Single-candidate DMTBR idea
El sáb., 12 de mar. de 2022 4:45 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 3/12/22 6:58 PM, Forest Simmons wrote:
Kristofer,
45 ABC
35 BCA
25 CAB
Each of the A and B factions has more than a third of the votes.
Candidate A defeats B pairwise.
Almost every respectable method except TACC (as well as most
non-respectable methods) agree that candidate A should have the greatest
winning probability.
But some nagging doubt persists ... whence the Condorcet Cycle?
A general scalene triangle has a longest side, and the endpoints of that
side are further from each other than they are from the vertex V
opposite that side, which means that the V faction favorite cannot be
the rational, sincere last choice of any of the three factions.
And yet the voted ballots in our above three faction example give each
candidate a turn at last place.
Somebody's lowest preference is either mis-triangulated or
mis-represented.
I don't get what you mean here. Certainly it's possible for honest
Condorcet cycles to exist. Warren gave an example ...
Warren and I gave an example an of a set of four factions in a single plane
that induces a cycle geometrically. This cannot happen with only three
factions ... if there is a cycle, then the preferences are inconsistent
with the lengths of the sides of the triangle.
A three dimensional issue space can give rise to a cycle but those
preferences are not based on metrics/distances between factions.
of candidates
evaluated on three issues, say corruption, domestic policy, and foreign
policy. Each faction cares primarily about one dimension, and the
candidates have positions on each issue that leads to a cyclical
majority (e.g. candidate A is incorruptible, has awful domestic policy,
and okayish foreign policy).
Honest cycles also exist in 2D spatial models, e.g. Poundstone's example
https://www.rangevoting.org/PoundstoneCondCyc.png.
But not when limited to 3 ballot factions.
If the cycle is false, though, then any faction could have done the
burial. You say the A faction is the only group that has anything to win
by conducting the burial, so they must have done it. However, there's a
bit of battle-of-wits logic here. Suppose that the method did elect C by
this logic. Then it's possible that the C voters, knowing this,
engineered the cycle (honest is C>B>A) in order to push their winner
from B (their second choice) to C (their first).
The burial of C by A isn't the only possibility ... just the most likely to
succeed ... so the one most in need of checking.
If the check confirms B>C as sincere, then B wins, so the burial of B did
not pay off for C.
Okay, so C can't win
because of second-order reasoning.
C can and will win if it was buried by A, because it is a finalist in the B
vs C honesty check.
Don't make this too hypothetical. The proposed implementation is tweaked
DMC:
The Implicit Approval order is A>B>C. The nominal DMC winner W is A,
because A is not pairwise defeated by any candidate with greater implicit
approval. The other finalist is C, the candidate that would be the DMC
winner if W did not defeat it pairwise.
The tweaked method elects the sincere winner between W=A and C, which is C
if C was the sincere CW, else A, the DMC winner with ballots taken at face
value.
The tweaked method changes the DMC winner only if it detects and elects a
better DMC winner (a sincere CW in this three faction example) while
exposing an insincere order reversal.
Of course, this tweaked version of DMC is not the only possibility for
exploiting the potential for a sincere, binding, binary choice between two
finalists.
I already suggested a tweaked version of TACC that makes use of the same
device.
Do any other applications come to mind?
-Forest
And A can't win because of
first-order reasoning. So B must win, right? But then it's possible that
the B faction knew this (honest: B>A>C) and buried A to make B win.
So my point would be that since there's a Condorcet cycle, any Condorcet
method (no matter who wins) will be open to burial. One could argue that
the sensible methods do the right thing and elect the candidate whose
defector coalition has to be the largest for this to be a successful
burial: a method that elects A is fooled by a faction of 45 voters
executing burial, but if the method were to elect C, it could be fooled
by a faction of 25 voters, which is worse.
In a way, that's what DMTBR says: there's no way for a Condorcet method
to be absolutely immune to burial, so the best thing we can do is to
make some set of candidates immune to being buried by candidates outside
of that set, and then try to make that set as small as possible. And I
suspect that 1/3 is the best possible...
At least without doing something clever with UD or repeated balloting.
I'm not sure how a second ballot question would help, because there's no
reason for an A>B>C burier to not also "bury" by indicating B>C where
honest is C>B... so I may be missing something. Duple rules (like Random
Pair) are IIRC only strategy-proof if the pair is decided independently
of the voters' input.
