FS
Forest Simmons
Wed, Jan 5, 2022 3:25 AM
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away all
of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Your Q&CBRS works well in a situation in which Approval Sorted Margins
does not: (due to Colin Champion):
Sincere:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
D is the beats-all Condorcet voter.
5 C-first voters bury D:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > B > E > D # Was C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
Pairwise array:
[-- 6. 2. 5. 8.]
[5. -- 4. 9. 9.]
[9. 7. -- 5. 7.]
[6. 2. 6. -- 4.]
[3. 2. 4. 7. --]
Now, if I understand your method correctly, we first find the Smith Set.
In the burial case, it is all 5 candidates, A, B, C, D, E.
Then we find the basic scores for each Smith candidate:
A: 8
B: 11
C: 10
D: 6
E: 9
The Smith candidate with the smallest basic score is D (our previous CW).
Smith candidates that defeat D are B and E. B has highest basic
score, therefore is the winner. C's strategy did not help C.
With Approval Sorted Margins, C is able to win using Burial.
The Basic Score needs to be tabulated separately from pairwise, and
depends on the other rankings on the ballot: each candidate gets a point
for any rating above minimum candidate rating, if using a ratings ballot.
In the case of many candidates, this could lead to basic score ties, since
each ballot would certainly have many candidates rated 0, so in those cases
I would recommend recounting after eliminating all candidates outside the
Smith Set.
I've been thinking about the impracticality of computing pairwise arrays
in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
were used, with rankings inferred, I would recommend a floating score
threshold, starting at 1% of maximum approval, but rising until at most 9
distinct candidate scores are above the threshold (allows score-tie
clusters), then recounting to get the reduced pairwise array and basic
scores. If the lowest basic score is tied, eliminate non-Smith candidates
and recount basic scores.
For elections with 9 or fewer candidates and no lowest-basic score ties
in the Smith set, this is summable, but requires recounts in the event of
more candidates or lowest basic score ties.
Looking back at Colin's burial example, what happens if basic score is
recalculated using ballots scoring candidate X above D?
A: 5
B: 9
C: 5
E: 7
B still wins. So your overall basic score as a proxy for ballots scoring
above the lowest Smith basic score candidate is a good proxy in this case.
On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Let's call it Q&CBRS.
Pre-requisite background:
Candidate X pairwise defeats candidate Y iff candidate X is ranked
above/before/ahead of candidate Y on more ballots than not.
A defeat chain is a sequence of candidates in which each candidate
pairwise defeats the subsequent member of the chain.
Using a "bubble sort" procedure to sort a list of candidates into
pairwise order produces a defeat chain of the listed candidates.
In this way we can easily find a defeat chain that includes all of the
candidates. The first candidate in such a chain is an example of a Smith
candidate. More generally, any candidate who has a defeat chain to any
other candidate is a member of the Smith Set.
Q&CBRS:
First, find the "basic score" for each candidate defined as the number of
ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with the
greatest basic score.
That's it! Quick & Clean!
Note that if there is only one Smith candidate X, then X will be the only
candidate not defeated by X, and therefore the one elected .
In general a candidate not defeated by X will defeat X, and thereby find
itself at the head of a defeat chain to every other candidate, i.e. it will
be a Smith candidate.
When there is only one Smith candidate, that candidate will not be
pairwise defeated by any other candidate.
I repeat the entire method procedure here:
First, find the "basic score" for each candidate defined as the number of
ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with the
greatest basic score.
Where else can you find such a simple, quick, and clean election method?
El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Good idea!
Although it seems to me that the highest approval candidate would have
to pairwise beat or tie the approval cutoff candidate X pairwise (which
would be impossible for a non-Smith candidate to do) ...I could be wrong
... and in any case redundancy reinforces communication and understanding.
El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
This sounds like an interesting method to me!
However, I would change the winning criteria to "Elect the most
approved member of the Smith Set".
On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I apologize for the defiant tone at the end of the previous message
... I must have gotten carried away with the "Dirty Dozen"' theme.
But isn't it frustrating to you when people use the 2nd law of
thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
justify their stubborn resistance to any kind of engineering progress?
In the previous message Q&D Burial Resistant Condorcet was formulate
in the typical "stitched together" form ... "Elect the CW if there is one,
Else ..."
In this message I would like to formulate a seamless version:
Let X be the Smith candidate who on the fewest ballots is ranked ahead
of any other Smith candidate. On each ballot approve all candidates down to
X, but include X only when no Smith candidate is ranked ahead of X.
Elect the candidate approved on the most ballots.
This method can be described as electing the approval winner when the
approval cutoff is (at the rank of) the weakest of the Smith candidates,
which itself is approved on (and only on) those ballots which do not
approve any other Smith candidate.
In other words, the approval cutoff is inclusive only when necessary
to ensure approval of at least one member of Smith.
Since a Smith member is approved on every ballot, the method satisfies
the Condorcet Criterion, i.e. it elects the only Smith member when Smith is
a singleton.
How does that grab you?
-FWS
El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Is there any burial resistant Condorcet method simpler than this?
The basic pre-requisite is to understand that whenever there is no
Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
candidates whose members are not defeated by any candidates outside of the
cycle, just as a Condorcet Winner is a candidate undefeated by any other
candidate.
Here's the Q&D burial resistant method:
Lacking a Condorcet Winner elect the candidate X having the greatest
pairwise victory over the top-cycle member Y that has the smallest ratio of
first to last place votes within the top cycle.
Two examples illustrate the method:
Example 1.
49 C
26 A>B
25 B (sincere B>A)
The top cycle is ABCA
Candidate A has the smallest ratio 26/74 of first to last place votes.
Candidate C is the only candidate with a pairwise victory over it, so
C wins.
Notice how our rule does not reward B for insincerely lowering A to
(equal) last?
Example 2.
45 A>B (sincere A>C)
35 B>C
25 C>A
Candidate C has the smallest ratio 25/45 of first to last.
Candidate B wins as the only candidate with a pairwise victory over
C. So A's burial of C backfires.
Typically, the faction A that buries or truncates a Condorcet Winner
C to create a top-cycle cannot by so doing become a pairwise victor over
the buried Condorcet Winner ... but must (in order to create a cycle) help
some other candidate B defeat C by insincerely voting B>C.
Our Quick and Dirty method insures that if the sincere CW's rightful
victory is subverted, it goes to B, not to A.
Is this method quick enough and dirty enough for the FairVote IRV
promoters?
What objections/criticisms might they have?
Do they have any counter proposal that rivals this one in any way?
If so, let them educate us ... we'll gladly join them if they can
show us a better way!
If not, then they should join us to educate the politicians, public,
and last but not least, the academics still stuck in the pre-EM era!
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away all
of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon@gmail.com>
escribió:
> Your Q&CBRS works well in a situation in which Approval Sorted Margins
> does not: (due to Colin Champion):
>
> Sincere:
>
> 1: B > D > A > E > C
> 1: B > D > E > C > A
> 5: C > A > D > B > E
> 1: D > A > C > B > E
> 1: D > C > B > A > E
> 2: E > B > D > C > A
>
> D is the beats-all Condorcet voter.
>
> 5 C-first voters bury D:
>
> 1: B > D > A > E > C
> 1: B > D > E > C > A
> 5: C > A > B > E > D # Was C > A > D > B > E
> 1: D > A > C > B > E
> 1: D > C > B > A > E
> 2: E > B > D > C > A
>
> Pairwise array:
> [-- 6. 2. 5. 8.]
> [5. -- 4. 9. 9.]
> [9. 7. -- 5. 7.]
> [6. 2. 6. -- 4.]
> [3. 2. 4. 7. --]
>
> Now, if I understand your method correctly, we first find the Smith Set.
> In the burial case, it is all 5 candidates, A, B, C, D, E.
>
> Then we find the basic scores for each Smith candidate:
>
> A: 8
> B: 11
> C: 10
> D: 6
> E: 9
>
> The Smith candidate with the smallest basic score is D (our previous CW).
>
> Smith candidates that defeat D are *B* and *E*. B has highest basic
> score, therefore is the winner. C's strategy did not help C.
>
> With Approval Sorted Margins, C is able to win using Burial.
>
> The Basic Score needs to be tabulated separately from pairwise, and
> depends on the other rankings on the ballot: each candidate gets a point
> for any rating above minimum candidate rating, if using a ratings ballot.
> In the case of many candidates, this could lead to basic score ties, since
> each ballot would certainly have many candidates rated 0, so in those cases
> I would recommend recounting after eliminating all candidates outside the
> Smith Set.
>
> I've been thinking about the impracticality of computing pairwise arrays
> in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
> were used, with rankings inferred, I would recommend a floating score
> threshold, starting at 1% of maximum approval, but rising until at most 9
> distinct candidate scores are above the threshold (allows score-tie
> clusters), then recounting to get the reduced pairwise array and basic
> scores. If the lowest basic score is tied, eliminate non-Smith candidates
> and recount basic scores.
>
> *For elections with 9 or fewer candidates and no lowest-basic score ties
> in the Smith set, this is summable, but requires recounts in the event of
> more candidates or lowest basic score ties.*
>
> Looking back at Colin's burial example, what happens if basic score is
> recalculated using ballots scoring candidate X above D?
>
> A: 5
> B: 9
> C: 5
> E: 7
>
> B still wins. So your overall basic score as a proxy for ballots scoring
> above the lowest Smith basic score candidate is a good proxy in this case.
>
> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <forest.simmons21@gmail.com>
> wrote:
>
>> Let's call it Q&CBRS.
>>
>> Pre-requisite background:
>>
>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>> above/before/ahead of candidate Y on more ballots than not.
>>
>> A defeat chain is a sequence of candidates in which each candidate
>> pairwise defeats the subsequent member of the chain.
>>
>> Using a "bubble sort" procedure to sort a list of candidates into
>> pairwise order produces a defeat chain of the listed candidates.
>>
>> In this way we can easily find a defeat chain that includes all of the
>> candidates. The first candidate in such a chain is an example of a Smith
>> candidate. More generally, any candidate who has a defeat chain to any
>> other candidate is a member of the Smith Set.
>>
>> Q&CBRS:
>>
>> First, find the "basic score" for each candidate defined as the number of
>> ballots on which it is ranked above one or more candidates.
>>
>> Then let X be the Smith candidate with the smallest basic score.
>>
>> Finally, among the candidates not defeated by X, elect the one with the
>> greatest basic score.
>>
>> That's it! Quick & Clean!
>>
>> Note that if there is only one Smith candidate X, then X will be the only
>> candidate not defeated by X, and therefore the one elected .
>>
>> In general a candidate not defeated by X will defeat X, and thereby find
>> itself at the head of a defeat chain to every other candidate, i.e. it will
>> be a Smith candidate.
>>
>> When there is only one Smith candidate, that candidate will not be
>> pairwise defeated by any other candidate.
>>
>> I repeat the entire method procedure here:
>>
>> First, find the "basic score" for each candidate defined as the number of
>> ballots on which it is ranked above one or more candidates.
>>
>> Then let X be the Smith candidate with the smallest basic score.
>>
>> Finally, among the candidates not defeated by X, elect the one with the
>> greatest basic score.
>>
>> Where else can you find such a simple, quick, and clean election method?
>>
>>
>>
>>
>>
>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>> forest.simmons21@gmail.com> escribió:
>>
>>> Good idea!
>>>
>>> Although it seems to me that the highest approval candidate would have
>>> to pairwise beat or tie the approval cutoff candidate X pairwise (which
>>> would be impossible for a non-Smith candidate to do) ...I could be wrong
>>> ... and in any case redundancy reinforces communication and understanding.
>>>
>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <dodecatheon@gmail.com>
>>> escribió:
>>>
>>>> Hi Forest,
>>>>
>>>> This sounds like an interesting method to me!
>>>>
>>>> However, I would change the winning criteria to "Elect the most
>>>> approved member of the Smith Set".
>>>>
>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>> forest.simmons21@gmail.com> wrote:
>>>>
>>>>> I apologize for the defiant tone at the end of the previous message
>>>>> ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>
>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>
>>>>> In the previous message Q&D Burial Resistant Condorcet was formulate
>>>>> in the typical "stitched together" form ... "Elect the CW if there is one,
>>>>> Else ..."
>>>>>
>>>>> In this message I would like to formulate a seamless version:
>>>>>
>>>>> Let X be the Smith candidate who on the fewest ballots is ranked ahead
>>>>> of any other Smith candidate. On each ballot approve all candidates down to
>>>>> X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>
>>>>> Elect the candidate approved on the most ballots.
>>>>>
>>>>> This method can be described as electing the approval winner when the
>>>>> approval cutoff is (at the rank of) the weakest of the Smith candidates,
>>>>> which itself is approved on (and only on) those ballots which do not
>>>>> approve any other Smith candidate.
>>>>>
>>>>> In other words, the approval cutoff is inclusive only when necessary
>>>>> to ensure approval of at least one member of Smith.
>>>>>
>>>>> Since a Smith member is approved on every ballot, the method satisfies
>>>>> the Condorcet Criterion, i.e. it elects the only Smith member when Smith is
>>>>> a singleton.
>>>>>
>>>>> How does that grab you?
>>>>>
>>>>> -FWS
>>>>>
>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>> forest.simmons21@gmail.com> escribió:
>>>>>
>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>
>>>>>> The basic pre-requisite is to understand that whenever there is no
>>>>>> Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>> candidate.
>>>>>>
>>>>>> Here's the Q&D burial resistant method:
>>>>>>
>>>>>> Lacking a Condorcet Winner elect the candidate X having the greatest
>>>>>> pairwise victory over the top-cycle member Y that has the smallest ratio of
>>>>>> first to last place votes within the top cycle.
>>>>>>
>>>>>> Two examples illustrate the method:
>>>>>>
>>>>>> Example 1.
>>>>>>
>>>>>> 49 C
>>>>>> 26 A>B
>>>>>> 25 B (sincere B>A)
>>>>>>
>>>>>> The top cycle is ABCA
>>>>>>
>>>>>> Candidate A has the smallest ratio 26/74 of first to last place votes.
>>>>>>
>>>>>> Candidate C is the only candidate with a pairwise victory over it, so
>>>>>> C wins.
>>>>>>
>>>>>> Notice how our rule does not reward B for insincerely lowering A to
>>>>>> (equal) last?
>>>>>>
>>>>>> Example 2.
>>>>>>
>>>>>> 45 A>B (sincere A>C)
>>>>>> 35 B>C
>>>>>> 25 C>A
>>>>>>
>>>>>> Candidate C has the smallest ratio 25/45 of first to last.
>>>>>>
>>>>>> Candidate B wins as the only candidate with a pairwise victory over
>>>>>> C. So A's burial of C backfires.
>>>>>>
>>>>>> Typically, the faction A that buries or truncates a Condorcet Winner
>>>>>> C to create a top-cycle cannot by so doing become a pairwise victor over
>>>>>> the buried Condorcet Winner ... but must (in order to create a cycle) help
>>>>>> some other candidate B defeat C by insincerely voting B>C.
>>>>>>
>>>>>> Our Quick and Dirty method insures that if the sincere CW's rightful
>>>>>> victory is subverted, it goes to B, not to A.
>>>>>>
>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>> promoters?
>>>>>>
>>>>>> What objections/criticisms might they have?
>>>>>>
>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>
>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>> show us a better way!
>>>>>>
>>>>>> If not, then they should join us to educate the politicians, public,
>>>>>> and last but not least, the academics still stuck in the pre-EM era!
>>>>>>
>>>>> ----
>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>> list info
>>>>>
>>>>
TS
Ted Stern
Wed, Jan 5, 2022 9:52 PM
Hi Forest,
Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B are
both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.
With approval cutoff at second rank, even Smith//Approval does better.
On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away all
of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Your Q&CBRS works well in a situation in which Approval Sorted Margins
does not: (due to Colin Champion):
Sincere:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
D is the beats-all Condorcet voter.
5 C-first voters bury D:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > B > E > D # Was C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
Pairwise array:
[-- 6. 2. 5. 8.]
[5. -- 4. 9. 9.]
[9. 7. -- 5. 7.]
[6. 2. 6. -- 4.]
[3. 2. 4. 7. --]
Now, if I understand your method correctly, we first find the Smith Set.
In the burial case, it is all 5 candidates, A, B, C, D, E.
Then we find the basic scores for each Smith candidate:
A: 8
B: 11
C: 10
D: 6
E: 9
The Smith candidate with the smallest basic score is D (our previous CW).
Smith candidates that defeat D are B and E. B has highest basic
score, therefore is the winner. C's strategy did not help C.
With Approval Sorted Margins, C is able to win using Burial.
The Basic Score needs to be tabulated separately from pairwise, and
depends on the other rankings on the ballot: each candidate gets a point
for any rating above minimum candidate rating, if using a ratings ballot.
In the case of many candidates, this could lead to basic score ties, since
each ballot would certainly have many candidates rated 0, so in those cases
I would recommend recounting after eliminating all candidates outside the
Smith Set.
I've been thinking about the impracticality of computing pairwise arrays
in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
were used, with rankings inferred, I would recommend a floating score
threshold, starting at 1% of maximum approval, but rising until at most 9
distinct candidate scores are above the threshold (allows score-tie
clusters), then recounting to get the reduced pairwise array and basic
scores. If the lowest basic score is tied, eliminate non-Smith candidates
and recount basic scores.
For elections with 9 or fewer candidates and no lowest-basic score ties
in the Smith set, this is summable, but requires recounts in the event of
more candidates or lowest basic score ties.
Looking back at Colin's burial example, what happens if basic score is
recalculated using ballots scoring candidate X above D?
A: 5
B: 9
C: 5
E: 7
B still wins. So your overall basic score as a proxy for ballots scoring
above the lowest Smith basic score candidate is a good proxy in this case.
On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Let's call it Q&CBRS.
Pre-requisite background:
Candidate X pairwise defeats candidate Y iff candidate X is ranked
above/before/ahead of candidate Y on more ballots than not.
A defeat chain is a sequence of candidates in which each candidate
pairwise defeats the subsequent member of the chain.
Using a "bubble sort" procedure to sort a list of candidates into
pairwise order produces a defeat chain of the listed candidates.
In this way we can easily find a defeat chain that includes all of the
candidates. The first candidate in such a chain is an example of a Smith
candidate. More generally, any candidate who has a defeat chain to any
other candidate is a member of the Smith Set.
Q&CBRS:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with the
greatest basic score.
That's it! Quick & Clean!
Note that if there is only one Smith candidate X, then X will be the
only candidate not defeated by X, and therefore the one elected .
In general a candidate not defeated by X will defeat X, and thereby find
itself at the head of a defeat chain to every other candidate, i.e. it will
be a Smith candidate.
