Empathy List Archives

KM

Kristofer Munsterhjelm

Mon, Jan 17, 2022 10:12 PM

On 17.01.2022 21:11, Kevin Venzke wrote:

Hi Kristofer,

Le samedi 8 janvier 2022, 17:21:22 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :

I have a hunch that if you put your "strategy-resistant Condorcet" hat on and
evaluate C//FPP, you will find it to be "good."

In my Monte Carlo (non-exhaustive) simulations, there are generally
three types of methods as far as strategy resistance goes: the type
that's susceptible >90% of the time whatever the number of candidates,
the type that's ~30% but increases with number of candidates to very
high levels with lots of candidates, and the type that's low and doesn't
increase.

A method is susceptible to strategy in a particular election if the
honest winner is A but voters who prefer some other B to A can conspire
to get B elected by changing their ballots.

C//FPP is the first type. MAM, Schulze, minmax, etc are of the second
type, and Smith-IRV, Benham, and fpA-fpC are of the third type.

Each election is a one-shot game (first some candidate wins, then
factions get to try to make other candidates win); there's no defensive
strategy. So it probably resembles your "never looks attractive in the
first place" setting.

Interesting. I suppose the obvious burial scenario for C//FPP is just every
single scenario where there's a CW who is not the FPW.

From scenarios I've been able to generate, it seems to me that fpA-fpC is purely
a middle ground between C//FPP and C//IRV. (Looking at three candidates.) I have
not, I think, seen a scenario where fpA-fpC, in a cycle, elects the FP loser, or
elects a non-FPW who didn't beat the FPW pairwise.

In other words, fpA-fpC usually likes the C//FPP winner, but sometimes it
prefers the second-place first preference candidate, provided they beat the
first preference winner. (Incidentally, my proposed expansion to fpA-fpC doesn't
maintain this pattern...)

It surprises me that you measure C//FPP to be in a whole other category, of
high manipulability, when in "method space" it is so close to those third type
methods.

It makes sense to me. I've later found out that fpA-fpC is very close to
Condorcet,Carey (as defined only for three candidates). And Carey is a
sort of compromise between IRV and Plurality; it's more like Plurality
so that monotonicity is satisfied, but not so much that it loses (single
candidate) DMTBR.

Suppose A is the Condorcet winner and has more than 1/3 of the first
preference votes. Then A is not eliminated in Carey: either both the
other two candidates are eliminated and A wins outright, or one
candidate is eliminated (say C without loss of generality) and then A
beats B pairwise due being the CW.

However, Plurality may choose another winner if the Plurality winner is B.

I wonder how you categorize BPW and SV? I find them to be quite
distant in method space from those third type methods. But by my
strategy measures they seem attractive, with less compromise and
burial incentive. The truncation performance is comparable/mixed.
It's just mono-raise and mono-add-top where BPW and SV look unusually
bad.

I would imagine they would be in the third category. I haven't checked
in detail because, since they're all nonmonotone, they don't seem to
provide a benefit over just using Smith,IRV. Perhaps unlike Smith,IRV
they could be made summable, but that would depend on the continuation.

I suspect that continuations would have to do something in the vicinity
of solid coalitions to retain DMTBR. Being disproved by chain climbing
retaining the base methods' strategic resistance would be great, of
course :-)