In the vein of DSV, imagine that I take some Condorcet method plus top
two and make the DSV procedure fill in the second ballot information so
that it's consistent with (or strategically advantageous given) the
first ballot ranking. Then either the combined method is not Condorcet
(and it's not surprising that it would resist burial better), or it's
subject to the same limitations as above, I would think...
I would guess the answer is that the combined method isn't Condorcet,
because there would be a tension between burying the honest CW so that
the second round consists of your favored candidate and someone who's
going to lose - and burying too far which means that someone intolerable
wins the second round. Perhaps most UD solutions are like Approval:
there may be a Nash (or core) equilibrium around the honest CW, but the
setting benefits whoever has got the most complete information, and the
potential backfire can get very unpleasant indeed.
-km
El sáb., 12 de mar. de 2022 10:02 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
El sáb., 12 de mar. de 2022 4:45 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 3/12/22 6:58 PM, Forest Simmons wrote:
Kristofer,
45 ABC
35 BCA
25 CAB
Each of the A and B factions has more than a third of the votes.
Candidate A defeats B pairwise.
Almost every respectable method except TACC (as well as most
non-respectable methods) agree that candidate A should have the
greatest
winning probability.
But some nagging doubt persists ... whence the Condorcet Cycle?
A general scalene triangle has a longest side, and the endpoints of
that
side are further from each other than they are from the vertex V
opposite that side, which means that the V faction favorite cannot be
the rational, sincere last choice of any of the three factions.
And yet the voted ballots in our above three faction example give each
candidate a turn at last place.
Somebody's lowest preference is either mis-triangulated or
mis-represented.
I don't get what you mean here. Certainly it's possible for honest
Condorcet cycles to exist. Warren gave an example ...
Warren and I gave an example an of a set of four factions in a single
plane that induces a cycle geometrically. This cannot happen with only
three factions ... if there is a cycle, then the preferences are
inconsistent with the lengths of the sides of the triangle.
A three dimensional issue space can give rise to a cycle but those
preferences are not based on metrics/distances between factions.
of candidates
evaluated on three issues, say corruption, domestic policy, and foreign
policy. Each faction cares primarily about one dimension, and the
candidates have positions on each issue that leads to a cyclical
majority (e.g. candidate A is incorruptible, has awful domestic policy,
and okayish foreign policy).
Honest cycles also exist in 2D spatial models, e.g. Poundstone's example
https://www.rangevoting.org/PoundstoneCondCyc.png.
But not when limited to 3 ballot factions.
If the cycle is false, though, then any faction could have done the
burial. You say the A faction is the only group that has anything to win
by conducting the burial, so they must have done it. However, there's a
bit of battle-of-wits logic here. Suppose that the method did elect C by
this logic. Then it's possible that the C voters, knowing this,
engineered the cycle (honest is C>B>A) in order to push their winner
from B (their second choice) to C (their first).
The burial of C by A isn't the only possibility ... just the most likely
to succeed ... so the one most in need of checking.
If the check confirms B>C as sincere, then B wins, so the burial of B did
not pay off for C.
Okay, so C can't win
because of second-order reasoning.
C can and will win if it was buried by A, because it is a finalist in the
B vs C honesty check.
Don't make this too hypothetical. The proposed implementation is tweaked
DMC:
The Implicit Approval order is A>B>C.
Whoops ... actually B>A>C is the IA order.
The nominal DMC winner W is A,
Not A, but B=W.
because A
[actually B]
is not pairwise defeated by any candidate with greater implicit approval.
The other finalist is C, the candidate that would be the DMC winner if W
did not defeat it pairwise.
The tweaked method elects the sincere winner between W and C, which is C
if C was the sincere CW, else A, the DMC winner with ballots taken at face
value.
The tweaked method changes the DMC winner only if it detects and elects a
better DMC winner (a sincere CW in this three faction example) while
exposing an insincere order reversal.
Of course, this tweaked version of DMC is not the only possibility for
exploiting the potential for a sincere, binding, binary choice between two
finalists.
I already suggested a tweaked version of TACC that makes use of the same
device.
Do any other applications come to mind?
-Forest
And A can't win because of
first-order reasoning. So B must win, right? But then it's possible that
the B faction knew this (honest: B>A>C) and buried A to make B win.