When there is only one Smith candidate, that candidate will not be
pairwise defeated by any other candidate.
I repeat the entire method procedure here:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with the
greatest basic score.
Where else can you find such a simple, quick, and clean election method?
El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Good idea!
Although it seems to me that the highest approval candidate would have
to pairwise beat or tie the approval cutoff candidate X pairwise (which
would be impossible for a non-Smith candidate to do) ...I could be wrong
... and in any case redundancy reinforces communication and understanding.
El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
This sounds like an interesting method to me!
However, I would change the winning criteria to "Elect the most
approved member of the Smith Set".
On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I apologize for the defiant tone at the end of the previous message
... I must have gotten carried away with the "Dirty Dozen"' theme.
But isn't it frustrating to you when people use the 2nd law of
thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
justify their stubborn resistance to any kind of engineering progress?
In the previous message Q&D Burial Resistant Condorcet was formulate
in the typical "stitched together" form ... "Elect the CW if there is one,
Else ..."
In this message I would like to formulate a seamless version:
Let X be the Smith candidate who on the fewest ballots is ranked
ahead of any other Smith candidate. On each ballot approve all candidates
down to X, but include X only when no Smith candidate is ranked ahead of X.
Elect the candidate approved on the most ballots.
This method can be described as electing the approval winner when the
approval cutoff is (at the rank of) the weakest of the Smith candidates,
which itself is approved on (and only on) those ballots which do not
approve any other Smith candidate.
In other words, the approval cutoff is inclusive only when necessary
to ensure approval of at least one member of Smith.
Since a Smith member is approved on every ballot, the method
satisfies the Condorcet Criterion, i.e. it elects the only Smith member
when Smith is a singleton.
How does that grab you?
-FWS
El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Is there any burial resistant Condorcet method simpler than this?
The basic pre-requisite is to understand that whenever there is no
Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
candidates whose members are not defeated by any candidates outside of the
cycle, just as a Condorcet Winner is a candidate undefeated by any other
candidate.
Here's the Q&D burial resistant method:
Lacking a Condorcet Winner elect the candidate X having the greatest
pairwise victory over the top-cycle member Y that has the smallest ratio of
first to last place votes within the top cycle.
Two examples illustrate the method:
Example 1.
49 C
26 A>B
25 B (sincere B>A)
The top cycle is ABCA
Candidate A has the smallest ratio 26/74 of first to last place
votes.
Candidate C is the only candidate with a pairwise victory over it,
so C wins.
Notice how our rule does not reward B for insincerely lowering A to
(equal) last?
Example 2.
45 A>B (sincere A>C)
35 B>C
25 C>A
Candidate C has the smallest ratio 25/45 of first to last.
Candidate B wins as the only candidate with a pairwise victory over
C. So A's burial of C backfires.
Typically, the faction A that buries or truncates a Condorcet Winner
C to create a top-cycle cannot by so doing become a pairwise victor over
the buried Condorcet Winner ... but must (in order to create a cycle) help
some other candidate B defeat C by insincerely voting B>C.
Our Quick and Dirty method insures that if the sincere CW's rightful
victory is subverted, it goes to B, not to A.
Is this method quick enough and dirty enough for the FairVote IRV
promoters?
What objections/criticisms might they have?
Do they have any counter proposal that rivals this one in any way?
If so, let them educate us ... we'll gladly join them if they can
show us a better way!
If not, then they should join us to educate the politicians, public,
and last but not least, the academics still stuck in the pre-EM era!
Hi Forest,
Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B are
both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.
With approval cutoff at second rank, even Smith//Approval does better.
On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <forest.simmons21@gmail.com>
wrote:
> Ted,
>
> Thanks for providing a great example that illustrates how the burial
> defense works.
>
> I agree that it's probably better, especially when there are many
> candidates, to eliminate the non-Smith candidates before counting the basic
> scores.
>
> In the case of public elections for political office we expect the Smith
> set to be small ... usually a singleton, and occasionally a triplet ...
> unless factions think they can get away with burial!
>
> In general, I like ratings information (as in your version of Approval
> Sorted Margins) better than rankings ... the Q&C and Q&D methods were
> created to see what a minimal acceptable rankings based burial resistant
> method would look like.
>
> With ratings and lots of candidates I would go back to Range Based (or
> Total Approval) Chain Climbing for a burial resistant method:
>
> While there is no pairwise undefeated candidate among the remaining
> candidates ... eliminate from the remaining candidates all of those that do
> not pairwise defeat the remaining candidate X that has the lowest Range
> score (or alternately .. lowest below midrange approval score ... with or
> without renormalization as candidates are eliminated).
>
> In the case of a three candidate Smith Set this method first eats away all
> of the non-Smith candidates ... then the lowest score Smith candidate X
> (the one that got buried) and finally Y, the one responsible for burying X,
> leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
> the sole survivor... someone not preferred over X by Y.
>
> It would be interesting to see how this method works on Colin Champion's
> five candidate example... especially to see if the renormalizations are
> worth the trouble ... and if perhaps the below midrange approval scores (or
> the ASM approval scores) work better than range scores.
>
> That's a lot of work! Do you have any students that need a project?
>
> My best,
>
> Forest
>
>
> El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon@gmail.com>
> escribió:
>
>> Your Q&CBRS works well in a situation in which Approval Sorted Margins
>> does not: (due to Colin Champion):
>>
>> Sincere:
>>
>> 1: B > D > A > E > C
>> 1: B > D > E > C > A
>> 5: C > A > D > B > E
>> 1: D > A > C > B > E
>> 1: D > C > B > A > E
>> 2: E > B > D > C > A
>>
>> D is the beats-all Condorcet voter.
>>
>> 5 C-first voters bury D:
>>
>> 1: B > D > A > E > C
>> 1: B > D > E > C > A
>> 5: C > A > B > E > D # Was C > A > D > B > E
>> 1: D > A > C > B > E
>> 1: D > C > B > A > E
>> 2: E > B > D > C > A
>>
>> Pairwise array:
>> [-- 6. 2. 5. 8.]
>> [5. -- 4. 9. 9.]
>> [9. 7. -- 5. 7.]
>> [6. 2. 6. -- 4.]
>> [3. 2. 4. 7. --]
>>
>> Now, if I understand your method correctly, we first find the Smith Set.
>> In the burial case, it is all 5 candidates, A, B, C, D, E.
>>
>> Then we find the basic scores for each Smith candidate:
>>
>> A: 8
>> B: 11
>> C: 10
>> D: 6
>> E: 9
>>
>> The Smith candidate with the smallest basic score is D (our previous CW).
>>
>> Smith candidates that defeat D are *B* and *E*. B has highest basic
>> score, therefore is the winner. C's strategy did not help C.
>>
>> With Approval Sorted Margins, C is able to win using Burial.
>>
>> The Basic Score needs to be tabulated separately from pairwise, and
>> depends on the other rankings on the ballot: each candidate gets a point
>> for any rating above minimum candidate rating, if using a ratings ballot.
>> In the case of many candidates, this could lead to basic score ties, since
>> each ballot would certainly have many candidates rated 0, so in those cases
>> I would recommend recounting after eliminating all candidates outside the
>> Smith Set.
>>
>> I've been thinking about the impracticality of computing pairwise arrays
>> in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
>> were used, with rankings inferred, I would recommend a floating score
>> threshold, starting at 1% of maximum approval, but rising until at most 9
>> distinct candidate scores are above the threshold (allows score-tie
>> clusters), then recounting to get the reduced pairwise array and basic
>> scores. If the lowest basic score is tied, eliminate non-Smith candidates
>> and recount basic scores.
>>
>> *For elections with 9 or fewer candidates and no lowest-basic score ties
>> in the Smith set, this is summable, but requires recounts in the event of
>> more candidates or lowest basic score ties.*
>>
>> Looking back at Colin's burial example, what happens if basic score is
>> recalculated using ballots scoring candidate X above D?
>>
>> A: 5
>> B: 9
>> C: 5
>> E: 7
>>
>> B still wins. So your overall basic score as a proxy for ballots scoring
>> above the lowest Smith basic score candidate is a good proxy in this case.
>>
>> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <forest.simmons21@gmail.com>
>> wrote:
>>
>>> Let's call it Q&CBRS.
>>>
>>> Pre-requisite background:
>>>
>>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>>> above/before/ahead of candidate Y on more ballots than not.
>>>
>>> A defeat chain is a sequence of candidates in which each candidate
>>> pairwise defeats the subsequent member of the chain.
>>>
>>> Using a "bubble sort" procedure to sort a list of candidates into
>>> pairwise order produces a defeat chain of the listed candidates.
>>>
>>> In this way we can easily find a defeat chain that includes all of the
>>> candidates. The first candidate in such a chain is an example of a Smith
>>> candidate. More generally, any candidate who has a defeat chain to any
>>> other candidate is a member of the Smith Set.
>>>
>>> Q&CBRS:
>>>
>>> First, find the "basic score" for each candidate defined as the number
>>> of ballots on which it is ranked above one or more candidates.
>>>
>>> Then let X be the Smith candidate with the smallest basic score.
>>>
>>> Finally, among the candidates not defeated by X, elect the one with the
>>> greatest basic score.
>>>
>>> That's it! Quick & Clean!
>>>
>>> Note that if there is only one Smith candidate X, then X will be the
>>> only candidate not defeated by X, and therefore the one elected .
>>>
>>> In general a candidate not defeated by X will defeat X, and thereby find
>>> itself at the head of a defeat chain to every other candidate, i.e. it will
>>> be a Smith candidate.
>>>
>>> When there is only one Smith candidate, that candidate will not be
>>> pairwise defeated by any other candidate.
>>>
>>> I repeat the entire method procedure here:
>>>
>>> First, find the "basic score" for each candidate defined as the number
>>> of ballots on which it is ranked above one or more candidates.
>>>
>>> Then let X be the Smith candidate with the smallest basic score.
>>>
>>> Finally, among the candidates not defeated by X, elect the one with the
>>> greatest basic score.
>>>
>>> Where else can you find such a simple, quick, and clean election method?
>>>
>>>
>>>
>>>
>>>
>>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>>> forest.simmons21@gmail.com> escribió:
>>>
>>>> Good idea!
>>>>
>>>> Although it seems to me that the highest approval candidate would have
>>>> to pairwise beat or tie the approval cutoff candidate X pairwise (which
>>>> would be impossible for a non-Smith candidate to do) ...I could be wrong
>>>> ... and in any case redundancy reinforces communication and understanding.
>>>>
>>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <dodecatheon@gmail.com>
>>>> escribió:
>>>>
>>>>> Hi Forest,
>>>>>
>>>>> This sounds like an interesting method to me!
>>>>>
>>>>> However, I would change the winning criteria to "Elect the most
>>>>> approved member of the Smith Set".
>>>>>
>>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>>> forest.simmons21@gmail.com> wrote:
>>>>>
>>>>>> I apologize for the defiant tone at the end of the previous message
>>>>>> ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>>
>>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>>
>>>>>> In the previous message Q&D Burial Resistant Condorcet was formulate
>>>>>> in the typical "stitched together" form ... "Elect the CW if there is one,
>>>>>> Else ..."
>>>>>>
>>>>>> In this message I would like to formulate a seamless version:
>>>>>>
>>>>>> Let X be the Smith candidate who on the fewest ballots is ranked
>>>>>> ahead of any other Smith candidate. On each ballot approve all candidates
>>>>>> down to X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>>
>>>>>> Elect the candidate approved on the most ballots.
>>>>>>
>>>>>> This method can be described as electing the approval winner when the
>>>>>> approval cutoff is (at the rank of) the weakest of the Smith candidates,
>>>>>> which itself is approved on (and only on) those ballots which do not
>>>>>> approve any other Smith candidate.
>>>>>>
>>>>>> In other words, the approval cutoff is inclusive only when necessary
>>>>>> to ensure approval of at least one member of Smith.
>>>>>>
>>>>>> Since a Smith member is approved on every ballot, the method
>>>>>> satisfies the Condorcet Criterion, i.e. it elects the only Smith member
>>>>>> when Smith is a singleton.
>>>>>>
>>>>>> How does that grab you?
>>>>>>
>>>>>> -FWS
>>>>>>
>>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>>> forest.simmons21@gmail.com> escribió:
>>>>>>
>>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>>
>>>>>>> The basic pre-requisite is to understand that whenever there is no
>>>>>>> Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>>> candidate.
>>>>>>>
>>>>>>> Here's the Q&D burial resistant method:
>>>>>>>
>>>>>>> Lacking a Condorcet Winner elect the candidate X having the greatest
>>>>>>> pairwise victory over the top-cycle member Y that has the smallest ratio of
>>>>>>> first to last place votes within the top cycle.
>>>>>>>
>>>>>>> Two examples illustrate the method:
>>>>>>>
>>>>>>> Example 1.
>>>>>>>
>>>>>>> 49 C
>>>>>>> 26 A>B
>>>>>>> 25 B (sincere B>A)
>>>>>>>
>>>>>>> The top cycle is ABCA
>>>>>>>
>>>>>>> Candidate A has the smallest ratio 26/74 of first to last place
>>>>>>> votes.
>>>>>>>
>>>>>>> Candidate C is the only candidate with a pairwise victory over it,
>>>>>>> so C wins.
>>>>>>>
>>>>>>> Notice how our rule does not reward B for insincerely lowering A to
>>>>>>> (equal) last?
>>>>>>>
>>>>>>> Example 2.
>>>>>>>
>>>>>>> 45 A>B (sincere A>C)
>>>>>>> 35 B>C
>>>>>>> 25 C>A
>>>>>>>
>>>>>>> Candidate C has the smallest ratio 25/45 of first to last.
>>>>>>>
>>>>>>> Candidate B wins as the only candidate with a pairwise victory over
>>>>>>> C. So A's burial of C backfires.
>>>>>>>
>>>>>>> Typically, the faction A that buries or truncates a Condorcet Winner
>>>>>>> C to create a top-cycle cannot by so doing become a pairwise victor over
>>>>>>> the buried Condorcet Winner ... but must (in order to create a cycle) help
>>>>>>> some other candidate B defeat C by insincerely voting B>C.
>>>>>>>
>>>>>>> Our Quick and Dirty method insures that if the sincere CW's rightful
>>>>>>> victory is subverted, it goes to B, not to A.
>>>>>>>
>>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>>> promoters?
>>>>>>>
>>>>>>> What objections/criticisms might they have?
>>>>>>>
>>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>>
>>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>>> show us a better way!
>>>>>>>
>>>>>>> If not, then they should join us to educate the politicians, public,
>>>>>>> and last but not least, the academics still stuck in the pre-EM era!
>>>>>>>
>>>>>> ----
>>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>>> list info
>>>>>>
>>>>>
FS
Forest Simmons
Thu, Jan 6, 2022 5:15 AM
Ted,
We may need a different example ... let's go through this one slowly and
carefully ...
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A gets 46+5=51 basic points from the first and third factions, resp.
B gets 46+44+5=95 points from the first, second, and fourth factions, resp.
C gets 44+5+5=54 points from the last three factions.
If this is right, A has the lowest score, and C wins as the only candidate
pairwise undefeated by A.
I don't doubt that this method could fail Chicken Defense, but it would
have to be a marginal failure that the attacked faction could easily
counter with a partial counter defection.
For a range of examples let's look at
p: C
q: A>B
r: B (sincere B>A)
Where r+q>p>q>r
A beats B because q>r
B beats C because r+q>p
C beats A because p>q
The respective basic scores for A, B, & C are q, p+q, & p.
The lowest of these is q which is A's score. So C wins, as the only
candidate pairwise undefeated by A.
So the method robustly punishes Chicken defection over a range of standard
examples.
Not bad for Quick and Dirty/Clean!
El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
are both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.
With approval cutoff at second rank, even Smith//Approval does better.
On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away
all of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Your Q&CBRS works well in a situation in which Approval Sorted Margins
does not: (due to Colin Champion):
Sincere:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
D is the beats-all Condorcet voter.
5 C-first voters bury D:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > B > E > D # Was C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
Pairwise array:
[-- 6. 2. 5. 8.]
[5. -- 4. 9. 9.]
[9. 7. -- 5. 7.]
[6. 2. 6. -- 4.]
[3. 2. 4. 7. --]
Now, if I understand your method correctly, we first find the Smith Set.
In the burial case, it is all 5 candidates, A, B, C, D, E.
Then we find the basic scores for each Smith candidate:
A: 8
B: 11
C: 10
D: 6
E: 9
The Smith candidate with the smallest basic score is D (our previous CW).
Smith candidates that defeat D are B and E. B has highest basic
score, therefore is the winner. C's strategy did not help C.
With Approval Sorted Margins, C is able to win using Burial.
The Basic Score needs to be tabulated separately from pairwise, and
depends on the other rankings on the ballot: each candidate gets a point
for any rating above minimum candidate rating, if using a ratings ballot.
In the case of many candidates, this could lead to basic score ties, since
each ballot would certainly have many candidates rated 0, so in those cases
I would recommend recounting after eliminating all candidates outside the
Smith Set.
I've been thinking about the impracticality of computing pairwise arrays
in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
were used, with rankings inferred, I would recommend a floating score
threshold, starting at 1% of maximum approval, but rising until at most 9
distinct candidate scores are above the threshold (allows score-tie
clusters), then recounting to get the reduced pairwise array and basic
scores. If the lowest basic score is tied, eliminate non-Smith candidates
and recount basic scores.
For elections with 9 or fewer candidates and no lowest-basic score ties
in the Smith set, this is summable, but requires recounts in the event of
more candidates or lowest basic score ties.
Looking back at Colin's burial example, what happens if basic score is
recalculated using ballots scoring candidate X above D?
A: 5
B: 9
C: 5
E: 7
B still wins. So your overall basic score as a proxy for ballots scoring
above the lowest Smith basic score candidate is a good proxy in this case.
On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
forest.simmons21@gmail.com> wrote:
Let's call it Q&CBRS.
Pre-requisite background:
Candidate X pairwise defeats candidate Y iff candidate X is ranked
above/before/ahead of candidate Y on more ballots than not.
A defeat chain is a sequence of candidates in which each candidate
pairwise defeats the subsequent member of the chain.
Using a "bubble sort" procedure to sort a list of candidates into
pairwise order produces a defeat chain of the listed candidates.
In this way we can easily find a defeat chain that includes all of the
candidates. The first candidate in such a chain is an example of a Smith
candidate. More generally, any candidate who has a defeat chain to any
other candidate is a member of the Smith Set.
Q&CBRS:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
That's it! Quick & Clean!
Note that if there is only one Smith candidate X, then X will be the
only candidate not defeated by X, and therefore the one elected .
