On 3/11/22 12:33 AM, Kristofer Munsterhjelm wrote:

I was thinking, perhaps this isn't monotone after all. The usual
nonmonotonicity problem (e.g. in IRV) has the pattern that you're in an
ABCA cycle, then some BAC voters rank A higher so that B drops below C
in first preferences, and then A is defeated by C in the second round.

But suppose A and B both have > 1/3 first preferences. Then upranking A
on a BAC ballot has no effect if B stays above 1/3 first preferences,
and if it pushes B below the threshold, then A wins outright as the sole
candidate with more than 1/3 fpp. Upranking A can't give C more first
preferences, and three candidates can't all have more than 1/3 of the
first preferences.

So is monotonicity preserved after all? If so, that's a nice trick!

Does it also apply to the generalization where you just take the two
candidates with the most first preferences? I'm not sure.

-km

Hi Kristofer,

In the three-candidate case, electing the pairwise winner between the top two
candidates is basically IRV.

Without the 1/3 limit it could happen that the FPW gets more votes and changes
who the second place candidate is. He might not beat the new one.

This is interesting though. The "obvious" way to expand IFPP to many candidates
is to eliminate candidates with a below-average vote count. But it seems like
the 1/3 rule was the important thing, as it's what enforces that always either
one or two candidates are eligible to win, and these candidates can't be harmed
by getting more votes.

Tricky, to reduce a scenario to this state without breaking mono-raise.

Kevin

On 3/11/22 6:04 AM, Kevin Venzke wrote:

Yes, that also explains where the "third" in dominant mutual third comes
from. Like with Droop proportionatliy, it's the smallest quota so that
only two candidates can exceed it. And that would also suggest that (at
least by this approach), third is the best we can do; there's no, say,
dominant mutual quarter for Condorcet.

We could of course just stitch something together, e.g. if there're
exactly two candidates above 1/3 fpp, elect the candidate who pairwise
beats the other, otherwise just elect the Plurality winner. This should
be monotone because raising A doesn't harm A when A has >1/3 fpp, and
raising B to >1/3 fpp gives him a second chance against A (if B beats A
pairwise).

It's not very elegant: the seams are very obvious. But perhaps elegance
can come later... or perhaps it will be induced by turning DMT candidate
BR into full DMTBR.

-km

On 3/11/22 6:04 AM, Kevin Venzke wrote:

Actually, now that I think about it, I think the rule I provided
combined with Condorcet is fpA-fpC (disregarding ties for now).

Suppose we have an ABCA cycle without pairwise ties. Since there are
three candidates, at least one of them must have more than 1/3 fpp,
unless there's an exact tie for first preferences. And in that case, the
election is a perfect tie too, because, since we have an ABCA cycle, the
ballots must be

x: A>B>C
x: B>C>A
x: C>A>B.

If there's only one candidate with more than 1/3 fpp, then that
candidate wins (both with the given rule and in fpA-fpC). So suppose the
candidates with more than 1/3 fpp are A and B (without loss of
generality; you can always go along the cycle to make it that way).

Then the rule given will elect A since he beats B pairwise. For fpA-fpC,
A's score is fpA-fpC and B's score is fpB - fpA. So let's try to
maximize B's score to produce an example where fpA-fpC and the rule
don't agree.

The margin between these is fpB - fpA - (fpA - fpC) = fpB - 2 fpA + fpC.
So to maximize the margin in favor of B, we should minimize A's number
of first preferences, and the distribution of first preferences between
B and C doesn't matter. So let A's fpp be 1/3 + epsilon. And for the
sake of convenience, set fpC = 0, so that B's first preference count is
2/3 - eps.

The margin in favor of B is then 2/3 - eps - 2/3 - 2 eps. For epsilon=0,
this is 0, but for any positive epsilon, the margin is negative and A
wins. Which was what was wanted.

(It's most likely impossible for an ABCA cycle to exist if B has 2/3 of
the first preferences, so the argument is generous to B.)

-km

Hi Kristofer,

Yes, I realized after posting that that would probably work, and was racing to
prove it. So you can have any number of candidates, but you can only have a
runoff if two candidates each have 1/3+ FPs.