-km

On 17.01.2022 21:11, Kevin Venzke wrote: > Hi Kristofer, > > Le samedi 8 janvier 2022, 17:21:22 UTC−6, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit : >>> I have a hunch that if you put your "strategy-resistant Condorcet" hat on and >>> evaluate C//FPP, you will find it to be "good." >> >> In my Monte Carlo (non-exhaustive) simulations, there are generally >> three types of methods as far as strategy resistance goes: the type >> that's susceptible >90% of the time whatever the number of candidates, >> the type that's ~30% but increases with number of candidates to very >> high levels with lots of candidates, and the type that's low and doesn't >> increase. >> >> A method is susceptible to strategy in a particular election if the >> honest winner is A but voters who prefer some other B to A can conspire >> to get B elected by changing their ballots. >> >> C//FPP is the first type. MAM, Schulze, minmax, etc are of the second >> type, and Smith-IRV, Benham, and fpA-fpC are of the third type. >> >> Each election is a one-shot game (first some candidate wins, then >> factions get to try to make other candidates win); there's no defensive >> strategy. So it probably resembles your "never looks attractive in the >> first place" setting. > > Interesting. I suppose the obvious burial scenario for C//FPP is just every > single scenario where there's a CW who is not the FPW. > > From scenarios I've been able to generate, it seems to me that fpA-fpC is purely > a middle ground between C//FPP and C//IRV. (Looking at three candidates.) I have > not, I think, seen a scenario where fpA-fpC, in a cycle, elects the FP loser, or > elects a non-FPW who didn't beat the FPW pairwise. > > In other words, fpA-fpC usually likes the C//FPP winner, but sometimes it > prefers the second-place first preference candidate, provided they beat the > first preference winner. (Incidentally, my proposed expansion to fpA-fpC doesn't > maintain this pattern...) > > It surprises me that you measure C//FPP to be in a whole other category, of > high manipulability, when in "method space" it is so close to those third type > methods. It makes sense to me. I've later found out that fpA-fpC is very close to Condorcet,Carey (as defined only for three candidates). And Carey is a sort of compromise between IRV and Plurality; it's more like Plurality so that monotonicity is satisfied, but not so much that it loses (single candidate) DMTBR. Suppose A is the Condorcet winner and has more than 1/3 of the first preference votes. Then A is not eliminated in Carey: either both the other two candidates are eliminated and A wins outright, or one candidate is eliminated (say C without loss of generality) and then A beats B pairwise due being the CW. However, Plurality may choose another winner if the Plurality winner is B. > I wonder how you categorize BPW and SV? I find them to be quite > distant in method space from those third type methods. But by my > strategy measures they seem attractive, with less compromise and > burial incentive. The truncation performance is comparable/mixed. > It's just mono-raise and mono-add-top where BPW and SV look unusually > bad. I would imagine they would be in the third category. I haven't checked in detail because, since they're all nonmonotone, they don't seem to provide a benefit over just using Smith,IRV. Perhaps unlike Smith,IRV they could be made summable, but that would depend on the continuation. I suspect that continuations would have to do something in the vicinity of solid coalitions to retain DMTBR. Being disproved by chain climbing retaining the base methods' strategic resistance would be great, of course :-) -km

DC

Daniel Carrera

Mon, Jan 17, 2022 11:10 PM

On Mon, Jan 17, 2022 at 6:00 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:

So the A-first voters have no incentive to
bury under the dark horse. Thus the ball doesn't get rolling and there's
no escalation into a chicken problem.

Warren then says "perhaps that won't save you because people will bury
blindly anyway". Which is something I don't agree with. Sometimes I get
the impression he draws too much experience from Borda, which is awful.

I get that Warren's scenario is a possibility, and I don't want to dismiss
it, but the evidence he cites doesn't seem persuasive:

Australian ranked-ballot voters use this sort of "maximal exaggeration" strategy.
90% of Nader-favorite voters voted, strategically, for somebody else in USA 2000

But those are opposing strategies (compromise toward the center vs
exaggeration). To me it seems that in Australia they exaggerate because
they use IRV and they know they can, but in the US they compromise, because
they use FPTP and they know they must.

As you noted in another email, Range expects voters to strategize to cover
its shortcomings. Last night I was scribbling DH3 scenarios and when I
tried one with Range I got the DH elected. I had to then go back and change
the votes to get Range to avoid the DH. That doesn't inspire confidence. At
the end of the day, if everyone tells the election system that the
extremist party is their #2 option because everyone assumes that nobody
else would be crazy enough to vote for those nutjobs ... well, what can you
do? I'd rather have a system that encourages honest voting than one that
encourages strategy but assumes that voters will never get the strategy
wrong.