In general a candidate not defeated by X will defeat X, and thereby
find itself at the head of a defeat chain to every other candidate, i.e. it
will be a Smith candidate.
When there is only one Smith candidate, that candidate will not be
pairwise defeated by any other candidate.
I repeat the entire method procedure here:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
Where else can you find such a simple, quick, and clean election method?
El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Good idea!
Although it seems to me that the highest approval candidate would have
to pairwise beat or tie the approval cutoff candidate X pairwise (which
would be impossible for a non-Smith candidate to do) ...I could be wrong
... and in any case redundancy reinforces communication and understanding.
El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
dodecatheon@gmail.com> escribió:
Hi Forest,
This sounds like an interesting method to me!
However, I would change the winning criteria to "Elect the most
approved member of the Smith Set".
On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I apologize for the defiant tone at the end of the previous message
... I must have gotten carried away with the "Dirty Dozen"' theme.
But isn't it frustrating to you when people use the 2nd law of
thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
justify their stubborn resistance to any kind of engineering progress?
In the previous message Q&D Burial Resistant Condorcet was formulate
in the typical "stitched together" form ... "Elect the CW if there is one,
Else ..."
In this message I would like to formulate a seamless version:
Let X be the Smith candidate who on the fewest ballots is ranked
ahead of any other Smith candidate. On each ballot approve all candidates
down to X, but include X only when no Smith candidate is ranked ahead of X.
Elect the candidate approved on the most ballots.
This method can be described as electing the approval winner when
the approval cutoff is (at the rank of) the weakest of the Smith
candidates, which itself is approved on (and only on) those ballots which
do not approve any other Smith candidate.
In other words, the approval cutoff is inclusive only when necessary
to ensure approval of at least one member of Smith.
Since a Smith member is approved on every ballot, the method
satisfies the Condorcet Criterion, i.e. it elects the only Smith member
when Smith is a singleton.
How does that grab you?
-FWS
El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Is there any burial resistant Condorcet method simpler than this?
The basic pre-requisite is to understand that whenever there is no
Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
candidates whose members are not defeated by any candidates outside of the
cycle, just as a Condorcet Winner is a candidate undefeated by any other
candidate.
Here's the Q&D burial resistant method:
Lacking a Condorcet Winner elect the candidate X having the
greatest pairwise victory over the top-cycle member Y that has the smallest
ratio of first to last place votes within the top cycle.
Two examples illustrate the method:
Example 1.
49 C
26 A>B
25 B (sincere B>A)
The top cycle is ABCA
Candidate A has the smallest ratio 26/74 of first to last place
votes.
Candidate C is the only candidate with a pairwise victory over it,
so C wins.
Notice how our rule does not reward B for insincerely lowering A to
(equal) last?
Example 2.
45 A>B (sincere A>C)
35 B>C
25 C>A
Candidate C has the smallest ratio 25/45 of first to last.
Candidate B wins as the only candidate with a pairwise victory over
C. So A's burial of C backfires.
Typically, the faction A that buries or truncates a Condorcet
Winner C to create a top-cycle cannot by so doing become a pairwise victor
over the buried Condorcet Winner ... but must (in order to create a cycle)
help some other candidate B defeat C by insincerely voting B>C.
Our Quick and Dirty method insures that if the sincere CW's
rightful victory is subverted, it goes to B, not to A.
Is this method quick enough and dirty enough for the FairVote IRV
promoters?
What objections/criticisms might they have?
Do they have any counter proposal that rivals this one in any way?
If so, let them educate us ... we'll gladly join them if they can
show us a better way!
If not, then they should join us to educate the politicians,
public, and last but not least, the academics still stuck in the pre-EM era!
Ted,
We may need a different example ... let's go through this one slowly and
carefully ...
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A gets 46+5=51 basic points from the first and third factions, resp.
B gets 46+44+5=95 points from the first, second, and fourth factions, resp.
C gets 44+5+5=54 points from the last three factions.
If this is right, A has the lowest score, and C wins as the only candidate
pairwise undefeated by A.
I don't doubt that this method could fail Chicken Defense, but it would
have to be a marginal failure that the attacked faction could easily
counter with a partial counter defection.
For a range of examples let's look at
p: C
q: A>B
r: B (sincere B>A)
Where r+q>p>q>r
A beats B because q>r
B beats C because r+q>p
C beats A because p>q
The respective basic scores for A, B, & C are q, p+q, & p.
The lowest of these is q which is A's score. So C wins, as the only
candidate pairwise undefeated by A.
So the method robustly punishes Chicken defection over a range of standard
examples.
Not bad for Quick and Dirty/Clean!
El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com>
escribió:
> Hi Forest,
>
> Unfortunately, your new method does not handle Chicken-Dilemma types of
> burial, which I am interested in. See for example Chris Benham's example:
>
> 46: A > B
> 44: B > C (sincere B or B > A)
> 05: C > A
> 05: C > B
>
> A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
> are both lowest basic score. If C is taken as lower, then C defeats A and B
> wins, rewarding burial.
>
> With approval cutoff at second rank, even Smith//Approval does better.
>
> On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <forest.simmons21@gmail.com>
> wrote:
>
>> Ted,
>>
>> Thanks for providing a great example that illustrates how the burial
>> defense works.
>>
>> I agree that it's probably better, especially when there are many
>> candidates, to eliminate the non-Smith candidates before counting the basic
>> scores.
>>
>> In the case of public elections for political office we expect the Smith
>> set to be small ... usually a singleton, and occasionally a triplet ...
>> unless factions think they can get away with burial!
>>
>> In general, I like ratings information (as in your version of Approval
>> Sorted Margins) better than rankings ... the Q&C and Q&D methods were
>> created to see what a minimal acceptable rankings based burial resistant
>> method would look like.
>>
>> With ratings and lots of candidates I would go back to Range Based (or
>> Total Approval) Chain Climbing for a burial resistant method:
>>
>> While there is no pairwise undefeated candidate among the remaining
>> candidates ... eliminate from the remaining candidates all of those that do
>> not pairwise defeat the remaining candidate X that has the lowest Range
>> score (or alternately .. lowest below midrange approval score ... with or
>> without renormalization as candidates are eliminated).
>>
>> In the case of a three candidate Smith Set this method first eats away
>> all of the non-Smith candidates ... then the lowest score Smith candidate X
>> (the one that got buried) and finally Y, the one responsible for burying X,
>> leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
>> the sole survivor... someone not preferred over X by Y.
>>
>> It would be interesting to see how this method works on Colin Champion's
>> five candidate example... especially to see if the renormalizations are
>> worth the trouble ... and if perhaps the below midrange approval scores (or
>> the ASM approval scores) work better than range scores.
>>
>> That's a lot of work! Do you have any students that need a project?
>>
>> My best,
>>
>> Forest
>>
>>
>> El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon@gmail.com>
>> escribió:
>>
>>> Your Q&CBRS works well in a situation in which Approval Sorted Margins
>>> does not: (due to Colin Champion):
>>>
>>> Sincere:
>>>
>>> 1: B > D > A > E > C
>>> 1: B > D > E > C > A
>>> 5: C > A > D > B > E
>>> 1: D > A > C > B > E
>>> 1: D > C > B > A > E
>>> 2: E > B > D > C > A
>>>
>>> D is the beats-all Condorcet voter.
>>>
>>> 5 C-first voters bury D:
>>>
>>> 1: B > D > A > E > C
>>> 1: B > D > E > C > A
>>> 5: C > A > B > E > D # Was C > A > D > B > E
>>> 1: D > A > C > B > E
>>> 1: D > C > B > A > E
>>> 2: E > B > D > C > A
>>>
>>> Pairwise array:
>>> [-- 6. 2. 5. 8.]
>>> [5. -- 4. 9. 9.]
>>> [9. 7. -- 5. 7.]
>>> [6. 2. 6. -- 4.]
>>> [3. 2. 4. 7. --]
>>>
>>> Now, if I understand your method correctly, we first find the Smith Set.
>>> In the burial case, it is all 5 candidates, A, B, C, D, E.
>>>
>>> Then we find the basic scores for each Smith candidate:
>>>
>>> A: 8
>>> B: 11
>>> C: 10
>>> D: 6
>>> E: 9
>>>
>>> The Smith candidate with the smallest basic score is D (our previous CW).
>>>
>>> Smith candidates that defeat D are *B* and *E*. B has highest basic
>>> score, therefore is the winner. C's strategy did not help C.
>>>
>>> With Approval Sorted Margins, C is able to win using Burial.
>>>
>>> The Basic Score needs to be tabulated separately from pairwise, and
>>> depends on the other rankings on the ballot: each candidate gets a point
>>> for any rating above minimum candidate rating, if using a ratings ballot.
>>> In the case of many candidates, this could lead to basic score ties, since
>>> each ballot would certainly have many candidates rated 0, so in those cases
>>> I would recommend recounting after eliminating all candidates outside the
>>> Smith Set.
>>>
>>> I've been thinking about the impracticality of computing pairwise arrays
>>> in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
>>> were used, with rankings inferred, I would recommend a floating score
>>> threshold, starting at 1% of maximum approval, but rising until at most 9
>>> distinct candidate scores are above the threshold (allows score-tie
>>> clusters), then recounting to get the reduced pairwise array and basic
>>> scores. If the lowest basic score is tied, eliminate non-Smith candidates
>>> and recount basic scores.
>>>
>>> *For elections with 9 or fewer candidates and no lowest-basic score ties
>>> in the Smith set, this is summable, but requires recounts in the event of
>>> more candidates or lowest basic score ties.*
>>>
>>> Looking back at Colin's burial example, what happens if basic score is
>>> recalculated using ballots scoring candidate X above D?
>>>
>>> A: 5
>>> B: 9
>>> C: 5
>>> E: 7
>>>
>>> B still wins. So your overall basic score as a proxy for ballots scoring
>>> above the lowest Smith basic score candidate is a good proxy in this case.
>>>
>>> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
>>> forest.simmons21@gmail.com> wrote:
>>>
>>>> Let's call it Q&CBRS.
>>>>
>>>> Pre-requisite background:
>>>>
>>>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>>>> above/before/ahead of candidate Y on more ballots than not.
>>>>
>>>> A defeat chain is a sequence of candidates in which each candidate
>>>> pairwise defeats the subsequent member of the chain.
>>>>
>>>> Using a "bubble sort" procedure to sort a list of candidates into
>>>> pairwise order produces a defeat chain of the listed candidates.
>>>>
>>>> In this way we can easily find a defeat chain that includes all of the
>>>> candidates. The first candidate in such a chain is an example of a Smith
>>>> candidate. More generally, any candidate who has a defeat chain to any
>>>> other candidate is a member of the Smith Set.
>>>>
>>>> Q&CBRS:
>>>>
>>>> First, find the "basic score" for each candidate defined as the number
>>>> of ballots on which it is ranked above one or more candidates.
>>>>
>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>
>>>> Finally, among the candidates not defeated by X, elect the one with
>>>> the greatest basic score.
>>>>
>>>> That's it! Quick & Clean!
>>>>
>>>> Note that if there is only one Smith candidate X, then X will be the
>>>> only candidate not defeated by X, and therefore the one elected .
>>>>
>>>> In general a candidate not defeated by X will defeat X, and thereby
>>>> find itself at the head of a defeat chain to every other candidate, i.e. it
>>>> will be a Smith candidate.
>>>>
>>>> When there is only one Smith candidate, that candidate will not be
>>>> pairwise defeated by any other candidate.
>>>>
>>>> I repeat the entire method procedure here:
>>>>
>>>> First, find the "basic score" for each candidate defined as the number
>>>> of ballots on which it is ranked above one or more candidates.
>>>>
>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>
>>>> Finally, among the candidates not defeated by X, elect the one with
>>>> the greatest basic score.
>>>>
>>>> Where else can you find such a simple, quick, and clean election method?
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>>>> forest.simmons21@gmail.com> escribió:
>>>>
>>>>> Good idea!
>>>>>
>>>>> Although it seems to me that the highest approval candidate would have
>>>>> to pairwise beat or tie the approval cutoff candidate X pairwise (which
>>>>> would be impossible for a non-Smith candidate to do) ...I could be wrong
>>>>> ... and in any case redundancy reinforces communication and understanding.
>>>>>
>>>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
>>>>> dodecatheon@gmail.com> escribió:
>>>>>
>>>>>> Hi Forest,
>>>>>>
>>>>>> This sounds like an interesting method to me!
>>>>>>
>>>>>> However, I would change the winning criteria to "Elect the most
>>>>>> approved member of the Smith Set".
>>>>>>
>>>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>>>> forest.simmons21@gmail.com> wrote:
>>>>>>
>>>>>>> I apologize for the defiant tone at the end of the previous message
>>>>>>> ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>>>
>>>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>>>
>>>>>>> In the previous message Q&D Burial Resistant Condorcet was formulate
>>>>>>> in the typical "stitched together" form ... "Elect the CW if there is one,
>>>>>>> Else ..."
>>>>>>>
>>>>>>> In this message I would like to formulate a seamless version:
>>>>>>>
>>>>>>> Let X be the Smith candidate who on the fewest ballots is ranked
>>>>>>> ahead of any other Smith candidate. On each ballot approve all candidates
>>>>>>> down to X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>>>
>>>>>>> Elect the candidate approved on the most ballots.
>>>>>>>
>>>>>>> This method can be described as electing the approval winner when
>>>>>>> the approval cutoff is (at the rank of) the weakest of the Smith
>>>>>>> candidates, which itself is approved on (and only on) those ballots which
>>>>>>> do not approve any other Smith candidate.
>>>>>>>
>>>>>>> In other words, the approval cutoff is inclusive only when necessary
>>>>>>> to ensure approval of at least one member of Smith.
>>>>>>>
>>>>>>> Since a Smith member is approved on every ballot, the method
>>>>>>> satisfies the Condorcet Criterion, i.e. it elects the only Smith member
>>>>>>> when Smith is a singleton.
>>>>>>>
>>>>>>> How does that grab you?
>>>>>>>
>>>>>>> -FWS
>>>>>>>
>>>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>>>> forest.simmons21@gmail.com> escribió:
>>>>>>>
>>>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>>>
>>>>>>>> The basic pre-requisite is to understand that whenever there is no
>>>>>>>> Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>>>> candidate.
>>>>>>>>
>>>>>>>> Here's the Q&D burial resistant method:
>>>>>>>>
>>>>>>>> Lacking a Condorcet Winner elect the candidate X having the
>>>>>>>> greatest pairwise victory over the top-cycle member Y that has the smallest
>>>>>>>> ratio of first to last place votes within the top cycle.
>>>>>>>>
>>>>>>>> Two examples illustrate the method:
>>>>>>>>
>>>>>>>> Example 1.
>>>>>>>>
>>>>>>>> 49 C
>>>>>>>> 26 A>B
>>>>>>>> 25 B (sincere B>A)
>>>>>>>>
>>>>>>>> The top cycle is ABCA
>>>>>>>>
>>>>>>>> Candidate A has the smallest ratio 26/74 of first to last place
>>>>>>>> votes.
>>>>>>>>
>>>>>>>> Candidate C is the only candidate with a pairwise victory over it,
>>>>>>>> so C wins.
>>>>>>>>
>>>>>>>> Notice how our rule does not reward B for insincerely lowering A to
>>>>>>>> (equal) last?
>>>>>>>>
>>>>>>>> Example 2.
>>>>>>>>
>>>>>>>> 45 A>B (sincere A>C)
>>>>>>>> 35 B>C
>>>>>>>> 25 C>A
>>>>>>>>
>>>>>>>> Candidate C has the smallest ratio 25/45 of first to last.
>>>>>>>>
>>>>>>>> Candidate B wins as the only candidate with a pairwise victory over
>>>>>>>> C. So A's burial of C backfires.
>>>>>>>>
>>>>>>>> Typically, the faction A that buries or truncates a Condorcet
>>>>>>>> Winner C to create a top-cycle cannot by so doing become a pairwise victor
>>>>>>>> over the buried Condorcet Winner ... but must (in order to create a cycle)
>>>>>>>> help some other candidate B defeat C by insincerely voting B>C.
>>>>>>>>
>>>>>>>> Our Quick and Dirty method insures that if the sincere CW's
>>>>>>>> rightful victory is subverted, it goes to B, not to A.
>>>>>>>>
>>>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>>>> promoters?
>>>>>>>>
>>>>>>>> What objections/criticisms might they have?
>>>>>>>>
>>>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>>>
>>>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>>>> show us a better way!
>>>>>>>>
>>>>>>>> If not, then they should join us to educate the politicians,
>>>>>>>> public, and last but not least, the academics still stuck in the pre-EM era!
>>>>>>>>
>>>>>>> ----
>>>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>>>> list info
>>>>>>>
>>>>>>
TS
Ted Stern
Thu, Jan 6, 2022 10:51 PM
Hi Forest,
thanks for correcting my error. I agree that B is not able to successfully
win this chicken dilemma scenario by defecting and using burial.
Overall, this has the flavor of a Smith//Approval variant, replacing
Approval with "Highest implicitly approved [Smith] candidate [HIASC]
defeating lowest implicitly approved Smith candidate [LIASC]".
To flesh it out, I'd like more details on the following:
- What is the motivation/rationale for excluding Smith candidates that
tie or are defeated by the LIASC? What makes the HIASC "better"? In the
most recent example, it is not clear to me that C is better than A, and
letting C win is merely supposed to provide motivation for B voters to not
defect and bury.
- What happens in the event of ties in lowest implicit approval among
Smith Candidates?
- Are there other tie situations to handle?
Other examples:
(Due to Chris Benham)
25 A>B
26 B>C
23 C>A
26 C
Basic scores:
A: 49, B: 51, C:75
A lowest basic score (implicit approval / above bottom) and defeats B;
C > A, therefore C wins Q&CBRS
C is the most top ranked and the most above-bottom ranked candidate.
WV, MMPO, IRV, Benham elect B
35 A
10 A=B
30 B>C
25 C
Basic scores: A:45, B40, C55. B is lowest basic score.
C55 > A45, A35>B30, B40>C25
C is defeated by B. A is remaining candidate defeating B, thus wins Q&CBRS.
A both pairwise-beats and positionally dominates B, but WV, Margins,
MMPO all elect B.
This is somewhat encouraging, however, I'd still like to get better
motivation for the earlier example as in the bullet point questions
above.
On Wed, Jan 5, 2022 at 9:15 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Ted,
We may need a different example ... let's go through this one slowly and
carefully ...
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A gets 46+5=51 basic points from the first and third factions, resp.
B gets 46+44+5=95 points from the first, second, and fourth factions, resp.
C gets 44+5+5=54 points from the last three factions.
If this is right, A has the lowest score, and C wins as the only candidate
pairwise undefeated by A.
