Left or right loser

[Mike Ossipoff]

DEMOREP1 at aol.com writes:
> 
> Yet another Condorcet circular tie with 15 voters--
> 
> 3 AC
> 4 AB
> 
> 5 BC
> 
> 2 CA
> 1 C
> 
> A v. B-- 9 to 5
> B v. C-- 9 to 6
> C v. A-- 8 to 7
> 
> According to Mr. Ossipoff, A should win by having the fewest votes against A
> in his/her worst (and only) defeat (8).  Note that all of the B and C voters
> combined thus could not defeat A (even if, horror of horrors, B and C were
> "progressives" against a "reactionary" A)

Nonsense. The B & C voters, together, number 8. That's a majority.
They have the power to ensure that A can't win.

But anyway, it isn't at all clear what your example is supposed to
be an example of. The A voters are evenly split between B & C,
suggesting that A is a middle candidate, if there is one. 
Most of the C voters, and all the ones who vote a 2nd choice,
vote A in 2nd place. But the B voters unanimously vote C in
2nd place. 

There are 2 possibilities: Either it's a natural ciruclar tie
or someone is order-reversing. That would appear to be the
B votes, and, if so, their attempt failed to defeat A.

> 
> However, B gets the fewest number of votes for him/her in his/her worst (and
> only) defeat (5). If B loses, then C wins by defeating A.

What's votes-for got to do with it? I fully admit that Condorcet(EM)
doesn't count votes-for, if that's what you're trying to show.
What's your point? And you seem to be making it into not only
a votes-for count, but a votes-for elimination. I have no idea
what you're trying to say.

> 
> If each vote is an approval vote, then there is-- 
> A 9, B 9, C 11. 

Again, nonsense. You don't know who'd give an approval vote
to their 2nd choice. If what follows below is based on that bad
assumption, then it has no validlity.

If not, I don't understand what's being said there anyway.

> 
> The general case again for 3 candidates is
> N1 A1 + N8 CA > N2 B1 + N9 CB
> N2 B1 + N4 AB > N3 C1 + N5 AC
> N3 C1 + N6 BC > N1 A1 + N7 BA
> 
> The "1" after A, B and C means first choice votes.  The  N amounts (N1, N2,
> ..., N9) are numbers of votes.  Assume there are minimums on both sides of
> the greater than symbol (>) and assumes no ties.
> The left side or right side minimum tiebreaker amount also applies when there
> are 4 or more candidates in a circular tie.  If the right side minimum is
> used, then after a loser is dropped the data is looked at to see if there is
> a Condorcet winner among the remaining candidates.  If there is no Condorcet
> winner, then the next right side minimum loser is dropped. 
> -----

Now it sounds like you're propsing yet another method (How many
does that make?). As I always say at this point, you need to
do 2 things when proposing a new method:

1. Express its choice rule briefly in language that can be understood.

2. Tell why it's supposed to be better (And if it isn't supposed
   to be better, then why propose it?)


> I repeat part of a July 27, 1996 posting--
> Condorcet's Theory of Voting by H. P. Young, 82 American Political Science
> Review 1231 (Dec. 1988), contains on page 1233----
> 
> In Condorcet's lexicon, an "opinion" is a series of pairwise comparisons on
> the alternatives. Each pairwise combination is called a "proposition" and
> written a>b, etc. An opinion is said to be "impossible", "contradictory", or
> "absurd" of some of the propositions composing it form a cycle, such as a>b,
> b>c, c>a. Normally, each individual voter is able to rank all of the
> candidates in a consistent order. *** To break such cyclic majorities,
> Condorcet proposed the following method.
> -----
> [Condorcet's comments translated (by Prof. Young ?) from his Essai sur
> l'application de l'analyse a la probabilite des decisions rendues a la
> probabilite des voix (1785), pp. 125-126 ]
> 1. All possible opinions that do not imply a contradiction reduce to an
> indication of the order of merit that one judges among the candidates .....
> therefore for n candidates one would have n(n-1).... 2 possibilities.
> 2. Each voter having thus given his or her opinion by indicating the
> candidates' order of worth, if none compares them two by two, one will have
> in each opinion n(n-1)/2 propositions to consider separately. Taking the
> number of times that each is contained in the opinion of one the q voters,
> one will have the number of voices who are for each proposition.
> 3. One forms an opinion from those n(n-1)/2 propositions that agree with the
> most voices. If this opinion is among the n(n-1)... 2 possible opinions, one
> regards as elected the subject to whom this opinion accords the preference.
> If this opinion is among the 2 (to the exponent n(n-1)/2) minus n(n-1) .....
> 2 impossible opinions, then one successively deletes from that impossible
> opinion the propositions that have the least plurality, and one adopts the
> opinion from those that remain.
> -----
> Prof. Young noted that Condorcet unfortunately did not give any math example
> with 4 or more candidates.
> 
> My comments----In view of Condorcet's definition of a "proposition" (i.e. a x
> versus y pairing), if there is a cycle (i.e. tie- such as the above a>b, b>c,
> c>a), then did Condorcet mean to drop the pairing(s) in the cycle with the
> least pluralities (the one or more lowest x minus y, if x>y) to produce a
> result without a tie ?   Note Condorcet's use of *successively deletes* and
> *the least plurality*.

What Condorcet suggested amounts to electing the alternative
for which the "majority" against it is the least, according to
the translation in Duncan Black's _Theory of Committees & Voting_
(If I've gotten the book's title right).

As I've often said, Condorcet didn't specify how those "majorities"
should be compared. If the word "majority" is taken literally,
at least as the word is used now, then that goes along with
the claim that Condorcet, like pretty much everyone, assumed
that everyone would rank all of the alternatives.

Really, we could go over various translations for a long time.
If you think you can show that your translation means that
the rest of us have been wrong all along, and that Condorcet
didn't say what amounts to a suggestion to elect the
candidate with smallest pairwise defeat, then specifically
show it. 

In answer to your question, if that's what you were asking,
Condorcet didn't say how to measure a defeat when truncation
has occurred. Therefore any method that elects the least-defeated,
however they measure defeats, is "Condorcet's method". We've
already thoroughly compared Condorcet(EM) to other versions.
Surely you aren't suggesting that we resume that issue?

> 
> However, Prof. Young notes that if one or more pairings is dropped, then one
> or more candidates may be undominated or tied for first.
> 
> .-
> 


--