I have other, more fundamental issues with Range. To me the premise seems
flawed. Yes, I get that technically the ballot has more information. But I
feel that its advocates skip a step when they assume that they know how to
convert those ballots into a winner. I don't believe that I or anyone can
convert their political views into a number. I don't believe that we'll
produce numbers the same way. I don't believe that adding those numbers is
the right thing to do with them. And I don't buy the argument that another
man's vote is worth more than mine because he feels more strongly about his.

Still better than IRV though.

Cheers,

Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University

On Mon, Jan 17, 2022 at 6:00 AM Kristofer Munsterhjelm <km_elmet@t-online.de> wrote: > So the A-first voters have no incentive to > bury under the dark horse. Thus the ball doesn't get rolling and there's > no escalation into a chicken problem. > > Warren then says "perhaps that won't save you because people will bury > blindly anyway". Which is something I don't agree with. Sometimes I get > the impression he draws too much experience from Borda, which *is* awful. > I get that Warren's scenario is a possibility, and I don't want to dismiss it, but the evidence he cites doesn't seem persuasive: 1) `Australian ranked-ballot voters use this sort of "maximal exaggeration" strategy`. 2) `90% of Nader-favorite voters voted, strategically, for somebody else in USA 2000` But those are opposing strategies (compromise toward the center vs exaggeration). To me it seems that in Australia they exaggerate because they use IRV and they know they can, but in the US they compromise, because they use FPTP and they know they must. As you noted in another email, Range expects voters to strategize to cover its shortcomings. Last night I was scribbling DH3 scenarios and when I tried one with Range I got the DH elected. I had to then go back and change the votes to get Range to avoid the DH. That doesn't inspire confidence. At the end of the day, if everyone tells the election system that the extremist party is their #2 option because everyone assumes that nobody else would be crazy enough to vote for those nutjobs ... well, what can you do? I'd rather have a system that encourages honest voting than one that encourages strategy but assumes that voters will never get the strategy wrong. I have other, more fundamental issues with Range. To me the premise seems flawed. Yes, I get that technically the ballot has more information. But I feel that its advocates skip a step when they assume that they know how to convert those ballots into a winner. I don't believe that I or anyone can convert their political views into a number. I don't believe that we'll produce numbers the same way. I don't believe that adding those numbers is the right thing to do with them. And I don't buy the argument that another man's vote is worth more than mine because he feels more strongly about his. Still better than IRV though. Cheers, -- Dr. Daniel Carrera Postdoctoral Research Associate Iowa State University

KV

Kevin Venzke

Tue, Jan 18, 2022 6:15 AM

Hi Kristofer,

Le lundi 17 janvier 2022, 06:00:34 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :

I'm surprised that STAR is such a poor performer, though! (How about
IRNR_p? This is IRV-style elimination of the loser, except after each
step, the original rated ballot has the eliminated candidates removed
and then the ballot is rescaled to have unit p-norm. Try p=1,2,infinity.)

I guess the hunt's still on for a cardinal method that really resists
strategy. There's Hay voting, which is a strategyproof method (under VNM
utilities), but it has predictably awful honest performance, just like
Random Ballot has in the ranked domain.

I was looking into IRNR_1 recently because it occurred to me I had no idea where
it plots in my method space. (I should say, that I have to infer ratings from
the rankings in order to do this at all.)

So I implemented it and found it not so similar to anything. It's kind of
strange when I really think about it. If the rating scale is 0 to some positive
number, to normalize the ratings means that you dilute your voting power more
by casting a bunch of high ratings than by a bunch of low ones. That reminds me
of cumulative voting. I can't think of why it ought to behave that way.

Brian Olson has a pdf suggesting he envisions the use of positive and negative
ratings. I can't quite picture the effect of that, except that it seems a lot
more likely to produce sums of zero so that you don't know how to normalize it.

Furthermore, unless I've totally misunderstood, given how the normalization
works, IRNR doesn't guarantee to elect the pairwise winner between the last two
candidates.