I don't doubt that this method could fail Chicken Defense, but it would
have to be a marginal failure that the attacked faction could easily
counter with a partial counter defection.
For a range of examples let's look at
p: C
q: A>B
r: B (sincere B>A)
Where r+q>p>q>r
A beats B because q>r
B beats C because r+q>p
C beats A because p>q
The respective basic scores for A, B, & C are q, p+q, & p.
The lowest of these is q which is A's score. So C wins, as the only
candidate pairwise undefeated by A.
So the method robustly punishes Chicken defection over a range of standard
examples.
Not bad for Quick and Dirty/Clean!
El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
are both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.
With approval cutoff at second rank, even Smith//Approval does better.
On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away
all of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Your Q&CBRS works well in a situation in which Approval Sorted Margins
does not: (due to Colin Champion):
Sincere:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
D is the beats-all Condorcet voter.
5 C-first voters bury D:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > B > E > D # Was C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
Pairwise array:
[-- 6. 2. 5. 8.]
[5. -- 4. 9. 9.]
[9. 7. -- 5. 7.]
[6. 2. 6. -- 4.]
[3. 2. 4. 7. --]
Now, if I understand your method correctly, we first find the Smith
Set. In the burial case, it is all 5 candidates, A, B, C, D, E.
Then we find the basic scores for each Smith candidate:
A: 8
B: 11
C: 10
D: 6
E: 9
The Smith candidate with the smallest basic score is D (our previous
CW).
Smith candidates that defeat D are B and E. B has highest basic
score, therefore is the winner. C's strategy did not help C.
With Approval Sorted Margins, C is able to win using Burial.
The Basic Score needs to be tabulated separately from pairwise, and
depends on the other rankings on the ballot: each candidate gets a point
for any rating above minimum candidate rating, if using a ratings ballot.
In the case of many candidates, this could lead to basic score ties, since
each ballot would certainly have many candidates rated 0, so in those cases
I would recommend recounting after eliminating all candidates outside the
Smith Set.
I've been thinking about the impracticality of computing pairwise
arrays in "jungle" elections, those with, say, >9 candidates. If a ratings
ballot were used, with rankings inferred, I would recommend a floating
score threshold, starting at 1% of maximum approval, but rising until at
most 9 distinct candidate scores are above the threshold (allows score-tie
clusters), then recounting to get the reduced pairwise array and basic
scores. If the lowest basic score is tied, eliminate non-Smith candidates
and recount basic scores.
For elections with 9 or fewer candidates and no lowest-basic score
ties in the Smith set, this is summable, but requires recounts in the event
of more candidates or lowest basic score ties.
Looking back at Colin's burial example, what happens if basic score is
recalculated using ballots scoring candidate X above D?
A: 5
B: 9
C: 5
E: 7
B still wins. So your overall basic score as a proxy for ballots
scoring above the lowest Smith basic score candidate is a good proxy in
this case.
On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
forest.simmons21@gmail.com> wrote:
Let's call it Q&CBRS.
Pre-requisite background:
Candidate X pairwise defeats candidate Y iff candidate X is ranked
above/before/ahead of candidate Y on more ballots than not.
A defeat chain is a sequence of candidates in which each candidate
pairwise defeats the subsequent member of the chain.
Using a "bubble sort" procedure to sort a list of candidates into
pairwise order produces a defeat chain of the listed candidates.
In this way we can easily find a defeat chain that includes all of the
candidates. The first candidate in such a chain is an example of a Smith
candidate. More generally, any candidate who has a defeat chain to any
other candidate is a member of the Smith Set.
Q&CBRS:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
That's it! Quick & Clean!
Note that if there is only one Smith candidate X, then X will be the
only candidate not defeated by X, and therefore the one elected .
In general a candidate not defeated by X will defeat X, and thereby
find itself at the head of a defeat chain to every other candidate, i.e. it
will be a Smith candidate.
When there is only one Smith candidate, that candidate will not be
pairwise defeated by any other candidate.
I repeat the entire method procedure here:
First, find the "basic score" for each candidate defined as the number
of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
Where else can you find such a simple, quick, and clean election
method?
El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Good idea!
Although it seems to me that the highest approval candidate would
have to pairwise beat or tie the approval cutoff candidate X pairwise
(which would be impossible for a non-Smith candidate to do) ...I could be
wrong ... and in any case redundancy reinforces communication and
understanding.
El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
dodecatheon@gmail.com> escribió:
Hi Forest,
This sounds like an interesting method to me!
However, I would change the winning criteria to "Elect the most
approved member of the Smith Set".
On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I apologize for the defiant tone at the end of the previous message
... I must have gotten carried away with the "Dirty Dozen"' theme.
But isn't it frustrating to you when people use the 2nd law of
thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
justify their stubborn resistance to any kind of engineering progress?
In the previous message Q&D Burial Resistant Condorcet was
formulate in the typical "stitched together" form ... "Elect the CW if
there is one, Else ..."
In this message I would like to formulate a seamless version:
Let X be the Smith candidate who on the fewest ballots is ranked
ahead of any other Smith candidate. On each ballot approve all candidates
down to X, but include X only when no Smith candidate is ranked ahead of X.
Elect the candidate approved on the most ballots.
This method can be described as electing the approval winner when
the approval cutoff is (at the rank of) the weakest of the Smith
candidates, which itself is approved on (and only on) those ballots which
do not approve any other Smith candidate.
In other words, the approval cutoff is inclusive only when
necessary to ensure approval of at least one member of Smith.
Since a Smith member is approved on every ballot, the method
satisfies the Condorcet Criterion, i.e. it elects the only Smith member
when Smith is a singleton.
How does that grab you?
-FWS
El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Is there any burial resistant Condorcet method simpler than this?
The basic pre-requisite is to understand that whenever there is no
Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
candidates whose members are not defeated by any candidates outside of the
cycle, just as a Condorcet Winner is a candidate undefeated by any other
candidate.
Here's the Q&D burial resistant method:
Lacking a Condorcet Winner elect the candidate X having the
greatest pairwise victory over the top-cycle member Y that has the smallest
ratio of first to last place votes within the top cycle.
Two examples illustrate the method:
Example 1.
49 C
26 A>B
25 B (sincere B>A)
The top cycle is ABCA
Candidate A has the smallest ratio 26/74 of first to last place
votes.
Candidate C is the only candidate with a pairwise victory over it,
so C wins.
Notice how our rule does not reward B for insincerely lowering A
to (equal) last?
Example 2.
45 A>B (sincere A>C)
35 B>C
25 C>A
Candidate C has the smallest ratio 25/45 of first to last.
Candidate B wins as the only candidate with a pairwise victory
over C. So A's burial of C backfires.
Typically, the faction A that buries or truncates a Condorcet
Winner C to create a top-cycle cannot by so doing become a pairwise victor
over the buried Condorcet Winner ... but must (in order to create a cycle)
help some other candidate B defeat C by insincerely voting B>C.
Our Quick and Dirty method insures that if the sincere CW's
rightful victory is subverted, it goes to B, not to A.
Is this method quick enough and dirty enough for the FairVote IRV
promoters?
What objections/criticisms might they have?
Do they have any counter proposal that rivals this one in any way?
If so, let them educate us ... we'll gladly join them if they can
show us a better way!
If not, then they should join us to educate the politicians,
public, and last but not least, the academics still stuck in the pre-EM era!
Hi Forest,
thanks for correcting my error. I agree that B is not able to successfully
win this chicken dilemma scenario by defecting and using burial.
Overall, this has the flavor of a Smith//Approval variant, replacing
Approval with "Highest implicitly approved [Smith] candidate [HIASC]
defeating lowest implicitly approved Smith candidate [LIASC]".
To flesh it out, I'd like more details on the following:
- What is the motivation/rationale for excluding Smith candidates that
tie or are defeated by the LIASC? What makes the HIASC "better"? In the
most recent example, it is not clear to me that C is better than A, and
letting C win is merely supposed to provide motivation for B voters to not
defect and bury.
- What happens in the event of ties in lowest implicit approval among
Smith Candidates?
- Are there other tie situations to handle?
Other examples:
-----------------
(Due to Chris Benham)
25 A>B
26 B>C
23 C>A
26 C
Basic scores:
A: 49, B: 51, C:75
A lowest basic score (implicit approval / above bottom) and defeats B;
C > A, therefore C wins Q&CBRS
C is the most top ranked and the most above-bottom ranked candidate.
WV, MMPO, IRV, Benham elect B
-----------------
35 A
10 A=B
30 B>C
25 C
Basic scores: A:45, B40, C55. B is lowest basic score.
C55 > A45, A35>B30, B40>C25
C is defeated by B. A is remaining candidate defeating B, thus wins Q&CBRS.
A both pairwise-beats and positionally dominates B, but WV, Margins,
MMPO all elect B.
This is somewhat encouraging, however, I'd still like to get better
motivation for the earlier example as in the bullet point questions
above.
On Wed, Jan 5, 2022 at 9:15 PM Forest Simmons <forest.simmons21@gmail.com>
wrote:
> Ted,
>
> We may need a different example ... let's go through this one slowly and
> carefully ...
> 46: A > B
> 44: B > C (sincere B or B > A)
> 05: C > A
> 05: C > B
>
> A gets 46+5=51 basic points from the first and third factions, resp.
>
> B gets 46+44+5=95 points from the first, second, and fourth factions, resp.
>
> C gets 44+5+5=54 points from the last three factions.
>
> If this is right, A has the lowest score, and C wins as the only candidate
> pairwise undefeated by A.
>
> I don't doubt that this method could fail Chicken Defense, but it would
> have to be a marginal failure that the attacked faction could easily
> counter with a partial counter defection.
>
> For a range of examples let's look at
>
> p: C
> q: A>B
> r: B (sincere B>A)
>
> Where r+q>p>q>r
>
> A beats B because q>r
> B beats C because r+q>p
> C beats A because p>q
>
> The respective basic scores for A, B, & C are q, p+q, & p.
>
> The lowest of these is q which is A's score. So C wins, as the only
> candidate pairwise undefeated by A.
>
> So the method robustly punishes Chicken defection over a range of standard
> examples.
>
> Not bad for Quick and Dirty/Clean!
>
>
>
>
> El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com>
> escribió:
>
>> Hi Forest,
>>
>> Unfortunately, your new method does not handle Chicken-Dilemma types of
>> burial, which I am interested in. See for example Chris Benham's example:
>>
>> 46: A > B
>> 44: B > C (sincere B or B > A)
>> 05: C > A
>> 05: C > B
>>
>> A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
>> are both lowest basic score. If C is taken as lower, then C defeats A and B
>> wins, rewarding burial.
>>
>> With approval cutoff at second rank, even Smith//Approval does better.
>>
>> On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <forest.simmons21@gmail.com>
>> wrote:
>>
>>> Ted,
>>>
>>> Thanks for providing a great example that illustrates how the burial
>>> defense works.
>>>
>>> I agree that it's probably better, especially when there are many
>>> candidates, to eliminate the non-Smith candidates before counting the basic
>>> scores.
>>>
>>> In the case of public elections for political office we expect the Smith
>>> set to be small ... usually a singleton, and occasionally a triplet ...
>>> unless factions think they can get away with burial!
>>>
>>> In general, I like ratings information (as in your version of Approval
>>> Sorted Margins) better than rankings ... the Q&C and Q&D methods were
>>> created to see what a minimal acceptable rankings based burial resistant
>>> method would look like.
>>>
>>> With ratings and lots of candidates I would go back to Range Based (or
>>> Total Approval) Chain Climbing for a burial resistant method:
>>>
>>> While there is no pairwise undefeated candidate among the remaining
>>> candidates ... eliminate from the remaining candidates all of those that do
>>> not pairwise defeat the remaining candidate X that has the lowest Range
>>> score (or alternately .. lowest below midrange approval score ... with or
>>> without renormalization as candidates are eliminated).
>>>
>>> In the case of a three candidate Smith Set this method first eats away
>>> all of the non-Smith candidates ... then the lowest score Smith candidate X
>>> (the one that got buried) and finally Y, the one responsible for burying X,
>>> leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
>>> the sole survivor... someone not preferred over X by Y.
>>>
>>> It would be interesting to see how this method works on Colin Champion's
>>> five candidate example... especially to see if the renormalizations are
>>> worth the trouble ... and if perhaps the below midrange approval scores (or
>>> the ASM approval scores) work better than range scores.
>>>
>>> That's a lot of work! Do you have any students that need a project?
>>>
>>> My best,
>>>
>>> Forest
>>>
>>>
>>> El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon@gmail.com>
>>> escribió:
>>>
>>>> Your Q&CBRS works well in a situation in which Approval Sorted Margins
>>>> does not: (due to Colin Champion):
>>>>
>>>> Sincere:
>>>>
>>>> 1: B > D > A > E > C
>>>> 1: B > D > E > C > A
>>>> 5: C > A > D > B > E
>>>> 1: D > A > C > B > E
>>>> 1: D > C > B > A > E
>>>> 2: E > B > D > C > A
>>>>
>>>> D is the beats-all Condorcet voter.
>>>>
>>>> 5 C-first voters bury D:
>>>>
>>>> 1: B > D > A > E > C
>>>> 1: B > D > E > C > A
>>>> 5: C > A > B > E > D # Was C > A > D > B > E
>>>> 1: D > A > C > B > E
>>>> 1: D > C > B > A > E
>>>> 2: E > B > D > C > A
>>>>
>>>> Pairwise array:
>>>> [-- 6. 2. 5. 8.]
>>>> [5. -- 4. 9. 9.]
>>>> [9. 7. -- 5. 7.]
>>>> [6. 2. 6. -- 4.]
>>>> [3. 2. 4. 7. --]
>>>>
>>>> Now, if I understand your method correctly, we first find the Smith
>>>> Set. In the burial case, it is all 5 candidates, A, B, C, D, E.
>>>>
>>>> Then we find the basic scores for each Smith candidate:
>>>>
>>>> A: 8
>>>> B: 11
>>>> C: 10
>>>> D: 6
>>>> E: 9
>>>>
>>>> The Smith candidate with the smallest basic score is D (our previous
>>>> CW).
>>>>
>>>> Smith candidates that defeat D are *B* and *E*. B has highest basic
>>>> score, therefore is the winner. C's strategy did not help C.
>>>>
>>>> With Approval Sorted Margins, C is able to win using Burial.
>>>>
>>>> The Basic Score needs to be tabulated separately from pairwise, and
>>>> depends on the other rankings on the ballot: each candidate gets a point
>>>> for any rating above minimum candidate rating, if using a ratings ballot.
>>>> In the case of many candidates, this could lead to basic score ties, since
>>>> each ballot would certainly have many candidates rated 0, so in those cases
>>>> I would recommend recounting after eliminating all candidates outside the
>>>> Smith Set.
>>>>
>>>> I've been thinking about the impracticality of computing pairwise
>>>> arrays in "jungle" elections, those with, say, >9 candidates. If a ratings
>>>> ballot were used, with rankings inferred, I would recommend a floating
>>>> score threshold, starting at 1% of maximum approval, but rising until at
>>>> most 9 distinct candidate scores are above the threshold (allows score-tie
>>>> clusters), then recounting to get the reduced pairwise array and basic
>>>> scores. If the lowest basic score is tied, eliminate non-Smith candidates
>>>> and recount basic scores.
>>>>
>>>> *For elections with 9 or fewer candidates and no lowest-basic score
>>>> ties in the Smith set, this is summable, but requires recounts in the event
>>>> of more candidates or lowest basic score ties.*
>>>>
>>>> Looking back at Colin's burial example, what happens if basic score is
>>>> recalculated using ballots scoring candidate X above D?
>>>>
>>>> A: 5
>>>> B: 9
>>>> C: 5
>>>> E: 7
>>>>
>>>> B still wins. So your overall basic score as a proxy for ballots
>>>> scoring above the lowest Smith basic score candidate is a good proxy in
>>>> this case.
>>>>
>>>> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
>>>> forest.simmons21@gmail.com> wrote:
>>>>
>>>>> Let's call it Q&CBRS.
>>>>>
>>>>> Pre-requisite background:
>>>>>
>>>>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>>>>> above/before/ahead of candidate Y on more ballots than not.
>>>>>
>>>>> A defeat chain is a sequence of candidates in which each candidate
>>>>> pairwise defeats the subsequent member of the chain.
>>>>>
>>>>> Using a "bubble sort" procedure to sort a list of candidates into
>>>>> pairwise order produces a defeat chain of the listed candidates.
>>>>>
>>>>> In this way we can easily find a defeat chain that includes all of the
>>>>> candidates. The first candidate in such a chain is an example of a Smith
>>>>> candidate. More generally, any candidate who has a defeat chain to any
>>>>> other candidate is a member of the Smith Set.
>>>>>
>>>>> Q&CBRS:
>>>>>
>>>>> First, find the "basic score" for each candidate defined as the number
>>>>> of ballots on which it is ranked above one or more candidates.
>>>>>
>>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>>
>>>>> Finally, among the candidates not defeated by X, elect the one with
>>>>> the greatest basic score.
>>>>>
>>>>> That's it! Quick & Clean!
>>>>>
>>>>> Note that if there is only one Smith candidate X, then X will be the
>>>>> only candidate not defeated by X, and therefore the one elected .
>>>>>
>>>>> In general a candidate not defeated by X will defeat X, and thereby
>>>>> find itself at the head of a defeat chain to every other candidate, i.e. it
>>>>> will be a Smith candidate.
>>>>>
>>>>> When there is only one Smith candidate, that candidate will not be
>>>>> pairwise defeated by any other candidate.
>>>>>
>>>>> I repeat the entire method procedure here:
>>>>>
>>>>> First, find the "basic score" for each candidate defined as the number
>>>>> of ballots on which it is ranked above one or more candidates.
>>>>>
>>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>>
>>>>> Finally, among the candidates not defeated by X, elect the one with
>>>>> the greatest basic score.
>>>>>
>>>>> Where else can you find such a simple, quick, and clean election
>>>>> method?
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>>>>> forest.simmons21@gmail.com> escribió:
>>>>>
>>>>>> Good idea!
>>>>>>
>>>>>> Although it seems to me that the highest approval candidate would
>>>>>> have to pairwise beat or tie the approval cutoff candidate X pairwise
>>>>>> (which would be impossible for a non-Smith candidate to do) ...I could be
>>>>>> wrong ... and in any case redundancy reinforces communication and
>>>>>> understanding.
>>>>>>
>>>>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
>>>>>> dodecatheon@gmail.com> escribió:
>>>>>>
>>>>>>> Hi Forest,
>>>>>>>
>>>>>>> This sounds like an interesting method to me!