So my point would be that since there's a Condorcet cycle, any Condorcet
method (no matter who wins) will be open to burial. One could argue that
the sensible methods do the right thing and elect the candidate whose
defector coalition has to be the largest for this to be a successful
burial: a method that elects A is fooled by a faction of 45 voters
executing burial, but if the method were to elect C, it could be fooled
by a faction of 25 voters, which is worse.
In a way, that's what DMTBR says: there's no way for a Condorcet method
to be absolutely immune to burial, so the best thing we can do is to
make some set of candidates immune to being buried by candidates outside
of that set, and then try to make that set as small as possible. And I
suspect that 1/3 is the best possible...
At least without doing something clever with UD or repeated balloting.
I'm not sure how a second ballot question would help, because there's no
reason for an A>B>C burier to not also "bury" by indicating B>C where
honest is C>B... so I may be missing something. Duple rules (like Random
Pair) are IIRC only strategy-proof if the pair is decided independently
of the voters' input.
In the vein of DSV, imagine that I take some Condorcet method plus top
two and make the DSV procedure fill in the second ballot information so
that it's consistent with (or strategically advantageous given) the
first ballot ranking. Then either the combined method is not Condorcet
(and it's not surprising that it would resist burial better), or it's
subject to the same limitations as above, I would think...
I would guess the answer is that the combined method isn't Condorcet,
because there would be a tension between burying the honest CW so that
the second round consists of your favored candidate and someone who's
going to lose - and burying too far which means that someone intolerable
wins the second round. Perhaps most UD solutions are like Approval:
there may be a Nash (or core) equilibrium around the honest CW, but the
setting benefits whoever has got the most complete information, and the
potential backfire can get very unpleasant indeed.
-km
On 13.03.2022 07:02, Forest Simmons wrote:
Warren and I gave an example an of a set of four factions in a single
plane that induces a cycle geometrically. This cannot happen with only
three factions ... if there is a cycle, then the preferences are
inconsistent with the lengths of the sides of the triangle.
A three dimensional issue space can give rise to a cycle but those
preferences are not based on metrics/distances between factions.
I think you're using a more constrained model than me. Here's a spatial
model example with three candidates and three faction centers that
should work (unless I've miscalculated):
Let A be at (0.85, 0.8), B at (0.2, 0.8), and C at (0.5, 0.4).
Then let voter faction 1 be at (1.1, 1.5). The distance to A is ~0.75.
The distance to B is ~1.14. The distance to C is ~1.25. So A>B>C.
Let voter faction 2 be at (-0.65, 0.9). The distance to B is ~0.86. The
distance to C is ~1.25. The distance to A is ~1.50. So B>C>A.
Let voter faction 3 be at (0.8, -0.1). The distance to C is ~0.58. The
distance to A is ~0.90. The distance to B is ~1.08. So C>A>B.
See also https://www.rangevoting.org/BlackSingle.html. Quoting:
Myth #2 busted: It also is not the case that, in two dimensions with
voters who prefer candidates closer to them in L2 (or L1) distance, a
Condorcet winner necessarily exists at all.
The page then proceeds to give an example with three voters (faction
centers in your example) for the Manhattan distance.
But my point is that even if it were impossible to create such examples
in the plane, then we couldn't as a consequence assume thet the voters
aren't voting in some more generalized space where it is possible.
That would be like saying that just because a cycle can't exist with
single-peaked preferences on a line, then all Condorcet cycles must be
the result of strategy.
In addition, there may be differential constraints that imply that if
33: A>B>C
33: B>C>A
33: C>A>B
is a perfect tie, then increasing the support for A and decreasing the
support for C should not make C win. It may be (though I haven't proven
so) that making the method elect C will induce some kind of
monotonicity failure as a consequence.
(From a spatial model, there's a similar argument: suppose we have 1/3
of the voters at each point, then we move some voters from the point
closest to C to the point closest to A. Changing the initial tie into a
C win as a result seems somewhat strange.)
So a choice to elect C would have to be about strategy resistance rather
than honest performance.
If the cycle is false, though, then any faction could have done the
burial. You say the A faction is the only group that has anything to
win
by conducting the burial, so they must have done it. However, there's a
bit of battle-of-wits logic here. Suppose that the method did elect
C by
this logic. Then it's possible that the C voters, knowing this,
engineered the cycle (honest is C>B>A) in order to push their winner
from B (their second choice) to C (their first).