I instead tried Instant Runoff "Rescaled" Ratings or "IRRR" where all you do is
adjust the ratings among remaining candidates so that they use the entire rating
scale. This plots in more familiar territory, about midway between the WV
methods and C//IRV. Though I doubt it's a method you're looking for.

As far as IRNR's properties, what I found was:
Low compromise incentive, though I think this may be illusory in some way. (For
example, a method that ignores all input will have zero compromise incentive.)
Lower burial incentive than C//IRV. So good.
Monotonicity about the same as IRV, but add some mono-add-top failures too.
Very high truncation incentive, can be even worse than Bucklin.
Poor Condorcet efficiency, sometimes worse than Bucklin.

Kevin

Hi Kristofer, Le lundi 17 janvier 2022, 06:00:34 UTC−6, Kristofer Munsterhjelm <km_elmet@t-online.de> a écrit : > I'm surprised that STAR is such a poor performer, though! (How about > IRNR_p? This is IRV-style elimination of the loser, except after each > step, the original rated ballot has the eliminated candidates removed > and then the ballot is rescaled to have unit p-norm. Try p=1,2,infinity.) > > I guess the hunt's still on for a cardinal method that really resists > strategy. There's Hay voting, which is a strategyproof method (under VNM > utilities), but it has predictably awful honest performance, just like > Random Ballot has in the ranked domain. I was looking into IRNR_1 recently because it occurred to me I had no idea where it plots in my method space. (I should say, that I have to infer ratings from the rankings in order to do this at all.) So I implemented it and found it not so similar to anything. It's kind of strange when I really think about it. If the rating scale is 0 to some positive number, to normalize the ratings means that you dilute your voting power more by casting a bunch of high ratings than by a bunch of low ones. That reminds me of cumulative voting. I can't think of why it ought to behave that way. Brian Olson has a pdf suggesting he envisions the use of positive and negative ratings. I can't quite picture the effect of that, except that it seems a lot more likely to produce sums of zero so that you don't know how to normalize it. Furthermore, unless I've totally misunderstood, given how the normalization works, IRNR doesn't guarantee to elect the pairwise winner between the last two candidates. I instead tried Instant Runoff "Rescaled" Ratings or "IRRR" where all you do is adjust the ratings among remaining candidates so that they use the entire rating scale. This plots in more familiar territory, about midway between the WV methods and C//IRV. Though I doubt it's a method you're looking for. As far as IRNR's properties, what I found was: Low compromise incentive, though I think this may be illusory in some way. (For example, a method that ignores all input will have zero compromise incentive.) Lower burial incentive than C//IRV. So good. Monotonicity about the same as IRV, but add some mono-add-top failures too. Very high truncation incentive, can be even worse than Bucklin. Poor Condorcet efficiency, sometimes worse than Bucklin. Kevin

RT

Richard, the VoteFair guy

Wed, Jan 26, 2022 7:30 PM

My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate
is too difficult to do as a visible process on an auditorium stage. As
Kristofer says:

On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:

But you still need to decide when to stop, which can be pretty
difficult. At first it'd seem to be obvious: if the bottom-most
candidate beats anyone else pairwise, he'd be in the Smith set, no?
But that doesn't quite work. Consider something like this: ...

It's much easier, and almost as "good," to eliminate pairwise losing
candidates as they occur during elimination rounds.

As a reminder, a "pairwise losing candidate" is the candidate who would
lose every one-on-one match against each and every remaining
(not-yet-eliminated) candidate.

That would eliminate MOST, but not all, non-Smith-set candidates in MOST
cases.

Thanks to those who helped get us close to a process that could be
understood by an audience of typical voters.

Richard Fobes
The VoteFair guy

On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:

On 15.01.2022 21:50, Richard, the VoteFair guy wrote:

Yet I believe that lots of audience members would give more trust to a
method that eliminates candidates one at a time. This is the simplicity
advantage that instant-runoff voting (IRV) has. In contrast, starting
with the Copeland winner might seem like magic to people who fear math.

Would eliminating from the bottom always work?