>>>>>>>
>>>>>>> However, I would change the winning criteria to "Elect the most
>>>>>>> approved member of the Smith Set".
>>>>>>>
>>>>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>>>>> forest.simmons21@gmail.com> wrote:
>>>>>>>
>>>>>>>> I apologize for the defiant tone at the end of the previous message
>>>>>>>> ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>>>>
>>>>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>>>>
>>>>>>>> In the previous message Q&D Burial Resistant Condorcet was
>>>>>>>> formulate in the typical "stitched together" form ... "Elect the CW if
>>>>>>>> there is one, Else ..."
>>>>>>>>
>>>>>>>> In this message I would like to formulate a seamless version:
>>>>>>>>
>>>>>>>> Let X be the Smith candidate who on the fewest ballots is ranked
>>>>>>>> ahead of any other Smith candidate. On each ballot approve all candidates
>>>>>>>> down to X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>>>>
>>>>>>>> Elect the candidate approved on the most ballots.
>>>>>>>>
>>>>>>>> This method can be described as electing the approval winner when
>>>>>>>> the approval cutoff is (at the rank of) the weakest of the Smith
>>>>>>>> candidates, which itself is approved on (and only on) those ballots which
>>>>>>>> do not approve any other Smith candidate.
>>>>>>>>
>>>>>>>> In other words, the approval cutoff is inclusive only when
>>>>>>>> necessary to ensure approval of at least one member of Smith.
>>>>>>>>
>>>>>>>> Since a Smith member is approved on every ballot, the method
>>>>>>>> satisfies the Condorcet Criterion, i.e. it elects the only Smith member
>>>>>>>> when Smith is a singleton.
>>>>>>>>
>>>>>>>> How does that grab you?
>>>>>>>>
>>>>>>>> -FWS
>>>>>>>>
>>>>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>>>>> forest.simmons21@gmail.com> escribió:
>>>>>>>>
>>>>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>>>>
>>>>>>>>> The basic pre-requisite is to understand that whenever there is no
>>>>>>>>> Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>>>>> candidate.
>>>>>>>>>
>>>>>>>>> Here's the Q&D burial resistant method:
>>>>>>>>>
>>>>>>>>> Lacking a Condorcet Winner elect the candidate X having the
>>>>>>>>> greatest pairwise victory over the top-cycle member Y that has the smallest
>>>>>>>>> ratio of first to last place votes within the top cycle.
>>>>>>>>>
>>>>>>>>> Two examples illustrate the method:
>>>>>>>>>
>>>>>>>>> Example 1.
>>>>>>>>>
>>>>>>>>> 49 C
>>>>>>>>> 26 A>B
>>>>>>>>> 25 B (sincere B>A)
>>>>>>>>>
>>>>>>>>> The top cycle is ABCA
>>>>>>>>>
>>>>>>>>> Candidate A has the smallest ratio 26/74 of first to last place
>>>>>>>>> votes.
>>>>>>>>>
>>>>>>>>> Candidate C is the only candidate with a pairwise victory over it,
>>>>>>>>> so C wins.
>>>>>>>>>
>>>>>>>>> Notice how our rule does not reward B for insincerely lowering A
>>>>>>>>> to (equal) last?
>>>>>>>>>
>>>>>>>>> Example 2.
>>>>>>>>>
>>>>>>>>> 45 A>B (sincere A>C)
>>>>>>>>> 35 B>C
>>>>>>>>> 25 C>A
>>>>>>>>>
>>>>>>>>> Candidate C has the smallest ratio 25/45 of first to last.
>>>>>>>>>
>>>>>>>>> Candidate B wins as the only candidate with a pairwise victory
>>>>>>>>> over C. So A's burial of C backfires.
>>>>>>>>>
>>>>>>>>> Typically, the faction A that buries or truncates a Condorcet
>>>>>>>>> Winner C to create a top-cycle cannot by so doing become a pairwise victor
>>>>>>>>> over the buried Condorcet Winner ... but must (in order to create a cycle)
>>>>>>>>> help some other candidate B defeat C by insincerely voting B>C.
>>>>>>>>>
>>>>>>>>> Our Quick and Dirty method insures that if the sincere CW's
>>>>>>>>> rightful victory is subverted, it goes to B, not to A.
>>>>>>>>>
>>>>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>>>>> promoters?
>>>>>>>>>
>>>>>>>>> What objections/criticisms might they have?
>>>>>>>>>
>>>>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>>>>
>>>>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>>>>> show us a better way!
>>>>>>>>>
>>>>>>>>> If not, then they should join us to educate the politicians,
>>>>>>>>> public, and last but not least, the academics still stuck in the pre-EM era!
>>>>>>>>>
>>>>>>>> ----
>>>>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>>>>> list info
>>>>>>>>
>>>>>>>
FS
Forest Simmons
Fri, Jan 7, 2022 6:05 AM
Most designers of Condorcet methods asume that the gentlemanly thing to do
is to give the votes a benefit of a doubt and assume that they must have
voted sincerely but cycles are a result of errores of judgement.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
I used to think that way, too.
But seasoned election observers are of the opinion that the vast majority
(more than 90 percent) of public elections for political office have a
sincere Condorcet candidate, and that when there is a defeat cycle, it is
more likely to be the result of intentional subversion of the Condorcet
candidate than of erroneous voter judgment.
This means that game theoretic considerations are more important than
statistical error correction considerations for resolving cycles.
So the point of Q&D/C is not to resolve non-existent honest cycles but to
irradicate dishonest cycles ... the ones created by insincere manipualtions
... the vast majority of cycles.
If less than ten percent of ballots currently have Smith cycles, and 90
percent of those are insincere cycles created opportunistically, then game
theoretic punishment of that kind of manipulation would make cycles
extremely rare ... less that 0.1 percent of elections.
What about the rare sincere cycle, not intentionally created by subversion
of the CW?
Well tha winner will be the member of the Smith set not beaten pairwise by
the weakest member of the Smith set.
Think of it this way: if the lowest approval Smith candidate X pairwise
beats Y, then what does Y have to brag about?
In fact, many people think of a defeat cycle as a kind of tie that could be
broken randomly.
The trouble with that idea is that it gives incentive to gamers to subvert
the CW in exchange for a positive chance of winning out right. It does not
help eradicate insincere cycles.
In sum, that's why the popular Condorcet methods like RP, MinMax, CSSD,
River, etc. do not thwart burial or Chicken attacks.
ASP, on the other hand, has the advantage of extra approval information not
available in ordinal ballots.
Q&D/C has the advantage of putting a fence at the top of the cliff instead
of an ambulance at the bottom
Now remember, I never pretended that C&D/C was a better method than ASM.
On the contrary, ASM is better because it gives the voters an extra lever
of expression/control, the cardinal/approval intensity dimension.
But it may well be the best Universal Domain method ... which is what I was
aiming for ... what can we achieve under the UD handicap?
Does that make sense?
El jue., 6 de ene. de 2022 2:51 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
thanks for correcting my error. I agree that B is not able to successfully
win this chicken dilemma scenario by defecting and using burial.
Overall, this has the flavor of a Smith//Approval variant, replacing
Approval with "Highest implicitly approved [Smith] candidate [HIASC]
defeating lowest implicitly approved Smith candidate [LIASC]".
To flesh it out, I'd like more details on the following:
- What is the motivation/rationale for excluding Smith candidates that
tie or are defeated by the LIASC? What makes the HIASC "better"? In the
most recent example, it is not clear to me that C is better than A, and
letting C win is merely supposed to provide motivation for B voters to not
defect and bury.
- What happens in the event of ties in lowest implicit approval among
Smith Candidates?
- Are there other tie situations to handle?
Other examples:
(Due to Chris Benham)
25 A>B
26 B>C
23 C>A
26 C
Basic scores:
A: 49, B: 51, C:75
A lowest basic score (implicit approval / above bottom) and defeats B;
C > A, therefore C wins Q&CBRS
C is the most top ranked and the most above-bottom ranked candidate.
WV, MMPO, IRV, Benham elect B
35 A
10 A=B
30 B>C
25 C
Basic scores: A:45, B40, C55. B is lowest basic score.
C55 > A45, A35>B30, B40>C25
C is defeated by B. A is remaining candidate defeating B, thus wins Q&CBRS.
A both pairwise-beats and positionally dominates B, but WV, Margins, MMPO all elect B.
This is somewhat encouraging, however, I'd still like to get better motivation for the earlier example as in the bullet point questions above.
On Wed, Jan 5, 2022 at 9:15 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Ted,
We may need a different example ... let's go through this one slowly and
carefully ...
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A gets 46+5=51 basic points from the first and third factions, resp.
B gets 46+44+5=95 points from the first, second, and fourth factions,
resp.
C gets 44+5+5=54 points from the last three factions.
If this is right, A has the lowest score, and C wins as the only
candidate pairwise undefeated by A.
I don't doubt that this method could fail Chicken Defense, but it would
have to be a marginal failure that the attacked faction could easily
counter with a partial counter defection.
For a range of examples let's look at
p: C
q: A>B
r: B (sincere B>A)
Where r+q>p>q>r
A beats B because q>r
B beats C because r+q>p
C beats A because p>q
The respective basic scores for A, B, & C are q, p+q, & p.
The lowest of these is q which is A's score. So C wins, as the only
candidate pairwise undefeated by A.
So the method robustly punishes Chicken defection over a range of
standard examples.
Not bad for Quick and Dirty/Clean!
El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Hi Forest,
Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:
46: A > B
44: B > C (sincere B or B > A)
05: C > A
05: C > B
A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
are both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.
With approval cutoff at second rank, even Smith//Approval does better.
On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <
forest.simmons21@gmail.com> wrote:
Ted,
Thanks for providing a great example that illustrates how the burial
defense works.
I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.
In the case of public elections for political office we expect the
Smith set to be small ... usually a singleton, and occasionally a triplet
... unless factions think they can get away with burial!
In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.
With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:
While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).
In the case of a three candidate Smith Set this method first eats away
all of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.
It would be interesting to see how this method works on Colin
Champion's five candidate example... especially to see if the
renormalizations are worth the trouble ... and if perhaps the below
midrange approval scores (or the ASM approval scores) work better than
range scores.
That's a lot of work! Do you have any students that need a project?
My best,
Forest
El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern dodecatheon@gmail.com
escribió:
Your Q&CBRS works well in a situation in which Approval Sorted Margins
does not: (due to Colin Champion):
Sincere:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
D is the beats-all Condorcet voter.
5 C-first voters bury D:
1: B > D > A > E > C
1: B > D > E > C > A
5: C > A > B > E > D # Was C > A > D > B > E
1: D > A > C > B > E
1: D > C > B > A > E
2: E > B > D > C > A
Pairwise array:
[-- 6. 2. 5. 8.]
[5. -- 4. 9. 9.]
[9. 7. -- 5. 7.]
[6. 2. 6. -- 4.]
[3. 2. 4. 7. --]
Now, if I understand your method correctly, we first find the Smith
Set. In the burial case, it is all 5 candidates, A, B, C, D, E.
Then we find the basic scores for each Smith candidate:
A: 8
B: 11
C: 10
D: 6
E: 9
The Smith candidate with the smallest basic score is D (our previous
CW).
Smith candidates that defeat D are B and E. B has highest basic
score, therefore is the winner. C's strategy did not help C.
With Approval Sorted Margins, C is able to win using Burial.
The Basic Score needs to be tabulated separately from pairwise, and
depends on the other rankings on the ballot: each candidate gets a point
for any rating above minimum candidate rating, if using a ratings ballot.
In the case of many candidates, this could lead to basic score ties, since
each ballot would certainly have many candidates rated 0, so in those cases
I would recommend recounting after eliminating all candidates outside the
Smith Set.
I've been thinking about the impracticality of computing pairwise
arrays in "jungle" elections, those with, say, >9 candidates. If a ratings
ballot were used, with rankings inferred, I would recommend a floating
score threshold, starting at 1% of maximum approval, but rising until at
most 9 distinct candidate scores are above the threshold (allows score-tie
clusters), then recounting to get the reduced pairwise array and basic
scores. If the lowest basic score is tied, eliminate non-Smith candidates
and recount basic scores.
For elections with 9 or fewer candidates and no lowest-basic score
ties in the Smith set, this is summable, but requires recounts in the event
of more candidates or lowest basic score ties.
Looking back at Colin's burial example, what happens if basic score is
recalculated using ballots scoring candidate X above D?
A: 5
B: 9
C: 5
E: 7
B still wins. So your overall basic score as a proxy for ballots
scoring above the lowest Smith basic score candidate is a good proxy in
this case.
On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
forest.simmons21@gmail.com> wrote:
Let's call it Q&CBRS.
Pre-requisite background:
Candidate X pairwise defeats candidate Y iff candidate X is ranked
above/before/ahead of candidate Y on more ballots than not.
A defeat chain is a sequence of candidates in which each candidate
pairwise defeats the subsequent member of the chain.
Using a "bubble sort" procedure to sort a list of candidates into
pairwise order produces a defeat chain of the listed candidates.
In this way we can easily find a defeat chain that includes all of
the candidates. The first candidate in such a chain is an example of a
Smith candidate. More generally, any candidate who has a defeat chain to
any other candidate is a member of the Smith Set.
Q&CBRS:
First, find the "basic score" for each candidate defined as the
number of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
That's it! Quick & Clean!
Note that if there is only one Smith candidate X, then X will be the
only candidate not defeated by X, and therefore the one elected .
In general a candidate not defeated by X will defeat X, and thereby
find itself at the head of a defeat chain to every other candidate, i.e. it
will be a Smith candidate.
When there is only one Smith candidate, that candidate will not be
pairwise defeated by any other candidate.
I repeat the entire method procedure here:
First, find the "basic score" for each candidate defined as the
number of ballots on which it is ranked above one or more candidates.
Then let X be the Smith candidate with the smallest basic score.
Finally, among the candidates not defeated by X, elect the one with
the greatest basic score.
Where else can you find such a simple, quick, and clean election
method?
El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Good idea!
Although it seems to me that the highest approval candidate would
have to pairwise beat or tie the approval cutoff candidate X pairwise
(which would be impossible for a non-Smith candidate to do) ...I could be
wrong ... and in any case redundancy reinforces communication and
understanding.
El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
dodecatheon@gmail.com> escribió:
Hi Forest,
This sounds like an interesting method to me!
However, I would change the winning criteria to "Elect the most
approved member of the Smith Set".
On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
forest.simmons21@gmail.com> wrote:
I apologize for the defiant tone at the end of the previous
message ... I must have gotten carried away with the "Dirty Dozen"' theme.
But isn't it frustrating to you when people use the 2nd law of
thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
justify their stubborn resistance to any kind of engineering progress?
In the previous message Q&D Burial Resistant Condorcet was
formulate in the typical "stitched together" form ... "Elect the CW if
there is one, Else ..."
In this message I would like to formulate a seamless version:
Let X be the Smith candidate who on the fewest ballots is ranked
ahead of any other Smith candidate. On each ballot approve all candidates
down to X, but include X only when no Smith candidate is ranked ahead of X.
Elect the candidate approved on the most ballots.
This method can be described as electing the approval winner when
the approval cutoff is (at the rank of) the weakest of the Smith
candidates, which itself is approved on (and only on) those ballots which
do not approve any other Smith candidate.
In other words, the approval cutoff is inclusive only when
necessary to ensure approval of at least one member of Smith.
Since a Smith member is approved on every ballot, the method
satisfies the Condorcet Criterion, i.e. it elects the only Smith member
when Smith is a singleton.
How does that grab you?
-FWS
El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Is there any burial resistant Condorcet method simpler than this?
The basic pre-requisite is to understand that whenever there is
no Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
candidates whose members are not defeated by any candidates outside of the
cycle, just as a Condorcet Winner is a candidate undefeated by any other
candidate.
Here's the Q&D burial resistant method:
Lacking a Condorcet Winner elect the candidate X having the
greatest pairwise victory over the top-cycle member Y that has the smallest
ratio of first to last place votes within the top cycle.
Two examples illustrate the method:
Example 1.
49 C
26 A>B
25 B (sincere B>A)
The top cycle is ABCA
Candidate A has the smallest ratio 26/74 of first to last place
votes.
Candidate C is the only candidate with a pairwise victory over
it, so C wins.
Notice how our rule does not reward B for insincerely lowering A
to (equal) last?
Example 2.
45 A>B (sincere A>C)
35 B>C
25 C>A
Candidate C has the smallest ratio 25/45 of first to last.
Candidate B wins as the only candidate with a pairwise victory
over C. So A's burial of C backfires.
Typically, the faction A that buries or truncates a Condorcet
Winner C to create a top-cycle cannot by so doing become a pairwise victor
over the buried Condorcet Winner ... but must (in order to create a cycle)
help some other candidate B defeat C by insincerely voting B>C.
Our Quick and Dirty method insures that if the sincere CW's
rightful victory is subverted, it goes to B, not to A.
Is this method quick enough and dirty enough for the FairVote IRV
promoters?
What objections/criticisms might they have?
Do they have any counter proposal that rivals this one in any way?
If so, let them educate us ... we'll gladly join them if they can
show us a better way!
If not, then they should join us to educate the politicians,
public, and last but not least, the academics still stuck in the pre-EM era!
Most designers of Condorcet methods asume that the gentlemanly thing to do
is to give the votes a benefit of a doubt and assume that they must have
voted sincerely but cycles are a result of errores of judgement.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
I used to think that way, too.
But seasoned election observers are of the opinion that the vast majority
(more than 90 percent) of public elections for political office have a
sincere Condorcet candidate, and that when there is a defeat cycle, it is
more likely to be the result of intentional subversion of the Condorcet
candidate than of erroneous voter judgment.
This means that game theoretic considerations are more important than
statistical error correction considerations for resolving cycles.
So the point of Q&D/C is not to resolve non-existent honest cycles but to
irradicate dishonest cycles ... the ones created by insincere manipualtions
... the vast majority of cycles.
If less than ten percent of ballots currently have Smith cycles, and 90
percent of those are insincere cycles created opportunistically, then game
theoretic punishment of that kind of manipulation would make cycles
extremely rare ... less that 0.1 percent of elections.
What about the rare sincere cycle, not intentionally created by subversion
of the CW?
Well tha winner will be the member of the Smith set not beaten pairwise by
the weakest member of the Smith set.
Think of it this way: if the lowest approval Smith candidate X pairwise
beats Y, then what does Y have to brag about?
In fact, many people think of a defeat cycle as a kind of tie that could be
broken randomly.
The trouble with that idea is that it gives incentive to gamers to subvert
the CW in exchange for a positive chance of winning out right. It does not
help eradicate insincere cycles.
In sum, that's why the popular Condorcet methods like RP, MinMax, CSSD,
River, etc. do not thwart burial or Chicken attacks.