The burial of C by A isn't the only possibility ... just the most likely
to succeed ... so the one most in need of checking.
If the check confirms B>C as sincere, then B wins, so the burial of B
did not pay off for C.
Okay, so C can't win
because of second-order reasoning.
C can and will win if it was buried by A, because it is a finalist in
the B vs C honesty check.
I was making a point about unrestricted domain. If the method passes
unrestricted domain, then there's no such check, and you can't determine
who the burying coalition is. Because it's a simple Condorcet cycle, no
matter who the winner ends up being, there could have been a burying
coalition who benefitted.
What threw me, I think, is that you were arguing that C should be
elected the winner of the single-winner ranked voting method, based on
the given ballots.
But on rereading, it seems that you're proposing that there's an
(essentially automated) top-two runoff after the ordinary ranked method
is finished, between the winner and the runner-up; and that the ranked
method should elect C so that C is one of the two finalists in the
second round, thus deterring first-round burial.
In that case, the first-round method is not trying to figure out who the
single winner of the election should be, but rather which set of two
candidates will be the finalists for the second round - for either a
manual runoff, or an automated one.
And I would suspect that a single-winner (ranked, UD) method has a
different objective than one that's intended to pick finalists for a
runoff. A runoff finalist method can prioritize covering all the bases
while a single-winner method is constrained to that the winner must be a
reasonable compromise, and that the runner-up has these properties in
case the winner is ineligible.
A note about the automatic runoff mechanism: it's interesting in that it
has different properties than Approval/Range, as far as breaking UD
goes. It shows that not all UD-failing methods have to be ambiguous (the
way ratings are) or lead to inherent instrumental voting even by honest
voters. And that, in turn, means that not all extensions beyond UD are
susceptible to the same problems as rated voting.[1]
I would think manual runoff is better, though, for two reasons. First,
it's a very subtle point that makes the honesty for the runoff ballot
incentive compatible, and ordinary voters might not get why it's safe to
be honest.
Second, having separate rounds makes it possible to have finalist
debates to examine their differences in more detail. The voters can then
change their minds between rounds based on what they see - e.g. a weak
CW could be exposed as a weak CW between the rounds if the runner-up's
platform withstands greater scrutiny. This might be part of the reason
why plurality + top two doesn't degrade into two-party rule the way
ordinary plurality, contingent vote, and IRV does. (Then again, the
necessary honest voting in the final round might be what makes TTR
better than IRV.)
Since the three-candidate method with a runoff is comparable to a "loser
election" (where the loser is disqualified), it should be possible to
rewrite my minimal manipulability finder to construct a minimally
manipulable method in this context. It would take some work, however.
-km
[1] Maybe the relevant distinction is whether the proposed
extension/breaking of UD would allow the method to pass IIA. If it does,
then the blurring of context and method may follow, as it does in
Range (e.g. IIA no longer implying that the winner doesn't change if
candidates enter or exit). But this is just an idea.
El mar., 15 de mar. de 2022 5:19 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 13.03.2022 07:02, Forest Simmons wrote:
Warren and I gave an example an of a set of four factions in a single
plane that induces a cycle geometrically. This cannot happen with only
three factions ... if there is a cycle, then the preferences are
inconsistent with the lengths of the sides of the triangle.
A three dimensional issue space can give rise to a cycle but those
preferences are not based on metrics/distances between factions.
I think you're using a more constrained model than me.
Yes, in your example, apparently the candidates are not voters, or if they
are, then there are six faction locations, not three.
My assumption is that the geometric proximity of the three factions (fo
each other) determines their preferences.
Here's a spatial
model example with three candidates and three faction centers that
should work (unless I've miscalculated):
Let A be at (0.85, 0.8), B at (0.2, 0.8), and C at (0.5, 0.4).
Then let voter faction 1 be at (1.1, 1.5). The distance to A is ~0.75.
The distance to B is ~1.14. The distance to C is ~1.25. So A>B>C.
Let voter faction 2 be at (-0.65, 0.9). The distance to B is ~0.86. The
distance to C is ~1.25. The distance to A is ~1.50. So B>C>A.
Let voter faction 3 be at (0.8, -0.1). The distance to C is ~0.58. The
distance to A is ~0.90. The distance to B is ~1.08. So C>A>B.
See also https://www.rangevoting.org/BlackSingle.html. Quoting:
Myth #2 busted: It also is not the case that, in two dimensions with
voters who prefer candidates closer to them in L2 (or L1) distance, a
Condorcet winner necessarily exists at all.