In the Wikipedia example below, it does work. It identifies B and E as
outside the Smith set, so the remaining candidates are in the Smith set.

5 wins for A

5 wins for D

4 wins for G

3.5 wins for C

2.5 wins for F

1 win for B

0 wins for E

(Half indicates a tie.)

So, my question is: Does elimination from the bottom always work?

My guess is the answer is yes.

I think so too, in theory. Suppose that A is in the Smith set and B is
not. Then A beats everybody who B beats and then some. So B will be
ranked below A by Copeland score. Hence the only way someone in the
Smith set will at the bottom of the list is if everybody is in it.

But you still need to decide when to stop, which can be pretty
difficult. At first it'd seem to be obvious: if the bottom-most
candidate beats anyone else pairwise, he'd be in the Smith set, no? But
that doesn't quite work. Consider something like this:

The candidates are ABCD. There's a Condorcet winner, A, who beats
everybody. But there's also a loser three-cycle, B>C>D>B. Thus whoever
you choose to eliminate first of B, C, and D, that candidate beats
someone else despite being outside the Smith set. And batch-eliminating
everybody with an identical least score fails in more complicated scenarios.

To justify the Copeland winner being the initial candidate, I would just
point to the reasoning above: any non-Smith set member beats or ties
fewer people than every Smith set member (almost by definition). So the
topmost by number of candidates beaten can't be outside the Smith set.

If the voters are expected to know what a Smith set is, that leap
shouldn't be too large. And if not, then every Smith//X method will look
like magic to some degree. (Better use Benham or BTR-STV.)

-km

My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate is too difficult to do as a visible process on an auditorium stage. As Kristofer says: On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote: > But you still need to decide when to stop, which can be pretty > difficult. At first it'd seem to be obvious: if the bottom-most > candidate beats anyone else pairwise, he'd be in the Smith set, no? > But that doesn't quite work. Consider something like this: ... It's much easier, and almost as "good," to eliminate pairwise losing candidates as they occur during elimination rounds. As a reminder, a "pairwise losing candidate" is the candidate who would lose every one-on-one match against each and every remaining (not-yet-eliminated) candidate. That would eliminate MOST, but not all, non-Smith-set candidates in MOST cases. Thanks to those who helped get us close to a process that could be understood by an audience of typical voters. Richard Fobes The VoteFair guy On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote: > On 15.01.2022 21:50, Richard, the VoteFair guy wrote: > >> Yet I believe that lots of audience members would give more trust to a >> method that eliminates candidates one at a time. This is the simplicity >> advantage that instant-runoff voting (IRV) has. In contrast, starting >> with the Copeland winner might seem like magic to people who fear math. >> >> Would eliminating from the bottom always work? >> >> In the Wikipedia example below, it does work. It identifies B and E as >> outside the Smith set, so the remaining candidates are in the Smith set. >> >> 5 wins for A >> >> 5 wins for D >> >> 4 wins for G >> >> 3.5 wins for C >> >> 2.5 wins for F >> >> 1 win for B >> >> 0 wins for E >> >> (Half indicates a tie.) >> >> So, my question is: Does elimination from the bottom always work? >> >> My guess is the answer is yes. > > I think so too, in theory. Suppose that A is in the Smith set and B is > not. Then A beats everybody who B beats and then some. So B will be > ranked below A by Copeland score. Hence the only way someone in the > Smith set will at the bottom of the list is if everybody is in it. > > But you still need to decide when to stop, which can be pretty > difficult. At first it'd seem to be obvious: if the bottom-most > candidate beats anyone else pairwise, he'd be in the Smith set, no? But > that doesn't quite work. Consider something like this: > > The candidates are ABCD. There's a Condorcet winner, A, who beats > everybody. But there's also a loser three-cycle, B>C>D>B. Thus whoever > you choose to eliminate first of B, C, and D, that candidate beats > someone else despite being outside the Smith set. And batch-eliminating > everybody with an identical least score fails in more complicated scenarios. > > To justify the Copeland winner being the initial candidate, I would just > point to the reasoning above: any non-Smith set member beats or ties > fewer people than every Smith set member (almost by definition). So the > topmost by number of candidates beaten can't be outside the Smith set. > > If the voters are expected to know what a Smith set is, that leap > shouldn't be too large. And if not, then every Smith//X method will look > like magic to some degree. (Better use Benham or BTR-STV.) > > -km >

