[EM] simple question (I think)
Hi, I haven't been around for a good while but some of you may remember me. I have recently been playing around with some stuff for scoring condorcet elections, and ran into a question that seemed obvious but maybe not: Is it possible, in any of the Condorcet election methods (beatpath, ranked pairs, etc), for someone who has fewer pairwise wins than another candidate, to win the election regardless? For instance, say there is no Condorcet winner. Candidates A, B and C all have 8 pairwise wins. D has 7. Could D still be chosen as the winner by any reasonable method? Thanks, -rob election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
On Wed, 2005-11-16 at 16:27 -0800, rob brown wrote: Hmmm, ok. Obviously though, if A has 8 pairwise wins, B has 7, and everyone else has less.A will be the winner no matter what, right? Regardless of how close A's wins are and how large B's majorities are? This is only guaranteed if there are 9 candidates, in which case A is the condorcet winner. Otherwise both A, B, and another could end up in the Smith set. -Scott Ritchie election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Scott Ritchie
Sent: Wednesday, November 16, 2005 6:49 PM
To: rob brown
Cc: election-methods@electorama.com
Subject: Re: [EM] simple question (I think)
On Wed, 2005-11-16 at 16:27 -0800, rob brown wrote:
Hmmm, ok.
Obviously though, if A has 8 pairwise wins, B has 7, and
everyone else
has less.A will be the winner no matter what, right? Regardless
of how close A's wins are and how large B's majorities are?
This is only guaranteed if there are 9 candidates, in which case A is
the condorcet winner. Otherwise both A, B, and another could
end up in
the Smith set.
But the original condition was that there was NO condorcet winner, so there
must be more than nine candidates.
In fact, this points out that the condition for D to win (the original
simple question) is that D must be in the Smith set. Presumably all of the
alternatives that have more pairwise wins than D are.
So the conditions that D win are:
1. {A B C D} pairwise beat everybody else, forming a Smith set
2. The method that selects the winner puts strength of support above number
of pairwise wins
election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
Ok, I see what you are saying. If there is a condorcet winner, that candidate will have more pairwise wins than any other candidate. But just because someone has more pairwise wins does not make that candidate the condorcet winner. Fair enough. Thanks.On 11/16/05, Scott Ritchie [EMAIL PROTECTED] wrote: This is only guaranteed if there are 9 candidates, in which case A is the condorcet winner.Otherwise both A, B, and another could end up inthe Smith set. -Scott Ritchie election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
On 11/16/05, Paul Kislanko [EMAIL PROTECTED] wrote: But the original condition was that there was NO condorcet winner, so theremust be more than nine candidates. Yes, Scott was replying to my simplified variant of the question. I was confusing a Condorcet tie with a tie for the number of pairwise victories. I blame this article for my confusion :) http://en.wikipedia.org/wiki/Copeland%27s_method ... where it seems to imply that they are the same thing (This method often leads to ties when there is no Condorcet winner might have been clearer if it just said This method often leads to ties). election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
On Wed, 2005-11-16 at 17:59 -0800, rob brown wrote: I was confusing a Condorcet tie with a tie for the number of pairwise victories. I blame this article for my confusion :) http://en.wikipedia.org/wiki/Copeland%27s_method ... where it seems to imply that they are the same thing (This method often leads to ties when there is no Condorcet winner might have been clearer if it just said This method often leads to ties). Hi Rob, The latter, while clearer, is not really true. Copeland always picks the Condorcet winner, when there is a Condorcet winner. Instances where there is not a Condorcet winner are rare. In those rare instances where there is not a Condorcet winner, there's usually also a tie in the Copeland result. Rob election-methods mailing list - see http://electorama.com/em for list info
Re: [EM] simple question (I think)
Rob, rob brown wrote: For instance, say there is no Condorcet winner. Candidates A, B and C all have 8 pairwise wins. D has 7. Could D still be chosen as the winner by any reasonable method? Yes. The method that just counts the number of pairwise wins is called Copeland. It hopelessly fails Clone Independence (Clone-Loser) and Rich Party. Imagine that that there are three candidates, each with the same number of pairwise wins, and the Condorcet method elects X. Say that the top cycle is XZYX Now say we add a clone of Y, that every voter ranks directly below Y. Now Y and Z will each have an extra pairwise win, one more than X and so now (by the Copeland criterion) X must lose to Z or Y. Adding a clone of a losing candidate (not to say adding a Pareto-dominated candidate) has changed the winner. Parties and factions that run more candidates will have an absurd and unfair advantage. Chris Benham election-methods mailing list - see http://electorama.com/em for list info