The page then proceeds to give an example with three voters (faction
centers in your example) for the Manhattan distance.
But my point is that even if it were impossible to create such examples
in the plane, then we couldn't as a consequence assume thet the voters
aren't voting in some more generalized space where it is possible.
That would be like saying that just because a cycle can't exist with
single-peaked preferences on a line, then all Condorcet cycles must be
the result of strategy.
In addition, there may be differential constraints that imply that if
33: A>B>C
33: B>C>A
33: C>A>B
is a perfect tie, then increasing the support for A and decreasing the
support for C should not make C win. It may be (though I haven't proven
so) that making the method elect C will induce some kind of
monotonicity failure as a consequence.
(From a spatial model, there's a similar argument: suppose we have 1/3
of the voters at each point, then we move some voters from the point
closest to C to the point closest to A. Changing the initial tie into a
C win as a result seems somewhat strange.)
So a choice to elect C would have to be about strategy resistance rather
than honest performance.
If the cycle is false, though, then any faction could have done the
burial. You say the A faction is the only group that has anything to
win
by conducting the burial, so they must have done it. However,
there's a
bit of battle-of-wits logic here. Suppose that the method did elect
C by
this logic. Then it's possible that the C voters, knowing this,
engineered the cycle (honest is C>B>A) in order to push their winner
from B (their second choice) to C (their first).
The burial of C by A isn't the only possibility ... just the most likely
to succeed ... so the one most in need of checking.
If the check confirms B>C as sincere, then B wins, so the burial of B
did not pay off for C.
Okay, so C can't win
because of second-order reasoning.
C can and will win if it was buried by A, because it is a finalist in
the B vs C honesty check.
I was making a point about unrestricted domain. If the method passes
unrestricted domain, then there's no such check, and you can't determine
who the burying coalition is. Because it's a simple Condorcet cycle, no
matter who the winner ends up being, there could have been a burying
coalition who benefitted.
What threw me, I think, is that you were arguing that C should be
elected the winner of the single-winner ranked voting method, based on
the given ballots.
But on rereading, it seems that you're proposing that there's an
(essentially automated) top-two runoff after the ordinary ranked method
is finished, between the winner and the runner-up; and that the ranked
method should elect C so that C is one of the two finalists in the
second round, thus deterring first-round burial.
In that case, the first-round method is not trying to figure out who the
single winner of the election should be, but rather which set of two
candidates will be the finalists for the second round - for either a
manual runoff, or an automated one.
And I would suspect that a single-winner (ranked, UD) method has a
different objective than one that's intended to pick finalists for a
runoff. A runoff finalist method can prioritize covering all the bases
while a single-winner method is constrained to that the winner must be a
reasonable compromise, and that the runner-up has these properties in
case the winner is ineligible.
A note about the automatic runoff mechanism: it's interesting in that it
has different properties than Approval/Range, as far as breaking UD
goes. It shows that not all UD-failing methods have to be ambiguous (the
way ratings are) or lead to inherent instrumental voting even by honest
voters. And that, in turn, means that not all extensions beyond UD are
susceptible to the same problems as rated voting.[1]
I would think manual runoff is better, though, for two reasons. First,
it's a very subtle point that makes the honesty for the runoff ballot
incentive compatible, and ordinary voters might not get why it's safe to
be honest.
Second, having separate rounds makes it possible to have finalist
debates to examine their differences in more detail. The voters can then
change their minds between rounds based on what they see - e.g. a weak
CW could be exposed as a weak CW between the rounds if the runner-up's
platform withstands greater scrutiny. This might be part of the reason
why plurality + top two doesn't degrade into two-party rule the way
ordinary plurality, contingent vote, and IRV does. (Then again, the
necessary honest voting in the final round might be what makes TTR
better than IRV.)
Since the three-candidate method with a runoff is comparable to a "loser
election" (where the loser is disqualified), it should be possible to
rewrite my minimal manipulability finder to construct a minimally
manipulable method in this context. It would take some work, however.
-km
[1] Maybe the relevant distinction is whether the proposed
extension/breaking of UD would allow the method to pass IIA. If it does,
then the blurring of context and method may follow, as it does in
Range (e.g. IIA no longer implying that the winner doesn't change if
candidates enter or exit). But this is just an idea.