KM

Kristofer Munsterhjelm

Thu, Jan 27, 2022 10:30 AM

On 26.01.2022 20:30, Richard, the VoteFair guy wrote:

My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate
is too difficult to do as a visible process on an auditorium stage. As
Kristofer says:

On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:

But you still need to decide when to stop, which can be pretty
difficult. At first it'd seem to be obvious: if the bottom-most
candidate beats anyone else pairwise, he'd be in the Smith set, no?
But that doesn't quite work. Consider something like this: ...

It's much easier, and almost as "good," to eliminate pairwise losing
candidates as they occur during elimination rounds.

As a reminder, a "pairwise losing candidate" is the candidate who would
lose every one-on-one match against each and every remaining
(not-yet-eliminated) candidate.

That would eliminate MOST, but not all, non-Smith-set candidates in MOST
cases.

Thanks to those who helped get us close to a process that could be
understood by an audience of typical voters.

The problem is, of course, that there's not always a Condorcet loser.
And so you may eliminate a Smith set member because the method's
fallback metric fails to exclude them from elimination.

Meanwhile, both the "move to the right of the line" and the sorting
proposals will succeed in identifying the Smith set. And they're both
simple, IMHO, so I would disagree with your conclusion.

If you need only a single winner, there are plenty of alternatives that
pass Smith. Copeland elimination seamlessly handles the pairwise loser
elimination step, because a Condorcet loser is also a Copeland loser;
and since Smith set members have above average Copeland score whenever
non-Smith members exist, you'll eliminate every non-Smith member before
you eliminate every Smith member. It shouldn't be too difficult to
phrase, either: instead of "eliminate the candidate who loses every
pairwise contests", it's "eliminate the candidate who loses the most
pairwise contests".

Or Forest's sequential pairwise elimination, which in a sense mimics
legislative procedure and so should also be pretty easy to explain.
First set up an ordering, then go up it and say "does the next proposal
beat my current proposal? If so, switch it out, otherwise keep my
current proposal".

Or even BTR. If your elimination procedure, instead of saying "eliminate
the worst by some measure X" says "eliminate the candidate of the second
worst and worst who beats the other one pairwise". This also passes Smith.

So I disagree. I don't think simplicity requires you to throw away Smith
compliance.

-km

On 26.01.2022 20:30, Richard, the VoteFair guy wrote: > My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate > is too difficult to do as a visible process on an auditorium stage. As > Kristofer says: > > On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote: >> But you still need to decide when to stop, which can be pretty >> difficult. At first it'd seem to be obvious: if the bottom-most >> candidate beats anyone else pairwise, he'd be in the Smith set, no? >> But that doesn't quite work. Consider something like this: ... > > It's much easier, and almost as "good," to eliminate pairwise losing > candidates as they occur during elimination rounds. > > As a reminder, a "pairwise losing candidate" is the candidate who would > lose every one-on-one match against each and every remaining > (not-yet-eliminated) candidate. > > That would eliminate MOST, but not all, non-Smith-set candidates in MOST > cases. > > Thanks to those who helped get us close to a process that could be > understood by an audience of typical voters. The problem is, of course, that there's not always a Condorcet loser. And so you may eliminate a Smith set member because the method's fallback metric fails to exclude them from elimination. Meanwhile, both the "move to the right of the line" and the sorting proposals will succeed in identifying the Smith set. And they're both simple, IMHO, so I would disagree with your conclusion. If you need only a single winner, there are plenty of alternatives that pass Smith. Copeland elimination seamlessly handles the pairwise loser elimination step, because a Condorcet loser is also a Copeland loser; and since Smith set members have above average Copeland score whenever non-Smith members exist, you'll eliminate every non-Smith member before you eliminate every Smith member. It shouldn't be too difficult to phrase, either: instead of "eliminate the candidate who loses every pairwise contests", it's "eliminate the candidate who loses the most pairwise contests". Or Forest's sequential pairwise elimination, which in a sense mimics legislative procedure and so should also be pretty easy to explain. First set up an ordering, then go up it and say "does the next proposal beat my current proposal? If so, switch it out, otherwise keep my current proposal". Or even BTR. If your elimination procedure, instead of saying "eliminate the worst by some measure X" says "eliminate the candidate of the second worst and worst who beats the other one pairwise". This also passes Smith. So I disagree. I don't think simplicity requires you to throw away Smith compliance. -km

