Smith//Score ?
Could Smith//Score be the ideal strategy-resistant Condorcet method?
The ballot would look like a Score ballot. To process the ballots, the
scores are converted into rankings (equal rankings allowed), and the
highest scoring candidate inside the Smith set is elected.
I'm hoping to make a Condorcet method that is very resistant to strategy.
It's not 100% resistant ---- you can devise an example where there is a
unique CW and a group of voters alter their ballots to produce a different
CW that they prefer. Fair enough. But in general the voter has to contend
with the fact that Score gives you a clear motivation to put your preferred
candidate on top and your least preferred candidate at the bottom. Trying
to alter the Smith set by ranking Y>X when you really prefer X>Y is a
strategy that could work, but it could also backfire if X was going to get
to the Smith set anyway. My intuition is that any strategy that works for
Smith//Score (and they do exist) should also work for any other Condorcet
method. So in that sense, this may be the most strategy-proof
Smith-efficient Condorcet method.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
Note that Smith//Score is the same as Smith,Score.
I would like to see how Smith//Score compares with Score Chain Climbing for
burial resistance, etc.
While Smith//Score can elect a covered candidate, Score Chain Climbing
cannot.
Score Chain Climbing is the closest clone free, monotonic method to
Banks//Score.
SCC:
While there remain two or more uneliminated candidates, from among those
remaining eliminate the highest score candidate that does not pairwise
defeat the lowest score candidate.
One reason this method is so burial resistant is because (at any stage
among the remaining candidates) the faction of the highest score candidate
that cannot defeat the lowest score candidate L is the most likely culprit
in the burial of L.
Since L does not pairwise beat itself, it will eliminate itself after all
of its potential buriers have been eliminated, if and only if there remains
another candidate that defeats it.
Note that the SCC winner is elected seamlessly from Banks without having to
compute Banks or Smith.
[This method is Banks efficient because the set of L's is a maximal chain
totally ordered by pairwise defeat.]
A Banks winner is always uncovered because every candidate is beaten by
some member L of the maximal chain, and the Banks winner is not beaten by
any of the L's.
El lun., 24 de ene. de 2022 12:51 p. m., Daniel Carrera dcarrera@gmail.com
escribió:
Could Smith//Score be the ideal strategy-resistant Condorcet method?
The ballot would look like a Score ballot. To process the ballots, the
scores are converted into rankings (equal rankings allowed), and the
highest scoring candidate inside the Smith set is elected.
I'm hoping to make a Condorcet method that is very resistant to strategy.
It's not 100% resistant ---- you can devise an example where there is a
unique CW and a group of voters alter their ballots to produce a different
CW that they prefer. Fair enough. But in general the voter has to contend
with the fact that Score gives you a clear motivation to put your preferred
candidate on top and your least preferred candidate at the bottom. Trying
to alter the Smith set by ranking Y>X when you really prefer X>Y is a
strategy that could work, but it could also backfire if X was going to get
to the Smith set anyway. My intuition is that any strategy that works for
Smith//Score (and they do exist) should also work for any other Condorcet
method. So in that sense, this may be the most strategy-proof
Smith-efficient Condorcet method.
Cheers,
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
Election-Methods mailing list - see https://electorama.com/em for list
info
On 24.01.2022 22:42, Forest Simmons wrote:
Note that Smith//Score is the same as Smith,Score.
That gives me an idea. How about Smith//Lp-cumulative?
That is, first remove everybody who's not part of the Smith set.
Renormalize all ballots to have unit p-norm. Then greatest score wins.
It probably isn't monotone, but the renormalization should mitigate at
least some of the Burr dilemma problems of plain Range.
-km
On 24.01.2022 21:51, Daniel Carrera wrote:
Could Smith//Score be the ideal strategy-resistant Condorcet method?
The ballot would look like a Score ballot. To process the ballots, the
scores are converted into rankings (equal rankings allowed), and the
highest scoring candidate inside the Smith set is elected.
I'm hoping to make a Condorcet method that is very resistant to
strategy. It's not 100% resistant ---- you can devise an example where
there is a unique CW and a group of voters alter their ballots to
produce a different CW that they prefer. Fair enough. But in general the
voter has to contend with the fact that Score gives you a clear
motivation to put your preferred candidate on top and your least
preferred candidate at the bottom. Trying to alter the Smith set by
ranking Y>X when you really prefer X>Y is a strategy that could work,
but it could also backfire if X was going to get to the Smith set
anyway. My intuition is that any strategy that works for Smith//Score
(and they do exist) should also work for any other Condorcet method. So
in that sense, this may be the most strategy-proof Smith-efficient
Condorcet method.
