smith/schwartz/landau
Condorcet methods can work also without measuring preference strengths. All fine as long as we have a Condorcet Winner (this is the most common outcome). After that, finding the correct winner becomes difficult. Maybe ties and lotteries are ok then.
BR, Juho
On 28 Mar 2018, at 21:08, Curt accounts@museworld.com wrote:
On Mar 28, 2018, at 1:26 AM, Juho Laatu <juho.laatu@gmail.com mailto:juho.laatu@gmail.com> wrote:
If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.
I view a 66:33 Smith Cycle as having no additional useful information over a 60:40 Smith Cycle or a 50:49 Smith Cycle. The only way to see a 66:33 Smith Cycle as being “stronger” than the other cycles is to imagine that it corresponds to a strength of preference or depth of division in the electorate, when there is no data to support that.
(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)
I would argue that they actually don’t behave that way, and should not be looked at this way. It assumes probability, from an imagined distribution that is not that predictable in reality. Electorate preference is not evenly distributed. It clusters all over the place in ways that are not predictable. There is nothing about a 66:33 cycle that makes “d” a more appropriate winner than a 60:40 or 50:49 cycle. To argue otherwise is to imagine a correspondence with utility.
Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.
Incidentally, this does point to what I believe is a major downside of a Condorcet method - it’s by definition entirely unsuitable for figuring proportional representation, because that again would imbuing vote margins with utility concepts.
Election-Methods mailing list - see http://electorama.com/em for list info
---------------------------- Original Message ----------------------------
Subject: Re: [EM] smith/schwartz/landau
From: "Curt" accounts@museworld.com
Date: Wed, March 28, 2018 1:08 pm
To: "EM" election-methods@lists.electorama.com
Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.
well, we have an ancillary concept in elections that we sometimes call "mandate".� seems to me
that margin has something to do with an election winner's mandate.
Incidentally, this does point to what I believe is a major downside of a Condorcet method - it’s by definition entirely unsuitable for figuring proportional representation, ...
are we discussing single winner
elections here or multi-winner elections?
... because that again would imbuing vote margins with utility concepts.
if we accept the premise of One-Person-One -Vote, which assigns equal utility to each voter's preference, seems to me that margins are directly related to overall
utility for the voting population.
--
r b-j� � � � � � � � � � � � �rbj@audioimagination.com
"Imagination is more important than knowledge."
�
�
�
�
On 03/28/2018 08:08 PM, Curt wrote:
Incidentally, this does point to what I believe is a major downside of
a Condorcet method - it’s by definition entirely unsuitable for figuring
proportional representation, because that again would imbuing vote
margins with utility concepts.
The Condorcet property (or even the majority property) is incompatible
with proportional representation. The usual example is something like:
51: X1>X2>X3>X4>X5>X6>X7
49: Y1>Y2>Y3>Y4>Y5>Y6>Y7
Suppose we want to elect four winners. The proportional outcome is to
have two Xes and two Ys, probably X1 X2 Y1 Y2. But applying the majority
criterion forces the election of X1, and after X1 is out of the picture,
it forces the election of X2, and so on until four X-es have been elected.
Majority and Condorcet simply are the wrong tools for the job of PR.
That said, there are methods that pass Condorcet when there's only one
candidate to elect, and pass the Droop proportionality criterion when
there's more than one to elect. One example is CPO-STV, and another is
Schulze STV.
Another proportional method that uses a Condorcet method (Kemeny) as its
foundation is VoteFair ranking.
Specifically VoteFair representation ranking handles two winners per
district, and VoteFair partial proportional ranking handles the filling
of (extra) nationwide (or statewide) seats.
The "partial" aspect can be adjusted to become fully proportional, just
by adding more nationwide seats.
Richard Fobes
On 3/28/2018 1:47 PM, Kristofer Munsterhjelm wrote:
On 03/28/2018 08:08 PM, Curt wrote:
Incidentally, this does point to what I believe is a major downside
of a Condorcet method - it’s by definition entirely unsuitable for
figuring proportional representation, because that again would imbuing
vote margins with utility concepts.
The Condorcet property (or even the majority property) is incompatible
with proportional representation. The usual example is something like:
51: X1>X2>X3>X4>X5>X6>X7
49: Y1>Y2>Y3>Y4>Y5>Y6>Y7
Suppose we want to elect four winners. The proportional outcome is to
have two Xes and two Ys, probably X1 X2 Y1 Y2. But applying the majority
criterion forces the election of X1, and after X1 is out of the picture,
it forces the election of X2, and so on until four X-es have been elected.
Majority and Condorcet simply are the wrong tools for the job of PR.
That said, there are methods that pass Condorcet when there's only one
candidate to elect, and pass the Droop proportionality criterion when
there's more than one to elect. One example is CPO-STV, and another is
Schulze STV.