CC

Colin Champion

Thu, Jan 27, 2022 10:46 AM

I agree with Kristofer and Forest that Smith compliance is easy to
attain, eg. through Copeland's method. I understood Richard as saying
that explicit reference to the Smith set in the definition of a voting
algorithm (eg. Smith/IRV, Smith/Minimax) puts it outside the class of
methods which most voters can be expected to grasp (and I agree with
that too).
CJC

I agree with Kristofer and Forest that Smith compliance is easy to attain, eg. through Copeland's method. I understood Richard as saying that explicit reference to the Smith set in the definition of a voting algorithm (eg. Smith/IRV, Smith/Minimax) puts it outside the class of methods which most voters can be expected to grasp (and I agree with that too). CJC

KM

Kristofer Munsterhjelm

Thu, Jan 27, 2022 11:05 AM

On 27.01.2022 11:46, Colin Champion wrote:

I agree with Kristofer and Forest that Smith compliance is easy to
attain, eg. through Copeland's method. I understood Richard as saying
that explicit reference to the Smith set in the definition of a voting
algorithm (eg. Smith/IRV, Smith/Minimax) puts it outside the class of
methods which most voters can be expected to grasp (and I agree with
that too).

Right. While I'm not sure it would be too difficult to calculate the
Smith set on stage, it would make the method considerably more complex
than one where Smith compliance happens more naturally.

So "on stage", Smith//IRV may not be a good method. I don't think that
IRV itself would be a good method for on-stage voting, but if you had to
do it then I would imagine either Benham (if you need strategy
resistance) or BTR (otherwise) would be better than Smith//IRV.

As for generally strategy resistant on-stage methods... I don't know,
it's a very tough constraint. UncAAO, perhaps, or something else making
use of implicit approval? I'm not sure. If it's not strategy resistant
then Copeland-elimination is as good as any. But if you need
monotonicity, perhaps minmax?

-km

On 27.01.2022 11:46, Colin Champion wrote: > I agree with Kristofer and Forest that Smith compliance is easy to > attain, eg. through Copeland's method. I understood Richard as saying > that explicit reference to the Smith set in the definition of a voting > algorithm (eg. Smith/IRV, Smith/Minimax) puts it outside the class of > methods which most voters can be expected to grasp (and I agree with > that too). Right. While I'm not sure it would be too difficult to calculate the Smith set on stage, it would make the method considerably more complex than one where Smith compliance happens more naturally. So "on stage", Smith//IRV may not be a good method. I don't think that IRV itself would be a good method for on-stage voting, but if you had to do it then I would imagine either Benham (if you need strategy resistance) or BTR (otherwise) would be better than Smith//IRV. As for generally strategy resistant on-stage methods... I don't know, it's a very tough constraint. UncAAO, perhaps, or something else making use of implicit approval? I'm not sure. If it's not strategy resistant then Copeland-elimination is as good as any. But if you need monotonicity, perhaps minmax? -km

KM

Kristofer Munsterhjelm

Thu, Jan 27, 2022 11:15 AM

On 27.01.2022 12:05, Kristofer Munsterhjelm wrote:

As for generally strategy resistant on-stage methods... I don't know,
it's a very tough constraint. UncAAO, perhaps, or something else making
use of implicit approval? I'm not sure. If it's not strategy resistant
then Copeland-elimination is as good as any. But if you need
monotonicity, perhaps minmax?