This means we have a 4+ candidate method that satisfies Mono-raise and both LNH,
and is not just FPP. So this should be an answer for Craig Carey here. I wonder
if he was aware of it, or what he would have thought about it.

I think so, yes.

I tried to compare C//(this IFPP expansion) with fpA-max(fpC). Note that the way
I define the latter doesn't block a Condorcet loser from winning. But it seems
like this method is better than using the IFPP expansion, as the overall burial
incentive is less.

Both methods appear to satisfy Mono-raise, Plurality, and DMTCBR.

Incidentally, do you know if there is a DMTBR (not just DMTCBR) satisfaction
proof known for any method? Some form of C//IRV maybe, but it doesn't seem like
an easy thing to show.

Yes. I've been sitting here trying to get a 25% rule to "work" (defining that
very generously), but there is a lot of trouble regulating who is allowed to
benefit from various vote changes when three candidates are eligible.

Kevin

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.

Suppose the smallest faction,  25CAB, to be the inaccurate one ... with
true preferences 25CBA. Then B would be the true CW, and they would be
kicking themselves for inadvertently reversing their B>A preference for
their ballots.

On the other hand, suppose the suspicious burial arose from a CB to BC swap
in the largest faction.  Then that faction would be congratulating itself
for its good fortune in converting their favorite A into a winner.

So, in the likely case that the ABCA ballot cycle was created artificially,
the only faction that actually improved its outcome by the burial is the
most likely culprit.

So A's win very likely was achieved by burial of C, which in turn, is the
likely true CW.

So, is this enough evidence to convict A and elect C?

No. This is like a Columbo episode where Columbo knows "who done it", on
the basis of compelling logic based on circumstantial evidence, but the
criminal is still taunting him for not having the kind of proof that will
hold up in court.

So what test would give the compelling evidence?

Only checking the B>C pairwise defeat has a chance of definitively
("dispositively") settling the case.

How can we check that supposed defeat without exhuming the victim's body
from the grave? (The Columbo equivalent of going back to the polls to check
the head-to-bead result between B and C.)

That's why we allow each voter an optional second ballot (paired with their
first) to be used in case (and only in case) a final binding, two-finalist
runoff is needed for definitive disposition of the election.

How would this work if the basic method were DMC (which nominally elects
the lowest implicit approval candidate that pairwise defeats every
candidate with greater IA)?

The two finalists are the nominal DMC winner W, and the candidate X that
would be the DMC winner if it pairwise defeated W, i.e. the candidate X
that defeats every candidate with greater IA, except possibly W itself (if
there is such an X).

It seems to me that this tweak of DMC would make it more resistant to
burial and chicken than any of the methods that elect A in our opening
example above ... better than any of the acceptable methods except TACC,
and even better than TACC because instead of merely punishing the burial of
C, it leads to vindication and election of C.

Thoughts?

-Forest

El vie., 11 de mar. de 2022 2:20 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:

Consider the ballot profile ...

p A>B=C
q B>A=C
r  C>B>A

Each ballot in the first faction has a Kendall-tau distance of 1.5 from
each ballot in the second faction and 2.5 from each ballot in the third
faction, while each ballot in the second faction has a distance of 1.5 from
each ballot in the third faction.

So geometrically the three factions form an isosceles triangle with the
longest side opposite the second faction.

So geometric proximity/affinity would require that candidate B be a clear
second choice in the first and third factions.

So the  geometrically corrected (consistent) factions would have to be ...

p A>B
q B
r C>B

In that case B would have to be the CW if neither A nor B was a majority
winner.

In short, a Condorcet cycle would be inconsistent with the geometric
affinities.

Let's look at Kemeny-Young:

The respective total distances from the three original factions to the
respective six finish orders would be ....

p/2+1.5q+3r to ABC,

p/2+2q+2r to ACB,

1.5p+g/2+2r to BAC,

2.5p+q/2+r to BCA,

1.5p+2.5q+r to CAB

1.5p+1.5q  to CBA

Suppose, for example p=q=30, and r=40.

Then the respective total costs (distances) would be [check my
arithmetic]...
180 for ABC,
155 for ACB,
140 for BAC,
130 for BCA,
160 for CAB, and
90 for CBA.

So in this case CBA would be the K-Y finish order, contrary to the
geometric affinities implied by the Kendall-tau geometry ...