I would guess that this would be vulnerable to burial, and that it would
be broadly similar to Smith,Borda -- although that's just a hunch. I'd
be curious to what your simulator would say about its strategy
resistance :-)
I think that an advanced Range-based method would need to acknowledge
the inherent ambiguity in VNM utilities (mainly, that we don't know how
the voters' utilities are affinely scaled to become ratings). Plain
Range does not, and this is what lets it get off with IIA compliance de
jure while not being independent of losers de facto.
-km
Hi Daniel,
Could Smith//Score be the ideal strategy-resistant Condorcet method?
The ballot would look like a Score ballot. To process the ballots, the scores
are converted into rankings (equal rankings allowed), and the highest scoring
candidate inside the Smith set is elected.
I'm hoping to make a Condorcet method that is very resistant to strategy. It's
not 100% resistant ---- you can devise an example where there is a unique CW and
a group of voters alter their ballots to produce a different CW that they
prefer. Fair enough. But in general the voter has to contend with the fact that
Score gives you a clear motivation to put your preferred candidate on top and
your least preferred candidate at the bottom. Trying to alter the Smith set by
ranking Y>X when you really prefer X>Y is a strategy that could work, but it
could also backfire if X was going to get to the Smith set anyway. My intuition
is that any strategy that works for Smith//Score (and they do exist) should also
work for any other Condorcet method. So in that sense, this may be the most
strategy-proof Smith-efficient Condorcet method.
I'm not sure what it is you're saying will differentiate Smith//Score.
My concern is this scenario:
A is the CW, but might not be the Score winner. B may also be the Score winner,
and they want to use a weaker candidate C as a "pawn" against A. So they vote
A 0/10, B 10/10, C 1/10. They may succeed in creating a C>A win, and which
should be a cycle. And they have done this without giving much support to C.
Possible outcomes:
- C is actually voted as CW, because the A voters use the same strategy as the
B voters. Or for whatever reason, C beats B pairwise. - A is actually the Score winner and still wins.
- B is the Score winner and B wins, strategy successful.
I prefer a version of this method where an X>Y preference implies you are giving
X the maximum rating, so that you can't create a cycle using a pawn candidate
without having a meaningful commitment to the pawn during the cycle resolution.
It might be interesting to look at Stensholt BPW, which is among the most
burial-resistant Condorcet methods. It's defined on first preferences, but a
Score adaptation should still work OK. In BPW, in a three-candidate cycle, the
winner is that candidate who defeats the "strongest" candidate. Of course, this
is not monotone, since it can penalize "strength" in whatever metric you are
using.
But the effect of this rule is that no matter whether it's an A>B>C>A or A>C>B>A
cycle, burial against the CW using a "pawn" candidate, will only give the
desired outcome to the strategists when the pawn candidate is the strongest one
of the three!
So in the above scenario that I described, if either A or B is the Score winner,
B will not win the (artificially created) cycle. The strategy either leaves the
win with A or gives it to C.
Kevin
Hi Kristofer,
Le lundi 24 janvier 2022, 16:46:06 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
That gives me an idea. How about Smith//Lp-cumulative?
That is, first remove everybody who's not part of the Smith set.
Renormalize all ballots to have unit p-norm. Then greatest score wins.
It probably isn't monotone, but the renormalization should mitigate at
least some of the Burr dilemma problems of plain Range.
I guess that you should rescale, not normalize. What if I rate both A and B
10/10 each and they are both in the Smith set? Smith//Score would treat me
better than that, I guess.
Kevin
On 25.01.2022 00:29, Kevin Venzke wrote:
Hi Kristofer,
Le lundi 24 janvier 2022, 16:46:06 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
That gives me an idea. How about Smith//Lp-cumulative?
That is, first remove everybody who's not part of the Smith set.
Renormalize all ballots to have unit p-norm. Then greatest score wins.
It probably isn't monotone, but the renormalization should mitigate at
least some of the Burr dilemma problems of plain Range.
I guess that you should rescale, not normalize. What if I rate both A and B
10/10 each and they are both in the Smith set? Smith//Score would treat me
better than that, I guess.