Possibly River. I vaguely remember Jobst saying it could be done
interactively (and manually) in a natural way, although I don't remember
the details. If so, that would be the best of the minmax class for
on-stage counting, since you'd automatically get ISDA and IPDA too.

-km

On 27.01.2022 12:05, Kristofer Munsterhjelm wrote: > As for generally strategy resistant on-stage methods... I don't know, > it's a very tough constraint. UncAAO, perhaps, or something else making > use of implicit approval? I'm not sure. If it's not strategy resistant > then Copeland-elimination is as good as any. But if you need > monotonicity, perhaps minmax? Possibly River. I vaguely remember Jobst saying it could be done interactively (and manually) in a natural way, although I don't remember the details. If so, that would be the best of the minmax class for on-stage counting, since you'd automatically get ISDA and IPDA too. -km

RT

Richard, the VoteFair guy

Thu, Jan 27, 2022 8:12 PM

On 1/27/2022 2:30 AM, Kristofer Munsterhjelm wrote:

So I disagree. I don't think simplicity requires you to
throw away Smith compliance.

I agree. Yet remember that my goal is to make it easy for voters to
watch vote counting on a stage and understand what's going on.

Kristofer suggests River and some other mathematically good methods, but
all of them require using the pairwise counts in ways that are not as
easy to understand as asking "which number is bigger?, that's the winner
in this pair."

Yes Copeland is simple, but ...

The popularity of IRV has shown that vote counting is easier to
understand when candidates are eliminated one at a time. Most watchers
won't trust a process that ends so quickly, and fails to also
(sequentially) reveal who was least popular, who was second-least
popular, etc.

In a different message ... Kristofer Munsterhjelm wrote:

So "on stage", Smith//IRV may not be a good method. I don't
think that IRV itself would be a good method for on-stage
voting, but if you had to do it then I would imagine either
Benham (if you need strategy resistance) or BTR (otherwise)
would be better than Smith//IRV.

Actually one of the big selling points of IRV is it can be done on stage
simply by piling ballots into stacks. The stacks are named by
candidate, and the ballot goes to the stack of that ballot's currently
highest-ranked candidate. The height of the stacks make some
comparisons very clear. When the heights are similar, simple counting
can confirm which stack has fewer ballots.

Near the end of the process the tallest stack can be counted to see if
it contains more than half the ballots. For added clarity, the ballots
in all the other stacks can be counted to show there are fewer ballots
in all those other stacks.

With IRV it's even easier -- but not fairer -- (than other methods)
because after each elimination the only ballot stack that needs to be
re-allocated is the stack that supports the candidate just eliminated.

I'll repeat that I strongly dislike IRV!!! So I'm not defending it!

I'm just trying to point to something better that also can be counted in
a way where watchers can easily verify the (relative) fairness of each step.

When a method requires subtraction -- such as for calculating margins --
or sorting pairwise counts, the ease of understanding disappears. (Yes
subtracting can be done by hand on a white board, but following the
subtraction process many times would make some watcher's head hurt. Yes
a calculator is usually trustworthy, but a watcher doesn't want to watch
lots of numbers being entered on the calculator and then transcribed
onto signs, and then named, etc. Yes sorting can be followed, but it's
not easy to trust if the process also involves something else going on too.)

Yes, BTR-IRV does fit the requirement of being better, but IMO it's not
good enough. It elects the Condorcet winner, but in a way that would
prompt lots of voters to ask "why are we protecting the Condorcet winner
when it has the shortest stack of ballots?"

In contrast, it's usually easy to follow eliminations that are based on
something that's easy to understand, such as eliminating a candidate who
loses every one-on-one match against the remaining candidates.

I believe such methods exist. They may not have established names.
They must be easy for non-math-savvy watchers to follow. Yes they won't
have superb mathematical characteristics. But IRV too easily yields
flawed results, so getting better results shouldn't be difficult to
achieve, even with the can-be-followed-on-stage constraint.