of FAVOR (FairVoteORegon) the organization
promoting IRV here in Oregon:
-- Forwarded message --
Date: Mon, 2
Here's a message that I forwarded to a friend of mine who is a prominent
and influentual member of FAVOR (FairVoteORegon) the organization
promoting IRV here in Oregon:
-- Forwarded message --
Date: Mon, 2 Dec 2002 14:28:47 -0800 (PST)
From: Forest Simmons [EMAIL PROTECTED
See below for my two cents worth.
On Thu, 28 Nov 2002, Douglas Greene wrote:
This is primarily directed to Mike, but I'd like to know why list members
support Condorcet over cardinal rankings/range voting. BTW, I've posted
Warren Smith's work on range voting to our Yahoo!Group.
There are at
On Wed, 20 Nov 2002 [EMAIL PROTECTED] wrote:
On 20 Nov 2002 at 14:11, Forest Simmons wrote:
Suppose that a voter ranks six candidates as follows in a three seat multiwinner
race:
A1B1B2A2A3B3 .
Which of the following two outcomes would this voter be most likely to prefer
Ollie, thanks for your insights and examples.
The main example below has led me to consider another, perhaps better, way
of scoring the head-to-head PR methods that I have been working on lately.
See below.
Forest
On Sun, 17 Nov 2002, Olli Salmi wrote in part:
snip
Cassel gives the
I'm coming around to the conclusion that reverse symmetry is not such a
good idea in multiwinner elections if one's aim is proportional
representation.
Consider the following examples:
Example 1:
25 ABC
25 BAC
25 CAB
25 CBA
In this example if we reverse the preferences, we get the same number
Scoring two candidate subsets relative to a ranked ballot can be
organized into a tableau as follows:
rank X1 X2 (X1-X2) Y1 Y2
-
100 0 00
201 -1 01
321 1 10
400 0 0
In this installment I would like to summarize (by example) the how to of
the current Condorcet Flavored Proportional Representation (CFPR) method
that takes into account the constructive criticism of Adam Tarr:
Here's information from one typical ballot:
The integral (for x from 0 to 1) of (1-x^(n+.5))/(1-x)
has the numerical value given by the finite sum
-ln(4) + 1/.5 + 1/1.5 + 1/2.5 + ... + 1/(n+.5).
Forest
On Thu, 7 Nov 2002, Forest Simmons wrote:
Now I'll try to tackle the second question:
On Thu, 7 Nov 2002, Adam Tarr
Adam, thanks for your interest and comments. I'll try to answer your
questions below.
On Thu, 7 Nov 2002, Adam Tarr wrote:
Forest, I finally got around to reading this series of posts. It's very
interesting stuff and you've obviously made a lot of progress on this. A
few comments:
- I'd
Now I'll try to tackle the second question:
On Thu, 7 Nov 2002, Adam Tarr wrote in part:
- I will admit this is the first election method I've dealt with where I
have trouble manipulating small examples. Here's a very small example that
was giving me trouble: say we are electing two
Here's why I believe that no voting method based on ranked ballots can
satisfy both the Favorite Betrayal Criterion and the Majority Criterion:
Suppose that sincere preferences are given by
x:ABC
y:BCA
z:CAB
and that none of the three factions has a majority.
Suppose (by way of contradiction)
Here's the instant version of CRAB:
Voters submit ranked preference ballots.
Suppose that there are N voters and K candidates.
Initialize a one by K array C by letting the j_th entry be the number of
first place votes of candidate j.
Then ...
While the maximum entry in C is less than N*K+1
Some Condorcet devotees disparage the Consistency Criterion only because
no Condorcet method can satisfy it. Others do not disparage it, but
reluctantly let go of it for the same reason.
But Condorcet (unlike IRV) methods are very close to the boundary of the
set of methods that do satisfy the
Random ballot does satisfy strong FBC.
I suspect that no majoritarian method absolutely satisfies strong FBC,
though some methods like the instant version of CRAB (Cumulative Repeated
Approval Balloting) satisfy it for all practical purposes.
I'll write more when my Internet Service Provider
Alex, it seems to me that if only the first two ranks get points, then in
a close race among several candidates if your favorite isn't among the top
three contenders with near equal chances, you may want to give the top to
slots on your ballot to your preferred among the top three contenders.
To
Good question. I wonder if Pareto Efficient means the same as
satisfying the Pareto Condition that we are all familiar with and that you
use in your proof sketch.
I know that the name Pareto is associated with various related but
distinct concepts in game theory because the referee of a paper
On Sun, 20 Oct 2002, Alex Small wrote:
MIKE OSSIPOFF said:
Of course that depends on how one defines IIAC. By the simple way that
I define it, Approval CR comply. But people have told me that they
believe that IIAC means something other than what I say it
means. But no one who has
On Mon, 30 Sep 2002 [EMAIL PROTECTED] wrote:
On 28 Sep 2002 at 16:17, Markus Schulze wrote:
Dear Forest,
you wrote (27 Sep 2002):
A Condorcet Flavored PR Method is an M-winner election method that
(1) compares candidate subsets of cardinality M head-to-head, and
(2) does the
When approval cutoff's are supplied (as they really should be, since
approval information is relatively cheap) there are several ways to
resolve the difficulty, and your way certainly gives the right answers to
the examples below.
I have some other ideas (along these lines) that I will post to
Suppose there are to be two winners in a PR election among several
candidates C1, C2, C3, C4, C5, ..., and in comparing subset {C1,C2} with
subset {C2, C3} the ballots show
170 C1C2C3
170 C1C3C2
330 C2C3C1
330 C3C2C1 .
Which of these two subsets provides better proportional representation?
On Fri, 20 Sep 2002, Elisabeth Varin/Stephane Rouillon wrote in part:
First why remove I and O if the ballots is presented like that?
I do not think there can be confusion...
I agree.
Next, I doubt we really need + and - signs because there is rarelly
26 candidates or more, but in case it
On Fri, 20 Sep 2002, Rob LeGrand wrote:
Steve asked:
Can you provide me with some examples where the IRV method elects
the Condorcet Candidate, but the 2-Stage Runoff does not?
Here's one:
40:ACDB
25:BCDA
20:CDBA
15:DCBA
Plurality picks A, top-two runoff picks B and IRV picks C,
On Wed, 18 Sep 2002, Narins, Josh wrote:
Joe, this statement is not rigorous.
Four levels are surely enough to
distinguish substantially distinct degrees of active approval.
From something I read back in college, 7 levels (much above, above, slightly
above, average, slightly below,
On Wed, 18 Sep 2002, Dave Ketchum wrote in part:
[Adam wrote]
At first I didn't like this idea, but its grown on me. The simplicity
to the voter of ABCD(E)F voting is worth it. The voters who are
interested and involved enough to actually need six distinct levels of
approval are the
On Tue, 10 Sep 2002 [EMAIL PROTECTED] wrote in part:
also, but not as a tie-breaker. Instead, I think that the approval cut-off should be
used to complete the ballot by placing the unvoted candidates between the
approved and unapproved candidates. Although this is imperfect, it seems to
Try the following web page
groups.yahoo.com/group/election-methods-list/messages
and then type PAV into the search box.
On Sat, 17 Aug 2002, James Gilmour wrote:
Forest Simmons wrote (in part)
Ordinary (i.e. non-sequential PAV) is a little harder to describe, but it
is easy to find
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