30 A>B
30 B
40 C>B,

which would make B the Condorcet Winner, and BCA the fish order.

So ordinary K-Y, which is based on the Kendall-tau metric, yields a result
inconsistent with preferences based on that same Kendal-tau distance
function.

That's one reason I reject K-Y ... the other reason is that the Kendall-tau
metric is clone dependent.

Those are the reasons I have suggested a de-cloned version of Kendall-tau
together with some geometrically consistent applications of that new metric
(other than de-cloned K-Y).

-Forest

El sáb., 12 de mar. de 2022 9:58 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:

On 3/12/22 12:15 PM, Kevin Venzke wrote:

Could you try to do Smith,(method) for both of these? That may be closer
to comparing apples to apples - it should keep the Condorcet loser from
winning, at least.

I'm not quite sure which method you mean by "this method", but if
fpA-max(fpC) is better than the plurality-runoff one, that would also
make sense as it's less obviously stitched together.

In passing, I could also say that if you have a pairwise tied two-set
(weak DMTC?), e.g. something like

33: A>B>C
1: A>C>B
8: B>A>C
32: B=A>C
26: C>B>A

then breaking the tie by first preferences should preserve both
mono-raise and DMTCBR. This is unlike fpA-fpC, which would just call
this a tie.

That would mean that if the smallest DMT set has >2/3 support and
consists of two candidates, then breaking the tie by first preferences
within that coalition should be okay. While that's not very useful (it
can only happen with an exact tie), maybe the pattern could provide some
ideas about how to extend DMTCBR to DMTBR.

IRV passes both LNHs and so burial has no effect. But to make the DMTBR
proof more explicit: there are two possibilities. Either two candidates
from the innermost DMT set last until the final round, or one of them
does and the other is eliminated. They can't both be eliminated because
when all but one candidate is, the remaining candidate has >1/3 fpp and
so will be protected from elimination until the final round. And in the
final round, whoever beats the other pairwise wins.

The DMT criterion says that a candidate from the innermost DMT set
should be elected. Suppose first that the innermost DMT set is two
candidates or more. Then two of these will be preserved until the final
round, and no matter who wins, the DMT criterion is satisfied. If there
is only one candidate (DMTC), then likewise that candidate is preserved
until the final round, and since by definition he beats the other one
pairwise, he wins.

DMTBR is satisfied because if you prefer A to W, then you're going to
rank A higher than W. If you now try to bury W under C, your ballot will
be of the form ...>A>...>C>...W. But the part of your ballot that gives
weight to C relative to W isn't exposed until A has been eliminated, so
if your burial changes anything, it's already too late for A.

(For IRV this holds even when C is in the innermost DMT set, but this
isn't true for Condorcet methods.)

I'm of the impression that if method M passes DMT and DMBTR, then
Smith,M and Condorcet//M also do. Obviously this is true for DMTCBR
because the DMT candidate is both the Condorcet winner and the winner of
method M, so engineering a cycle won't switch the winner away from the
original CW to someone else.

For Smith,IRV (and Condorcet//IRV even moreso), I need to show that when
the IRV winner is some candidate inside the innermost DMT set but this
candidate is not the CW or not in the Smith set, then the voters who
prefer the IRV winner to the combined method's winner can't change the
latter to the former. And it's not completely clear to me just how to do
that. Maybe I'll find something out after exploring a bit with my linear
programming solver, but I feel there's a simple argument that I'm missing.

(I think the LCR scenario shows that full burial immunity is impossible.
But the innermost DMT set in that election is {L,C,R} so there's nobody
to bury under.)

Since the trick about 33% is that you can only have two, perhaps there's
an impossibility proof along the lines of IIA, that you have three >1/4
fpp candidates and no matter who wins, it's possible to make someone
else the CW through burial?

I'd have to think more about it. A method like plain IRV has no problem
ensuring burial immunity for such a group because it doesn't care about
burial at all. So the problem must be induced by Condorcet compliance
itself.

-km

On 3/12/22 6:58 PM, Forest Simmons wrote:

I don't get what you mean here. Certainly it's possible for honest
Condorcet cycles to exist. Warren gave an example 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.

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). Okay, so C can't win
because of second-order reasoning. 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