If normalizing is meant in the sense of subtracting the mean, then I
agree, it should better be called rescaling. The "normalization" I'm
thinking of is the one cumulative voting imposes on the voter, or IRNR
does between rounds.
So e.g. if p->infty, you would scale the least preferred Smith set
member to zero and the most preferred to the max of the Range ballot.
(If you have no preference between any candidate in the Smith set, then
what value you set them to doesn't matter as long as each candidate gets
the same rating.)
If p=1, you get ordinary cumulative voting, which tends to have a
bullet-voting incentive. For p=2, there may be strategy resistance to be
had since (IIRC) the optimal zero information ballot for l2-cumulative
voting rates candidates proportional to the utilities.
In any case, the ballot is transformed so that no score is outside of
the range of the Range ballot (e.g. 0-10) and that the p-norm is equal
to the same constant for every voter, unless that voter rates every
candidate equal.
-km
Alright. It looks like I have misunderstood IRNR, and that what I am calling
IRRR actually is IRNR. Well, that at least lets me drop an extremely strange
method out of my simulations.
Le lundi 24 janvier 2022, 17:37:42 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
In any case, the ballot is transformed so that no score is outside of
the range of the Range ballot (e.g. 0-10) and that the p-norm is equal
to the same constant for every voter, unless that voter rates every
candidate equal.
Kevin
El lun., 24 de ene. de 2022 2:46 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 24.01.2022 22:42, Forest Simmons wrote:
Note that Smith//Score is the same as Smith,Score.
That gives me an idea. How about Smith//Lp-cumulative?
That is, first remove everybody who's not part of the Smith set.
Renormalize all ballots to have unit p-norm. Then greatest score wins.
It probably isn't monotone, but the renormalization should mitigate at
least some of the Burr dilemma problems of plain Range.
Score Chain Climbing generally disappoints both burial and Burr dilemma
defectors.
That's why it is becoming my favorite method.
SCC
While more than one candidate remains eliminate the highest score candidate
that does not pairwise defeat the lowest score remaining candidate.
The Burr defector, like the burial culprit is typically a fairly strong
candidate that sees a chance to bury or truncate an opponent that he does
not defeat pairwise, but might well come out ahead of if the opponent's
score is lowered.
SCC is practically tailor made to disappoint this kind of manipulation ...
the lowered score candidate still defeats her detractor pairwise, and her
lowered score makes her the pairwise eliminator at some early stage ... the
lower her score, the earlier her chance for revenge!
-km
On 25.01.2022 08:29, Forest Simmons wrote:
El lun., 24 de ene. de 2022 2:46 p. m., Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> escribió:
On 24.01.2022 22:42, Forest Simmons wrote:
Note that Smith//Score is the same as Smith,Score.
That gives me an idea. How about Smith//Lp-cumulative?
That is, first remove everybody who's not part of the Smith set.
Renormalize all ballots to have unit p-norm. Then greatest score wins.
It probably isn't monotone, but the renormalization should mitigate at
least some of the Burr dilemma problems of plain Range.
Score Chain Climbing generally disappoints both burial and Burr dilemma
defectors.
That's why it is becoming my favorite method.
SCC
While more than one candidate remains eliminate the highest score
candidate that does not pairwise defeat the lowest score remaining
candidate.
The Burr defector, like the burial culprit is typically a fairly strong
candidate that sees a chance to bury or truncate an opponent that he
does not defeat pairwise, but might well come out ahead of if the
opponent's score is lowered.
I'll have to check the performance of SCC when/if I make a simulator to
quick-test methods. I had the impression, though, that it produced some
strange honest results? That might have been the Borda variant, though,
so I'm not going to say it's bad on such a weak memory. Or I might be
misremembering altogether.
By the way, I usually consider the Approval/Range Burr dilemma fallout
to be mostly about honest miscalculation. E.g. suppose you want to vote
Perfect > Good > Bad in Approval. You misjudge the polls or vote early
and so you approve Perfect alone. Then Bad wins because Good doesn't
have enough support.
If there were only one honest ballot, then deliberately strengthening
Perfect>others at the expense of weakening Good>Bad would be a strategy.
But since Approval has multiple honest votes, even honest voters are
faced with the dilemma. And so they're the ones who have to deal with
the fallout.
It's kind of like monotonicity that way. Sure, you can strategize with
it, but that's not why it's bad :-)
-km