ASP, on the other hand, has the advantage of extra approval information not
available in ordinal ballots.
Q&D/C has the advantage of putting a fence at the top of the cliff instead
of an ambulance at the bottom
Now remember, I never pretended that C&D/C was a better method than ASM.
On the contrary, ASM is better because it gives the voters an extra lever
of expression/control, the cardinal/approval intensity dimension.
But it may well be the best Universal Domain method ... which is what I was
aiming for ... what can we achieve under the UD handicap?
Does that make sense?
El jue., 6 de ene. de 2022 2:51 p. m., Ted Stern <dodecatheon@gmail.com>
escribió:
> Hi Forest,
>
> thanks for correcting my error. I agree that B is not able to successfully
> win this chicken dilemma scenario by defecting and using burial.
>
> Overall, this has the flavor of a Smith//Approval variant, replacing
> Approval with "Highest implicitly approved [Smith] candidate [HIASC]
> defeating lowest implicitly approved Smith candidate [LIASC]".
>
> To flesh it out, I'd like more details on the following:
>
> - What is the motivation/rationale for excluding Smith candidates that
> tie or are defeated by the LIASC? What makes the HIASC "better"? In the
> most recent example, it is not clear to me that C is better than A, and
> letting C win is merely supposed to provide motivation for B voters to not
> defect and bury.
> - What happens in the event of ties in lowest implicit approval among
> Smith Candidates?
> - Are there other tie situations to handle?
>
>
> Other examples:
> -----------------
> (Due to Chris Benham)
>
> 25 A>B
> 26 B>C
> 23 C>A
> 26 C
>
> Basic scores:
> A: 49, B: 51, C:75
>
> A lowest basic score (implicit approval / above bottom) and defeats B;
>
> C > A, therefore C wins Q&CBRS
>
> C is the most top ranked and the most above-bottom ranked candidate.
>
> WV, MMPO, IRV, Benham elect B
> -----------------
>
>
> 35 A
> 10 A=B
> 30 B>C
> 25 C
>
> Basic scores: A:45, B40, C55. B is lowest basic score.
> C55 > A45, A35>B30, B40>C25
> C is defeated by B. A is remaining candidate defeating B, thus wins Q&CBRS.
>
> A both pairwise-beats and positionally dominates B, but WV, Margins, MMPO all elect B.
>
>
> This is somewhat encouraging, however, I'd still like to get better motivation for the earlier example as in the bullet point questions above.
>
>
> On Wed, Jan 5, 2022 at 9:15 PM Forest Simmons <forest.simmons21@gmail.com>
> wrote:
>
>> Ted,
>>
>> We may need a different example ... let's go through this one slowly and
>> carefully ...
>> 46: A > B
>> 44: B > C (sincere B or B > A)
>> 05: C > A
>> 05: C > B
>>
>> A gets 46+5=51 basic points from the first and third factions, resp.
>>
>> B gets 46+44+5=95 points from the first, second, and fourth factions,
>> resp.
>>
>> C gets 44+5+5=54 points from the last three factions.
>>
>> If this is right, A has the lowest score, and C wins as the only
>> candidate pairwise undefeated by A.
>>
>> I don't doubt that this method could fail Chicken Defense, but it would
>> have to be a marginal failure that the attacked faction could easily
>> counter with a partial counter defection.
>>
>> For a range of examples let's look at
>>
>> p: C
>> q: A>B
>> r: B (sincere B>A)
>>
>> Where r+q>p>q>r
>>
>> A beats B because q>r
>> B beats C because r+q>p
>> C beats A because p>q
>>
>> The respective basic scores for A, B, & C are q, p+q, & p.
>>
>> The lowest of these is q which is A's score. So C wins, as the only
>> candidate pairwise undefeated by A.
>>
>> So the method robustly punishes Chicken defection over a range of
>> standard examples.
>>
>> Not bad for Quick and Dirty/Clean!
>>
>>
>>
>>
>> El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern <dodecatheon@gmail.com>
>> escribió:
>>
>>> Hi Forest,
>>>
>>> Unfortunately, your new method does not handle Chicken-Dilemma types of
>>> burial, which I am interested in. See for example Chris Benham's example:
>>>
>>> 46: A > B
>>> 44: B > C (sincere B or B > A)
>>> 05: C > A
>>> 05: C > B
>>>
>>> A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B
>>> are both lowest basic score. If C is taken as lower, then C defeats A and B
>>> wins, rewarding burial.
>>>
>>> With approval cutoff at second rank, even Smith//Approval does better.
>>>
>>> On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <
>>> forest.simmons21@gmail.com> wrote:
>>>
>>>> Ted,
>>>>
>>>> Thanks for providing a great example that illustrates how the burial
>>>> defense works.
>>>>
>>>> I agree that it's probably better, especially when there are many
>>>> candidates, to eliminate the non-Smith candidates before counting the basic
>>>> scores.
>>>>
>>>> In the case of public elections for political office we expect the
>>>> Smith set to be small ... usually a singleton, and occasionally a triplet
>>>> ... unless factions think they can get away with burial!
>>>>
>>>> In general, I like ratings information (as in your version of Approval
>>>> Sorted Margins) better than rankings ... the Q&C and Q&D methods were
>>>> created to see what a minimal acceptable rankings based burial resistant
>>>> method would look like.
>>>>
>>>> With ratings and lots of candidates I would go back to Range Based (or
>>>> Total Approval) Chain Climbing for a burial resistant method:
>>>>
>>>> While there is no pairwise undefeated candidate among the remaining
>>>> candidates ... eliminate from the remaining candidates all of those that do
>>>> not pairwise defeat the remaining candidate X that has the lowest Range
>>>> score (or alternately .. lowest below midrange approval score ... with or
>>>> without renormalization as candidates are eliminated).
>>>>
>>>> In the case of a three candidate Smith Set this method first eats away
>>>> all of the non-Smith candidates ... then the lowest score Smith candidate X
>>>> (the one that got buried) and finally Y, the one responsible for burying X,
>>>> leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
>>>> the sole survivor... someone not preferred over X by Y.
>>>>
>>>> It would be interesting to see how this method works on Colin
>>>> Champion's five candidate example... especially to see if the
>>>> renormalizations are worth the trouble ... and if perhaps the below
>>>> midrange approval scores (or the ASM approval scores) work better than
>>>> range scores.
>>>>
>>>> That's a lot of work! Do you have any students that need a project?
>>>>
>>>> My best,
>>>>
>>>> Forest
>>>>
>>>>
>>>> El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon@gmail.com>
>>>> escribió:
>>>>
>>>>> Your Q&CBRS works well in a situation in which Approval Sorted Margins
>>>>> does not: (due to Colin Champion):
>>>>>
>>>>> Sincere:
>>>>>
>>>>> 1: B > D > A > E > C
>>>>> 1: B > D > E > C > A
>>>>> 5: C > A > D > B > E
>>>>> 1: D > A > C > B > E
>>>>> 1: D > C > B > A > E
>>>>> 2: E > B > D > C > A
>>>>>
>>>>> D is the beats-all Condorcet voter.
>>>>>
>>>>> 5 C-first voters bury D:
>>>>>
>>>>> 1: B > D > A > E > C
>>>>> 1: B > D > E > C > A
>>>>> 5: C > A > B > E > D # Was C > A > D > B > E
>>>>> 1: D > A > C > B > E
>>>>> 1: D > C > B > A > E
>>>>> 2: E > B > D > C > A
>>>>>
>>>>> Pairwise array:
>>>>> [-- 6. 2. 5. 8.]
>>>>> [5. -- 4. 9. 9.]
>>>>> [9. 7. -- 5. 7.]
>>>>> [6. 2. 6. -- 4.]
>>>>> [3. 2. 4. 7. --]
>>>>>
>>>>> Now, if I understand your method correctly, we first find the Smith
>>>>> Set. In the burial case, it is all 5 candidates, A, B, C, D, E.
>>>>>
>>>>> Then we find the basic scores for each Smith candidate:
>>>>>
>>>>> A: 8
>>>>> B: 11
>>>>> C: 10
>>>>> D: 6
>>>>> E: 9
>>>>>
>>>>> The Smith candidate with the smallest basic score is D (our previous
>>>>> CW).
>>>>>
>>>>> Smith candidates that defeat D are *B* and *E*. B has highest basic
>>>>> score, therefore is the winner. C's strategy did not help C.
>>>>>
>>>>> With Approval Sorted Margins, C is able to win using Burial.
>>>>>
>>>>> The Basic Score needs to be tabulated separately from pairwise, and
>>>>> depends on the other rankings on the ballot: each candidate gets a point
>>>>> for any rating above minimum candidate rating, if using a ratings ballot.
>>>>> In the case of many candidates, this could lead to basic score ties, since
>>>>> each ballot would certainly have many candidates rated 0, so in those cases
>>>>> I would recommend recounting after eliminating all candidates outside the
>>>>> Smith Set.
>>>>>
>>>>> I've been thinking about the impracticality of computing pairwise
>>>>> arrays in "jungle" elections, those with, say, >9 candidates. If a ratings
>>>>> ballot were used, with rankings inferred, I would recommend a floating
>>>>> score threshold, starting at 1% of maximum approval, but rising until at
>>>>> most 9 distinct candidate scores are above the threshold (allows score-tie
>>>>> clusters), then recounting to get the reduced pairwise array and basic
>>>>> scores. If the lowest basic score is tied, eliminate non-Smith candidates
>>>>> and recount basic scores.
>>>>>
>>>>> *For elections with 9 or fewer candidates and no lowest-basic score
>>>>> ties in the Smith set, this is summable, but requires recounts in the event
>>>>> of more candidates or lowest basic score ties.*
>>>>>
>>>>> Looking back at Colin's burial example, what happens if basic score is
>>>>> recalculated using ballots scoring candidate X above D?
>>>>>
>>>>> A: 5
>>>>> B: 9
>>>>> C: 5
>>>>> E: 7
>>>>>
>>>>> B still wins. So your overall basic score as a proxy for ballots
>>>>> scoring above the lowest Smith basic score candidate is a good proxy in
>>>>> this case.
>>>>>
>>>>> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <
>>>>> forest.simmons21@gmail.com> wrote:
>>>>>
>>>>>> Let's call it Q&CBRS.
>>>>>>
>>>>>> Pre-requisite background:
>>>>>>
>>>>>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>>>>>> above/before/ahead of candidate Y on more ballots than not.
>>>>>>
>>>>>> A defeat chain is a sequence of candidates in which each candidate
>>>>>> pairwise defeats the subsequent member of the chain.
>>>>>>
>>>>>> Using a "bubble sort" procedure to sort a list of candidates into
>>>>>> pairwise order produces a defeat chain of the listed candidates.
>>>>>>
>>>>>> In this way we can easily find a defeat chain that includes all of
>>>>>> the candidates. The first candidate in such a chain is an example of a
>>>>>> Smith candidate. More generally, any candidate who has a defeat chain to
>>>>>> any other candidate is a member of the Smith Set.
>>>>>>
>>>>>> Q&CBRS:
>>>>>>
>>>>>> First, find the "basic score" for each candidate defined as the
>>>>>> number of ballots on which it is ranked above one or more candidates.
>>>>>>
>>>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>>>
>>>>>> Finally, among the candidates not defeated by X, elect the one with
>>>>>> the greatest basic score.
>>>>>>
>>>>>> That's it! Quick & Clean!
>>>>>>
>>>>>> Note that if there is only one Smith candidate X, then X will be the
>>>>>> only candidate not defeated by X, and therefore the one elected .
>>>>>>
>>>>>> In general a candidate not defeated by X will defeat X, and thereby
>>>>>> find itself at the head of a defeat chain to every other candidate, i.e. it
>>>>>> will be a Smith candidate.
>>>>>>
>>>>>> When there is only one Smith candidate, that candidate will not be
>>>>>> pairwise defeated by any other candidate.
>>>>>>
>>>>>> I repeat the entire method procedure here:
>>>>>>
>>>>>> First, find the "basic score" for each candidate defined as the
>>>>>> number of ballots on which it is ranked above one or more candidates.
>>>>>>
>>>>>> Then let X be the Smith candidate with the smallest basic score.
>>>>>>
>>>>>> Finally, among the candidates not defeated by X, elect the one with
>>>>>> the greatest basic score.
>>>>>>
>>>>>> Where else can you find such a simple, quick, and clean election
>>>>>> method?
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>>>>>> forest.simmons21@gmail.com> escribió:
>>>>>>
>>>>>>> Good idea!
>>>>>>>
>>>>>>> Although it seems to me that the highest approval candidate would
>>>>>>> have to pairwise beat or tie the approval cutoff candidate X pairwise
>>>>>>> (which would be impossible for a non-Smith candidate to do) ...I could be
>>>>>>> wrong ... and in any case redundancy reinforces communication and
>>>>>>> understanding.
>>>>>>>
>>>>>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <
>>>>>>> dodecatheon@gmail.com> escribió:
>>>>>>>
>>>>>>>> Hi Forest,
>>>>>>>>
>>>>>>>> This sounds like an interesting method to me!
>>>>>>>>
>>>>>>>> However, I would change the winning criteria to "Elect the most
>>>>>>>> approved member of the Smith Set".
>>>>>>>>
>>>>>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>>>>>> forest.simmons21@gmail.com> wrote:
>>>>>>>>
>>>>>>>>> I apologize for the defiant tone at the end of the previous
>>>>>>>>> message ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>>>>>
>>>>>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>>>>>
>>>>>>>>> In the previous message Q&D Burial Resistant Condorcet was
>>>>>>>>> formulate in the typical "stitched together" form ... "Elect the CW if
>>>>>>>>> there is one, Else ..."
>>>>>>>>>
>>>>>>>>> In this message I would like to formulate a seamless version:
>>>>>>>>>
>>>>>>>>> Let X be the Smith candidate who on the fewest ballots is ranked
>>>>>>>>> ahead of any other Smith candidate. On each ballot approve all candidates
>>>>>>>>> down to X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>>>>>
>>>>>>>>> Elect the candidate approved on the most ballots.
>>>>>>>>>
>>>>>>>>> This method can be described as electing the approval winner when
>>>>>>>>> the approval cutoff is (at the rank of) the weakest of the Smith
>>>>>>>>> candidates, which itself is approved on (and only on) those ballots which
>>>>>>>>> do not approve any other Smith candidate.
>>>>>>>>>
>>>>>>>>> In other words, the approval cutoff is inclusive only when
>>>>>>>>> necessary to ensure approval of at least one member of Smith.
>>>>>>>>>
>>>>>>>>> Since a Smith member is approved on every ballot, the method
>>>>>>>>> satisfies the Condorcet Criterion, i.e. it elects the only Smith member
>>>>>>>>> when Smith is a singleton.
>>>>>>>>>
>>>>>>>>> How does that grab you?
>>>>>>>>>
>>>>>>>>> -FWS
>>>>>>>>>
>>>>>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>>>>>> forest.simmons21@gmail.com> escribió:
>>>>>>>>>
>>>>>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>>>>>
>>>>>>>>>> The basic pre-requisite is to understand that whenever there is
>>>>>>>>>> no Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>>>>>> candidate.
>>>>>>>>>>
>>>>>>>>>> Here's the Q&D burial resistant method:
>>>>>>>>>>
>>>>>>>>>> Lacking a Condorcet Winner elect the candidate X having the
>>>>>>>>>> greatest pairwise victory over the top-cycle member Y that has the smallest
>>>>>>>>>> ratio of first to last place votes within the top cycle.
>>>>>>>>>>
>>>>>>>>>> Two examples illustrate the method:
>>>>>>>>>>
>>>>>>>>>> Example 1.
>>>>>>>>>>
>>>>>>>>>> 49 C
>>>>>>>>>> 26 A>B
>>>>>>>>>> 25 B (sincere B>A)
>>>>>>>>>>
>>>>>>>>>> The top cycle is ABCA
>>>>>>>>>>
>>>>>>>>>> Candidate A has the smallest ratio 26/74 of first to last place
>>>>>>>>>> votes.
>>>>>>>>>>
>>>>>>>>>> Candidate C is the only candidate with a pairwise victory over
>>>>>>>>>> it, so C wins.
>>>>>>>>>>
>>>>>>>>>> Notice how our rule does not reward B for insincerely lowering A
>>>>>>>>>> to (equal) last?
>>>>>>>>>>
>>>>>>>>>> Example 2.
>>>>>>>>>>
>>>>>>>>>> 45 A>B (sincere A>C)
>>>>>>>>>> 35 B>C
>>>>>>>>>> 25 C>A
>>>>>>>>>>
>>>>>>>>>> Candidate C has the smallest ratio 25/45 of first to last.
>>>>>>>>>>
>>>>>>>>>> Candidate B wins as the only candidate with a pairwise victory
>>>>>>>>>> over C. So A's burial of C backfires.
>>>>>>>>>>
>>>>>>>>>> Typically, the faction A that buries or truncates a Condorcet
>>>>>>>>>> Winner C to create a top-cycle cannot by so doing become a pairwise victor
>>>>>>>>>> over the buried Condorcet Winner ... but must (in order to create a cycle)
>>>>>>>>>> help some other candidate B defeat C by insincerely voting B>C.
>>>>>>>>>>
>>>>>>>>>> Our Quick and Dirty method insures that if the sincere CW's
>>>>>>>>>> rightful victory is subverted, it goes to B, not to A.
>>>>>>>>>>
>>>>>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>>>>>> promoters?
>>>>>>>>>>
>>>>>>>>>> What objections/criticisms might they have?
>>>>>>>>>>
>>>>>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>>>>>
>>>>>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>>>>>> show us a better way!
>>>>>>>>>>
>>>>>>>>>> If not, then they should join us to educate the politicians,
>>>>>>>>>> public, and last but not least, the academics still stuck in the pre-EM era!
>>>>>>>>>>
>>>>>>>>> ----
>>>>>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>>>>>> list info
>>>>>>>>>
>>>>>>>>
RB
robert bristow-johnson
Fri, Jan 7, 2022 8:15 AM
On 01/07/2022 1:05 AM Forest Simmons forest.simmons21@gmail.com wrote:
Most designers of Condorcet methods asume that the gentlemanly thing to do is to give the votes a benefit of a doubt and assume that they must have voted sincerely
I just think that, without some other information to suggest otherwise, the marked ballot should be assumed to represent the voter's sincere preferences.
but cycles are a result of errores of judgement.
A preference cycle would need a very close 3-way race and and somewhat schizoid electorate. "If I can't have my favorite Bernie Sanders, then I'm voting for T****."
Probably Schulze or RP is the best thing to do for those cases when there is no Condorcet winner. But getting that into legislative language is difficult, which is why I have advocated for BTR-STV.