Election-Methods mailing list - see http://electorama.com/em for list info
On 03/28/2018 04:36 PM, Juho Laatu wrote:
On 28 Mar 2018, at 14:05, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
But it's not that implausible; and if it's true, that means that
whatever makes a Condorcet loser deserve to win, if anything, must
come from information not provided by the pairwise matrix.
Note that the problems between matrix and ballot information mainly
emerge from the clone criterion. One could say that pure clone
independence / existence of clone candidates can not be measured from
the matrix only (without doing the "overkill"). The non Smith Set
arguments are more neutral (e.g. minmax style arguments) with respect to
using the matrix only vs also the ballots (matrix is enough). The
"overkill" is the problem that forces d not to be elected also when
there are no technical clones. (Smith Set criterion is close to the
clone independence criterion.)
It's not just clones. Since the example's collapsed ballots are of the form
m: A>d
n: d>A
with m>n, majority implies that A should be elected. In the uncollapsed
example, that means that the set {A, B, C} is first on a majority of the
ballots, so any method that passes mutual majority must elect from this set.
That's perhaps a stronger example of how a generalization of majority
forces one of {A, B, C} to be elected, since the point of mutual
majority (as I see it, at least) is that a majority can get the
candidate they want to be elected, elected, without having to coordinate
beforehand to rank the candidates in the same order.
It's hard to be opposed to such a property. I imagine it's easier to say
"okay, the pairwise matrix doesn't supply enough information" and
require that the method use more than just the pairwise matrix to decide
the winner.
On 02 Apr 2018, at 15:13, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
On 03/28/2018 04:36 PM, Juho Laatu wrote:
On 28 Mar 2018, at 14:05, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.
Note that the problems between matrix and ballot information mainly
emerge from the clone criterion. One could say that pure clone
independence / existence of clone candidates can not be measured from
the matrix only (without doing the "overkill"). The non Smith Set
arguments are more neutral (e.g. minmax style arguments) with respect to
using the matrix only vs also the ballots (matrix is enough). The
"overkill" is the problem that forces d not to be elected also when
there are no technical clones. (Smith Set criterion is close to the
clone independence criterion.)
It's not just clones. Since the example's collapsed ballots are of the form
m: A>d
n: d>A
with m>n, majority implies that A should be elected. In the uncollapsed example, that means that the set {A, B, C} is first on a majority of the ballots, so any method that passes mutual majority must elect from this set.
That's perhaps a stronger example of how a generalization of majority forces one of {A, B, C} to be elected, since the point of mutual majority (as I see it, at least) is that a majority can get the candidate they want to be elected, elected, without having to coordinate beforehand to rank the candidates in the same order.
It's hard to be opposed to such a property. I imagine it's easier to say "okay, the pairwise matrix doesn't supply enough information" and require that the method use more than just the pairwise matrix to decide the winner.
In the case that we have mutual majority the mutual majority candidates could nominate only one of them, and that candidate would win (if voter preferences would stay the same). But if they all run, then there may be also defeats among them, and those defeats could be considered worse than the defeats of some candidates outside that set. The alternative strategy is to rank A, B and C in the same order. If all ABC supporters implement this strategy (100%), they could break the ABC loop that other voters maybe generated.
Having the (theoretical) possibility to implement a working strategy should not be seen as a right to win without implementing that strategy. Implementing the first strategy could be difficult since voters may not understand if some of the competing candidates would withdraw. In the second strategy many voters might not follow the recommended strategy. In both cases it could be difficult to know if the strategic opportunity exists, and all three candidates to agree that all of them will benefit of this strategy. If d seems to have only minority in any case, then why let one of the other mutual majority candidates win for free (assuming that these three are from competing parties).
You already mentioned the problem that If one wants mutual majority without doing an overkill, then one must use the ballots instead of the matrix. In practice one would however probably use the matrix and a Smith Set compatible method (i.e. exactly measured mutual majority methods may not be very practical).
We know that some sincere winner or strategy related criteria will be violated in any case when preferences are circular. If one likes methods that may elect outside the Smith Set, then making a matrix based method compatible with mutual majority criterion probably violates the "outside the Smith Set target". If you want both and you can't, then you abandon one of them, or (maybe preferably) violate both of them but only lightly. (Just noting these facts.)
The mutual majority ballots could be as follows.
50: circular_mix_of_A_B_C > d > e > f
49: d > e > f > circular_mix_of_A_B_C
The idea is just to show that d could still be quite popular despite of the mutual majority. If one thinks that sometimes candidates outside the Smith Set could win, this might be one of those cases (despite of the mutual majority). Note that candidates d, e and f are not very far from having mutual majority, and among them it is clear that d is the best.
In this example the majority related key facts (of sincere opinions) are. 1) d loses marginally to A, B and C, 2) ABC is marginally better in mutual majority measurements than def, 3) A, B and C lose quite a lot, each to one of the others. Is the marginal mutual majority result more important than the strong (majority) defeats of A, B and C?
BR, Juho