Because of these assumptions they attempt to filter out the erroneous preferences statistically
I just think that the method tries to make the best thing out of a confusing situation that will rarely happen.
.. the main heuristic is that larger majorities are less apt to hold erroneous opinions than smaller ones ... hence cycles are broken by annulling the defeats with the smallest majorities.
I used to think that way, too.
But seasoned election observers are of the opinion that the vast majority (more than 90 percent) of public elections for political office have a sincere Condorcet candidate, and that when there is a defeat cycle, it is more likely to be the result of intentional subversion of the Condorcet candidate than of erroneous voter judgment.
The thing is more like 99.5% have Condorcet winner. Right now, at least with the 440 RCV elections that FairVote says they analyzed, that all had Condorcet winners and all but one succeeded at electing the Condorcet winner.
I just sorta wanna get any Condorcet method. The simpler language the better. I think cycles will be rare. If we elect the plurality winner in case of a cycle, that might be an indication of preference. It's not Schulze. It might not elect the bestest candidate that disincentivizes certain tactical voting. But if simple language get a Condorcet method understood, it has a better chance of maybe someday getting legislated.
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
> On 01/07/2022 1:05 AM Forest Simmons <forest.simmons21@gmail.com> wrote:
>
>
> Most designers of Condorcet methods asume that the gentlemanly thing to do is to give the votes a benefit of a doubt and assume that they must have voted sincerely
I just think that, without some other information to suggest otherwise, the marked ballot should be assumed to represent the voter's sincere preferences.
> but cycles are a result of errores of judgement.
>
A preference cycle would need a very close 3-way race *and* and somewhat schizoid electorate. "If I can't have my favorite Bernie Sanders, then I'm voting for T****."
Probably Schulze or RP is the best thing to do for those cases when there is no Condorcet winner. But getting that into legislative language is difficult, which is why I have advocated for BTR-STV.
> Because of these assumptions they attempt to filter out the erroneous preferences statistically
I just think that the method tries to make the best thing out of a confusing situation that will rarely happen.
> .. the main heuristic is that larger majorities are less apt to hold erroneous opinions than smaller ones ... hence cycles are broken by annulling the defeats with the smallest majorities.
>
That's one way to do it.
> I used to think that way, too.
>
> But seasoned election observers are of the opinion that the vast majority (more than 90 percent) of public elections for political office have a sincere Condorcet candidate, and that when there is a defeat cycle, it is more likely to be the result of intentional subversion of the Condorcet candidate than of erroneous voter judgment.
>
The thing is more like 99.5% have Condorcet winner. Right now, at least with the 440 RCV elections that FairVote says they analyzed, that all had Condorcet winners and all but one succeeded at electing the Condorcet winner.
I just sorta wanna get any Condorcet method. The simpler language the better. I think cycles will be rare. If we elect the plurality winner in case of a cycle, that might be an indication of preference. It's not Schulze. It might not elect the bestest candidate that disincentivizes certain tactical voting. But if simple language get a Condorcet method understood, it has a better chance of maybe someday getting legislated.
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
FS
Forest Simmons
Fri, Jan 7, 2022 9:53 AM
Esteemed & Indefatigable Robert:
I used to think like you about Condorcet Completion, but after years of
designing methods based on the assumption of innocent cycle creation, you
brought up the extreme rarity of Condorcet cycles among native voters ...
now you are quoting 99.5 percent cycle free ...which reinforces my about
face even more!
So here's the thing:
"Decisiones based on false assumptions lead to undesireable consequences."
For example, designing Condorcet completion methods on the false assumption
that cycles are usually created innocently/honestly/unintentionally has the
undesireable consequence of encouraging dishonest cycle creators, because
that false assumption allows the culprits to act with impunity
If you let them get away with it, they will do it more and more, making
ballot CW's less likely.
But if you punish it, they will stop doing it so that there will always be
a CW .... which, in turn, will always be elected by Q&C, because it
satisfies Independence from Smith Dominated Alternatives.
In summary, the best Condorcet completion method is the one that is best at
discouraging cycle creation.
Because Schulze is more tolerant of cycle creation, it is less likely to
find a Condorcet winner ... Schulze is no better than the next method at
resurrecting buried CW's.
"An ounce of prevention ..."
It's better to install a guard rail at the top of a cliff than to park an
ambulance at the bottom of the cliff.
The whole point of Q&D/C is to show how simple it is to install an
effective guard rail compared to the academically reputable but complex and
ineffective ambulances like Schulze, RP, etc that naively assume that
cycles are created innocently ... just bad judgment, not intentional
dishonest manipulaion.
[I used to admire those methods, too, but not any more.They are more
complex but less effective than Quick&Dirty/Clean. Why? Because they put
all of their focus on reconstructing an intentionally contaminated signal
instead of preventing the contamination by making it back-fire. They were
designed before the importance of game theory was fully appreciated.]
If we don't think game theoretically, then we are at the mercy of the
gamers.
"Be as wise as serpents, but as harmless as doves."
I hope you know that I admire and appreciate your activism, and I
understand that Top Two Runoff may be the best politically feasible cycle
resolution method for now, but hope you can see the importance of a simple
manipulation resistant cycle resolution method in the long run.
Q&D/C guard rail at the top of the cliff trumps an elaborate RP/Schulze
hospital at the bottom!
My Best To You!
Forest
El vie., 7 de ene. de 2022 12:15 a. m., robert bristow-johnson <
rbj@audioimagination.com> escribió:
On 01/07/2022 1:05 AM Forest Simmons forest.simmons21@gmail.com wrote:
Most designers of Condorcet methods asume that the gentlemanly thing to
do is to give the votes a benefit of a doubt and assume that they must have
voted sincerely
I just think that, without some other information to suggest otherwise,
the marked ballot should be assumed to represent the voter's sincere
preferences.
but cycles are a result of errores of judgement.
A preference cycle would need a very close 3-way race and and somewhat
schizoid electorate. "If I can't have my favorite Bernie Sanders, then I'm
voting for T****."
Probably Schulze or RP is the best thing to do for those cases when there
is no Condorcet winner. But getting that into legislative language is
difficult, which is why I have advocated for BTR-STV.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
I just think that the method tries to make the best thing out of a
confusing situation that will rarely happen.
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
I used to think that way, too.
But seasoned election observers are of the opinion that the vast
majority (more than 90 percent) of public elections for political office
have a sincere Condorcet candidate, and that when there is a defeat cycle,
it is more likely to be the result of intentional subversion of the
Condorcet candidate than of erroneous voter judgment.
The thing is more like 99.5% have Condorcet winner. Right now, at least
with the 440 RCV elections that FairVote says they analyzed, that all had
Condorcet winners and all but one succeeded at electing the Condorcet
winner.
I just sorta wanna get any Condorcet method. The simpler language the
better. I think cycles will be rare. If we elect the plurality winner in
case of a cycle, that might be an indication of preference. It's not
Schulze. It might not elect the bestest candidate that disincentivizes
certain tactical voting. But if simple language get a Condorcet method
understood, it has a better chance of maybe someday getting legislated.
--
r b-j . _ . _ . _ . _ rbj@audioimagination.com
"Imagination is more important than knowledge."
.
.
.
Election-Methods mailing list - see https://electorama.com/em for list
info
Esteemed & Indefatigable Robert:
I used to think like you about Condorcet Completion, but after years of
designing methods based on the assumption of innocent cycle creation, you
brought up the extreme rarity of Condorcet cycles among native voters ...
now you are quoting 99.5 percent cycle free ...which reinforces my about
face even more!
So here's the thing:
"Decisiones based on false assumptions lead to undesireable consequences."
For example, designing Condorcet completion methods on the false assumption
that cycles are usually created innocently/honestly/unintentionally has the
undesireable consequence of encouraging dishonest cycle creators, because
that false assumption allows the culprits to act with impunity
If you let them get away with it, they will do it more and more, making
ballot CW's less likely.
But if you punish it, they will stop doing it so that there will always be
a CW .... which, in turn, will always be elected by Q&C, because it
satisfies Independence from Smith Dominated Alternatives.
In summary, the best Condorcet completion method is the one that is best at
discouraging cycle creation.
Because Schulze is more tolerant of cycle creation, it is less likely to
find a Condorcet winner ... Schulze is no better than the next method at
resurrecting buried CW's.
"An ounce of prevention ..."
It's better to install a guard rail at the top of a cliff than to park an
ambulance at the bottom of the cliff.
The whole point of Q&D/C is to show how simple it is to install an
effective guard rail compared to the academically reputable but complex and
ineffective ambulances like Schulze, RP, etc that naively assume that
cycles are created innocently ... just bad judgment, not intentional
dishonest manipulaion.
[I used to admire those methods, too, but not any more.They are more
complex but less effective than Quick&Dirty/Clean. Why? Because they put
all of their focus on reconstructing an intentionally contaminated signal
instead of preventing the contamination by making it back-fire. They were
designed before the importance of game theory was fully appreciated.]
If we don't think game theoretically, then we are at the mercy of the
gamers.
"Be as wise as serpents, but as harmless as doves."
I hope you know that I admire and appreciate your activism, and I
understand that Top Two Runoff may be the best politically feasible cycle
resolution method for now, but hope you can see the importance of a simple
manipulation resistant cycle resolution method in the long run.
Q&D/C guard rail at the top of the cliff trumps an elaborate RP/Schulze
hospital at the bottom!
My Best To You!
Forest
El vie., 7 de ene. de 2022 12:15 a. m., robert bristow-johnson <
rbj@audioimagination.com> escribió:
>
>
> > On 01/07/2022 1:05 AM Forest Simmons <forest.simmons21@gmail.com> wrote:
> >
> >
> > Most designers of Condorcet methods asume that the gentlemanly thing to
> do is to give the votes a benefit of a doubt and assume that they must have
> voted sincerely
>
> I just think that, without some other information to suggest otherwise,
> the marked ballot should be assumed to represent the voter's sincere
> preferences.
>
> > but cycles are a result of errores of judgement.
> >
>
> A preference cycle would need a very close 3-way race *and* and somewhat
> schizoid electorate. "If I can't have my favorite Bernie Sanders, then I'm
> voting for T****."
>
> Probably Schulze or RP is the best thing to do for those cases when there
> is no Condorcet winner. But getting that into legislative language is
> difficult, which is why I have advocated for BTR-STV.
>
> > Because of these assumptions they attempt to filter out the erroneous
> preferences statistically
>
> I just think that the method tries to make the best thing out of a
> confusing situation that will rarely happen.
>
> > .. the main heuristic is that larger majorities are less apt to hold
> erroneous opinions than smaller ones ... hence cycles are broken by
> annulling the defeats with the smallest majorities.
> >
>
> That's one way to do it.
>
> > I used to think that way, too.
> >
> > But seasoned election observers are of the opinion that the vast
> majority (more than 90 percent) of public elections for political office
> have a sincere Condorcet candidate, and that when there is a defeat cycle,
> it is more likely to be the result of intentional subversion of the
> Condorcet candidate than of erroneous voter judgment.
> >
>
> The thing is more like 99.5% have Condorcet winner. Right now, at least
> with the 440 RCV elections that FairVote says they analyzed, that all had
> Condorcet winners and all but one succeeded at electing the Condorcet
> winner.
>
> I just sorta wanna get any Condorcet method. The simpler language the
> better. I think cycles will be rare. If we elect the plurality winner in
> case of a cycle, that might be an indication of preference. It's not
> Schulze. It might not elect the bestest candidate that disincentivizes
> certain tactical voting. But if simple language get a Condorcet method
> understood, it has a better chance of maybe someday getting legislated.
>
> --
>
> r b-j . _ . _ . _ . _ rbj@audioimagination.com
>
> "Imagination is more important than knowledge."
>
> .
> .
> .
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
KM
Kristofer Munsterhjelm
Fri, Jan 7, 2022 12:38 PM
On 07.01.2022 07:05, Forest Simmons wrote:
Most designers of Condorcet methods asume that the gentlemanly thing to
do is to give the votes a benefit of a doubt and assume that they must
have voted sincerely but cycles are a result of errores of judgement.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
There is probably a tradeoff between strategic resistance and honest
VSE. The more you want one, the less you get the other, and at the
extremes you completely disregard one or the other.
So if we could construct methods to spec, the best approach would be to
somehow infer just how much strategy the method needs to resist, and
then maximize VSE in a suitable model (probably spatial) subject to this
constraint.
But we don't really know how strong the barrier has to be against
strategy. As I've mentioned before, I think it differs based on culture:
IIRC Ireland saw much less vote management under STV than did New York.
If we only have one shot, it's reasonable to err on the side of strategy
resistance. That's not to say that the honesty-favoring methods don't
have their place, though: Debian seems to do pretty well with Schulze,
for instance.
As for methods like Plurality and (probably) IRV -- well, they're just
Pareto-dominated. You can find methods with better VSE and the same
level of strategic resistance, or methods that handle strategy better
while providing the same VSE.
-km
On 07.01.2022 07:05, Forest Simmons wrote:
> Most designers of Condorcet methods asume that the gentlemanly thing to
> do is to give the votes a benefit of a doubt and assume that they must
> have voted sincerely but cycles are a result of errores of judgement.
>
> Because of these assumptions they attempt to filter out the erroneous
> preferences statistically
> .. the main heuristic is that larger majorities are less apt to hold
> erroneous opinions than smaller ones ... hence cycles are broken by
> annulling the defeats with the smallest majorities.
There is probably a tradeoff between strategic resistance and honest
VSE. The more you want one, the less you get the other, and at the
extremes you completely disregard one or the other.
So if we could construct methods to spec, the best approach would be to
somehow infer just how much strategy the method needs to resist, and
then maximize VSE in a suitable model (probably spatial) subject to this
constraint.
But we don't really know how strong the barrier has to be against
strategy. As I've mentioned before, I think it differs based on culture:
IIRC Ireland saw much less vote management under STV than did New York.
If we only have one shot, it's reasonable to err on the side of strategy
resistance. That's not to say that the honesty-favoring methods don't
have their place, though: Debian seems to do pretty well with Schulze,
for instance.
As for methods like Plurality and (probably) IRV -- well, they're just
Pareto-dominated. You can find methods with better VSE and the same
level of strategic resistance, or methods that handle strategy better
while providing the same VSE.
-km
FS
Forest Simmons
Fri, Jan 7, 2022 9:57 PM
Very true, and I would have said err on the side of VSE until Robert B-J
convinced me that sincere cycles are practically non-existent with an
occurrence of less than 0.5 percent.
He got that statistic from Fair Vote's analysis of over 400 elections, and
they have no particular reason to exaggerate that statistic. So suppose
they're right, then do we conclude that any old Condorcet method is as good
as another?
You know the answer to that, Kristofer, but I elaborate for the benefit of
the typical EM List reader:
No, because, as Kristofer pointed out, it depends on the "neighborhood."
Back in the fifties when I was a kid in rural/small town eastern Washington
state nobody locked their doors. In fact, they usually left their keys in
the ignition for convenience. But was that the prudent/sustainable thing to
do?
Nowadays sixty years later, in those same neighborhoods people have learned
by experience to adopt city slicker habits of door locking.
When 99.5 percent of election polls reveal the existence of sincere
Condorcet candidates, the election method with the greatest Condorcet
efficiency will come the closest to electing a CW 99.5 percent of the time.
But don't all Condorcet methods have equal Condorcet efficiency? Isn't the
very definition of "Condorcet method" a method that always elects the
Condorcet candidate when one exists?
That would be nice if there were such a method, but Gibbard-Satterwaithe
shows the impossibility of that ideal in any tough neiborhood.
Then what is the real definition of "Condorcet Method"?
It is a method that elects a ballot Condorcet candidate. In tough
neighborhoods there will often be sincere Condorcet candidates that are not
ballot Condorcet Candidates ... because they are ranked insincerely on the
ballots.
So how to ensure that sincere Condorcet candidates retain their Condorcet
status on the ballots?
Only methods that take this question seriously can have sustainable
Condorcet efficiency.
Let's do a thought experiment to compare the Condorcet efficiency of Ranked
Pairs, Benham, Schulze, and other Condorcet methods:
Suppose that sincere preferences are given...
40 A>C
35 B>C
25 C>A
Like 99.5 percent of electorates this one has a sincere Condorcet
candidate, namely C, which is preferred over A by a 60 percent majority,
and preferred over B by a 65 percent majority.
The question of Condorcet efficiency in this example is "Which Condorcet
methods are most likely to elect C ?"
But that depends on the neighborhood. In Grant County, Washington of the
fifties, any Condorcet method would elect C.
But which Condorcet method would robustly elect C even in Brooklyn, New
York?
That depends on which methods take the possibility of subversion
seriously... subversion of the sincere CW by intentional cycle creation.
Under Schulze, RP, River, MinMax, etc. candidate A would likely win because
the A faction could confidently bury A under B, creating an insincere
defeat cycle with weakest defeat being the 60 percent C over A compared
with the 75 percent B over C, and the 65 percent A over B.
All of these standard Condorcet methods are built on the assumption that
the smallest majority (60% C over A) is the one most likely to be "wrong".
So they elect A, unwittingly rewarding the A faction for subverting the
sincere CW.
But how about Q&D/C?
It would elect C because it is a ballot Condorcet method intolerant of the
kind of subversion that rewarded the A faction under Schulze, etc. Since
the subversion would just backfire, no faction would be dumb enough to try
it.
Let's look at how it would backfire were the A faction stupid enough to
attempt the burial of C under B:
The candidate with the lowest basic score would then be C, so B would win
as the only candidate pairwise undefeated by C.
So A's gambit would just change the winner from its second choice C to its
last choice B.
Do you think any major faction in Brooklyn would gamble on a sure loss like
that?
To be clear, what this example shows is that our Quick and Dirty/Clean
method has Condorcet efficiency superior to that of Schulze, etc.
And why is it more Condorcet efficient? Because it was designed on a
realistic game theoretic principle, rather than a statistical error
correcting technique designed mainly for filtering out inadvertent
"mistakes" in judgment (or minority opinion).
This example could be (and has been) multiplied indefinitely with the same
pattern of results:
Q&D/C is much more Condorcet efficient than Schulze et al in rough
neighborhoods, as well as 100 percent efficient in easy neighborhoods (like
any and every Condorcet method is).
So here's the best advice for Condorcet method design: focus on frustrating
cycle creators. Make sure that all attempts at cycle creation backfire
automatically. Beyond that accommodate any (possible but vanishingly rare)
sincere cycles by making sure the method always elects simply,
monotonically, and clone indepently from the ballot Smith set, as does
Q&D/C, a method that Pareto dominates all of the old Condorcet methods on
these criteria.
To put it bluntly, all of those older methods are pretty much passe ...
uniformly dominated by Q&D/C when it comes to single winner elections for
public office.
I'm sorry to talk so bluntly, but subtle words don't seem to register in
this context, especially among the self-styled movers and shakers [who
probably won't even read them anyway🤔]
On a technical note ... to make sure that Q&D/C is ISDA, use a version that
immediately restricts to Smith, or at least counts the "basic scores"
relative to the Smith Set candidates.
Forest
El vie., 7 de ene. de 2022 4:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 07.01.2022 07:05, Forest Simmons wrote:
Most designers of Condorcet methods asume that the gentlemanly thing to
do is to give the votes a benefit of a doubt and assume that they must
have voted sincerely but cycles are a result of errores of judgement.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
There is probably a tradeoff between strategic resistance and honest
VSE. The more you want one, the less you get the other, and at the
extremes you completely disregard one or the other.
So if we could construct methods to spec, the best approach would be to
somehow infer just how much strategy the method needs to resist, and
then maximize VSE in a suitable model (probably spatial) subject to this
constraint.
But we don't really know how strong the barrier has to be against
strategy. As I've mentioned before, I think it differs based on culture:
IIRC Ireland saw much less vote management under STV than did New York.
If we only have one shot, it's reasonable to err on the side of strategy
resistance. That's not to say that the honesty-favoring methods don't
have their place, though: Debian seems to do pretty well with Schulze,
for instance.
As for methods like Plurality and (probably) IRV -- well, they're just
Pareto-dominated. You can find methods with better VSE and the same
level of strategic resistance, or methods that handle strategy better
while providing the same VSE.
-km
Very true, and I would have said err on the side of VSE until Robert B-J
convinced me that sincere cycles are practically non-existent with an
occurrence of less than 0.5 percent.
He got that statistic from Fair Vote's analysis of over 400 elections, and
they have no particular reason to exaggerate that statistic. So suppose
they're right, then do we conclude that any old Condorcet method is as good
as another?
You know the answer to that, Kristofer, but I elaborate for the benefit of
the typical EM List reader:
No, because, as Kristofer pointed out, it depends on the "neighborhood."
Back in the fifties when I was a kid in rural/small town eastern Washington
state nobody locked their doors. In fact, they usually left their keys in
the ignition for convenience. But was that the prudent/sustainable thing to
do?
Nowadays sixty years later, in those same neighborhoods people have learned
by experience to adopt city slicker habits of door locking.
When 99.5 percent of election polls reveal the existence of sincere
Condorcet candidates, the election method with the greatest Condorcet
efficiency will come the closest to electing a CW 99.5 percent of the time.
But don't all Condorcet methods have equal Condorcet efficiency? Isn't the
very definition of "Condorcet method" a method that always elects the
Condorcet candidate when one exists?
That would be nice if there were such a method, but Gibbard-Satterwaithe
shows the impossibility of that ideal in any tough neiborhood.
Then what is the real definition of "Condorcet Method"?
It is a method that elects a ballot Condorcet candidate. In tough
neighborhoods there will often be sincere Condorcet candidates that are not
ballot Condorcet Candidates ... because they are ranked insincerely on the
ballots.
So how to ensure that sincere Condorcet candidates retain their Condorcet
status on the ballots?
Only methods that take this question seriously can have sustainable
Condorcet efficiency.
Let's do a thought experiment to compare the Condorcet efficiency of Ranked
Pairs, Benham, Schulze, and other Condorcet methods:
Suppose that sincere preferences are given...
40 A>C
35 B>C
25 C>A
Like 99.5 percent of electorates this one has a sincere Condorcet
candidate, namely C, which is preferred over A by a 60 percent majority,
and preferred over B by a 65 percent majority.
The question of Condorcet efficiency in this example is "Which Condorcet
methods are most likely to elect C ?"
But that depends on the neighborhood. In Grant County, Washington of the
fifties, any Condorcet method would elect C.
But which Condorcet method would robustly elect C even in Brooklyn, New
York?
That depends on which methods take the possibility of subversion
seriously... subversion of the sincere CW by intentional cycle creation.
Under Schulze, RP, River, MinMax, etc. candidate A would likely win because
the A faction could confidently bury A under B, creating an insincere
defeat cycle with weakest defeat being the 60 percent C over A compared
with the 75 percent B over C, and the 65 percent A over B.
All of these standard Condorcet methods are built on the assumption that
the smallest majority (60% C over A) is the one most likely to be "wrong".
So they elect A, unwittingly rewarding the A faction for subverting the
sincere CW.
But how about Q&D/C?
It would elect C because it is a ballot Condorcet method intolerant of the
kind of subversion that rewarded the A faction under Schulze, etc. Since
the subversion would just backfire, no faction would be dumb enough to try
it.
Let's look at how it would backfire were the A faction stupid enough to
attempt the burial of C under B:
The candidate with the lowest basic score would then be C, so B would win
as the only candidate pairwise undefeated by C.
So A's gambit would just change the winner from its second choice C to its
last choice B.
Do you think any major faction in Brooklyn would gamble on a sure loss like
that?
To be clear, what this example shows is that our Quick and Dirty/Clean
method has Condorcet efficiency superior to that of Schulze, etc.
And why is it more Condorcet efficient? Because it was designed on a
realistic game theoretic principle, rather than a statistical error
correcting technique designed mainly for filtering out inadvertent
"mistakes" in judgment (or minority opinion).
This example could be (and has been) multiplied indefinitely with the same
pattern of results:
Q&D/C is much more Condorcet efficient than Schulze et al in rough
neighborhoods, as well as 100 percent efficient in easy neighborhoods (like
any and every Condorcet method is).
So here's the best advice for Condorcet method design: focus on frustrating
cycle creators. Make sure that all attempts at cycle creation backfire
automatically. Beyond that accommodate any (possible but vanishingly rare)
sincere cycles by making sure the method always elects simply,
monotonically, and clone indepently from the ballot Smith set, as does
Q&D/C, a method that Pareto dominates all of the old Condorcet methods on
these criteria.
To put it bluntly, all of those older methods are pretty much passe ...
uniformly dominated by Q&D/C when it comes to single winner elections for
public office.
I'm sorry to talk so bluntly, but subtle words don't seem to register in
this context, especially among the self-styled movers and shakers [who
probably won't even read them anyway🤔]
On a technical note ... to make sure that Q&D/C is ISDA, use a version that
immediately restricts to Smith, or at least counts the "basic scores"
relative to the Smith Set candidates.
Forest
El vie., 7 de ene. de 2022 4:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
> On 07.01.2022 07:05, Forest Simmons wrote:
> > Most designers of Condorcet methods asume that the gentlemanly thing to
> > do is to give the votes a benefit of a doubt and assume that they must
> > have voted sincerely but cycles are a result of errores of judgement.
> >
> > Because of these assumptions they attempt to filter out the erroneous
> > preferences statistically
> > .. the main heuristic is that larger majorities are less apt to hold
> > erroneous opinions than smaller ones ... hence cycles are broken by
> > annulling the defeats with the smallest majorities.
>
> There is probably a tradeoff between strategic resistance and honest
> VSE. The more you want one, the less you get the other, and at the
> extremes you completely disregard one or the other.
>
> So if we could construct methods to spec, the best approach would be to
> somehow infer just how much strategy the method needs to resist, and
> then maximize VSE in a suitable model (probably spatial) subject to this
> constraint.
>
> But we don't really know how strong the barrier has to be against
> strategy. As I've mentioned before, I think it differs based on culture:
> IIRC Ireland saw much less vote management under STV than did New York.
>
> If we only have one shot, it's reasonable to err on the side of strategy
> resistance. That's not to say that the honesty-favoring methods don't
> have their place, though: Debian seems to do pretty well with Schulze,
> for instance.
>
> As for methods like Plurality and (probably) IRV -- well, they're just
> Pareto-dominated. You can find methods with better VSE and the same
> level of strategic resistance, or methods that handle strategy better
> while providing the same VSE.
>
> -km
>
TS
Ted Stern
Fri, Jan 7, 2022 10:17 PM
[ISDA = Independence of Smith-dominated alternatives]
ISDA worthy criterion to satisfy. Unfortunately, you lose summability by
requiring a recount. Is there any way around having to eliminate non-Smith
candidates and recount?
I was going to suggest calling your method "Practical Ranked Approval", to
avoid having to include terms like Smith, Game-resistant, Burial, etc. But
requiring a recount might not be considered Practical. So the best you
could say would be Strategy-resistant Ranked Approval.
Thinking along the lines of practicality, I have been mulling how best to
keep pairwise arrays to a reasonable size for practical summability. If
pre-election polling is available, one could accumulate pairwise counts
explicitly for any candidate with more than, say, 2% approval, and lump
pairwise counts for other candidates under "Other". Compute the Smith Set.
If Other is in the Smith Set, then reduce the threshold and recount.
On Fri, Jan 7, 2022 at 1:57 PM Forest Simmons forest.simmons21@gmail.com
wrote:
Very true, and I would have said err on the side of VSE until Robert B-J
convinced me that sincere cycles are practically non-existent with an
occurrence of less than 0.5 percent.
He got that statistic from Fair Vote's analysis of over 400 elections, and
they have no particular reason to exaggerate that statistic. So suppose
they're right, then do we conclude that any old Condorcet method is as good
as another?
You know the answer to that, Kristofer, but I elaborate for the benefit of
the typical EM List reader:
No, because, as Kristofer pointed out, it depends on the "neighborhood."
Back in the fifties when I was a kid in rural/small town eastern
Washington state nobody locked their doors. In fact, they usually left
their keys in the ignition for convenience. But was that the
prudent/sustainable thing to do?
Nowadays sixty years later, in those same neighborhoods people have
learned by experience to adopt city slicker habits of door locking.
When 99.5 percent of election polls reveal the existence of sincere
Condorcet candidates, the election method with the greatest Condorcet
efficiency will come the closest to electing a CW 99.5 percent of the time.
But don't all Condorcet methods have equal Condorcet efficiency? Isn't the
very definition of "Condorcet method" a method that always elects the
Condorcet candidate when one exists?
That would be nice if there were such a method, but Gibbard-Satterwaithe
shows the impossibility of that ideal in any tough neiborhood.
Then what is the real definition of "Condorcet Method"?
It is a method that elects a ballot Condorcet candidate. In tough
neighborhoods there will often be sincere Condorcet candidates that are not
ballot Condorcet Candidates ... because they are ranked insincerely on the
ballots.
So how to ensure that sincere Condorcet candidates retain their Condorcet
status on the ballots?
Only methods that take this question seriously can have sustainable
Condorcet efficiency.
Let's do a thought experiment to compare the Condorcet efficiency of
Ranked Pairs, Benham, Schulze, and other Condorcet methods:
Suppose that sincere preferences are given...
40 A>C
35 B>C
25 C>A
Like 99.5 percent of electorates this one has a sincere Condorcet
candidate, namely C, which is preferred over A by a 60 percent majority,
and preferred over B by a 65 percent majority.
The question of Condorcet efficiency in this example is "Which Condorcet
methods are most likely to elect C ?"
But that depends on the neighborhood. In Grant County, Washington of the
fifties, any Condorcet method would elect C.
But which Condorcet method would robustly elect C even in Brooklyn, New
York?
That depends on which methods take the possibility of subversion
seriously... subversion of the sincere CW by intentional cycle creation.
Under Schulze, RP, River, MinMax, etc. candidate A would likely win
because the A faction could confidently bury A under B, creating an
insincere defeat cycle with weakest defeat being the 60 percent C over A
compared with the 75 percent B over C, and the 65 percent A over B.
All of these standard Condorcet methods are built on the assumption that
the smallest majority (60% C over A) is the one most likely to be "wrong".
So they elect A, unwittingly rewarding the A faction for subverting the
sincere CW.
But how about Q&D/C?
It would elect C because it is a ballot Condorcet method intolerant of the
kind of subversion that rewarded the A faction under Schulze, etc. Since
the subversion would just backfire, no faction would be dumb enough to try
it.
Let's look at how it would backfire were the A faction stupid enough to
attempt the burial of C under B:
The candidate with the lowest basic score would then be C, so B would win
as the only candidate pairwise undefeated by C.
So A's gambit would just change the winner from its second choice C to its
last choice B.
Do you think any major faction in Brooklyn would gamble on a sure loss
like that?
To be clear, what this example shows is that our Quick and Dirty/Clean
method has Condorcet efficiency superior to that of Schulze, etc.
And why is it more Condorcet efficient? Because it was designed on a
realistic game theoretic principle, rather than a statistical error
correcting technique designed mainly for filtering out inadvertent
"mistakes" in judgment (or minority opinion).
This example could be (and has been) multiplied indefinitely with the same
pattern of results:
Q&D/C is much more Condorcet efficient than Schulze et al in rough
neighborhoods, as well as 100 percent efficient in easy neighborhoods (like
any and every Condorcet method is).
So here's the best advice for Condorcet method design: focus on
frustrating cycle creators. Make sure that all attempts at cycle creation
backfire automatically. Beyond that accommodate any (possible but
vanishingly rare) sincere cycles by making sure the method always elects
simply, monotonically, and clone indepently from the ballot Smith set, as
does Q&D/C, a method that Pareto dominates all of the old Condorcet methods
on these criteria.
To put it bluntly, all of those older methods are pretty much passe ...
uniformly dominated by Q&D/C when it comes to single winner elections for
public office.
I'm sorry to talk so bluntly, but subtle words don't seem to register in
this context, especially among the self-styled movers and shakers [who
probably won't even read them anyway🤔]
On a technical note ... to make sure that Q&D/C is ISDA, use a version
that immediately restricts to Smith, or at least counts the "basic scores"
relative to the Smith Set candidates.
Forest
El vie., 7 de ene. de 2022 4:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 07.01.2022 07:05, Forest Simmons wrote:
Most designers of Condorcet methods asume that the gentlemanly thing to
do is to give the votes a benefit of a doubt and assume that they must
have voted sincerely but cycles are a result of errores of judgement.
Because of these assumptions they attempt to filter out the erroneous
preferences statistically
.. the main heuristic is that larger majorities are less apt to hold
erroneous opinions than smaller ones ... hence cycles are broken by
annulling the defeats with the smallest majorities.
There is probably a tradeoff between strategic resistance and honest
VSE. The more you want one, the less you get the other, and at the
extremes you completely disregard one or the other.
So if we could construct methods to spec, the best approach would be to
somehow infer just how much strategy the method needs to resist, and
then maximize VSE in a suitable model (probably spatial) subject to this
constraint.
But we don't really know how strong the barrier has to be against
strategy. As I've mentioned before, I think it differs based on culture:
IIRC Ireland saw much less vote management under STV than did New York.
If we only have one shot, it's reasonable to err on the side of strategy
resistance. That's not to say that the honesty-favoring methods don't
have their place, though: Debian seems to do pretty well with Schulze,
for instance.
As for methods like Plurality and (probably) IRV -- well, they're just
Pareto-dominated. You can find methods with better VSE and the same
level of strategic resistance, or methods that handle strategy better
while providing the same VSE.
-km
[ISDA = Independence of Smith-dominated alternatives]
ISDA worthy criterion to satisfy. Unfortunately, you lose summability by
requiring a recount. Is there any way around having to eliminate non-Smith
candidates and recount?
I was going to suggest calling your method "Practical Ranked Approval", to
avoid having to include terms like Smith, Game-resistant, Burial, etc. But
requiring a recount might not be considered Practical. So the best you
could say would be Strategy-resistant Ranked Approval.
Thinking along the lines of practicality, I have been mulling how best to
keep pairwise arrays to a reasonable size for practical summability. If
pre-election polling is available, one could accumulate pairwise counts
explicitly for any candidate with more than, say, 2% approval, and lump
pairwise counts for other candidates under "Other". Compute the Smith Set.
If Other is in the Smith Set, then reduce the threshold and recount.
On Fri, Jan 7, 2022 at 1:57 PM Forest Simmons <forest.simmons21@gmail.com>
wrote:
> Very true, and I would have said err on the side of VSE until Robert B-J
> convinced me that sincere cycles are practically non-existent with an
> occurrence of less than 0.5 percent.
>
> He got that statistic from Fair Vote's analysis of over 400 elections, and
> they have no particular reason to exaggerate that statistic. So suppose
> they're right, then do we conclude that any old Condorcet method is as good
> as another?
>
> You know the answer to that, Kristofer, but I elaborate for the benefit of
> the typical EM List reader:
>
> No, because, as Kristofer pointed out, it depends on the "neighborhood."
>
> Back in the fifties when I was a kid in rural/small town eastern
> Washington state nobody locked their doors. In fact, they usually left
> their keys in the ignition for convenience. But was that the
> prudent/sustainable thing to do?
>
> Nowadays sixty years later, in those same neighborhoods people have
> learned by experience to adopt city slicker habits of door locking.
>
> When 99.5 percent of election polls reveal the existence of sincere
> Condorcet candidates, the election method with the greatest Condorcet
> efficiency will come the closest to electing a CW 99.5 percent of the time.
>
> But don't all Condorcet methods have equal Condorcet efficiency? Isn't the
> very definition of "Condorcet method" a method that always elects the
> Condorcet candidate when one exists?
>
> That would be nice if there were such a method, but Gibbard-Satterwaithe
> shows the impossibility of that ideal in any tough neiborhood.
>
> Then what is the real definition of "Condorcet Method"?
>
> It is a method that elects a ballot Condorcet candidate. In tough
> neighborhoods there will often be sincere Condorcet candidates that are not
> ballot Condorcet Candidates ... because they are ranked insincerely on the
> ballots.
>
> So how to ensure that sincere Condorcet candidates retain their Condorcet
> status on the ballots?
>
> Only methods that take this question seriously can have sustainable
> Condorcet efficiency.
>
> Let's do a thought experiment to compare the Condorcet efficiency of
> Ranked Pairs, Benham, Schulze, and other Condorcet methods:
>
> Suppose that sincere preferences are given...
>
> 40 A>C
> 35 B>C
> 25 C>A
>
> Like 99.5 percent of electorates this one has a sincere Condorcet
> candidate, namely C, which is preferred over A by a 60 percent majority,
> and preferred over B by a 65 percent majority.
>
> The question of Condorcet efficiency in this example is "Which Condorcet
> methods are most likely to elect C ?"
>
> But that depends on the neighborhood. In Grant County, Washington of the
> fifties, any Condorcet method would elect C.
>
> But which Condorcet method would robustly elect C even in Brooklyn, New
> York?
>
> That depends on which methods take the possibility of subversion
> seriously... subversion of the sincere CW by intentional cycle creation.
>
> Under Schulze, RP, River, MinMax, etc. candidate A would likely win
> because the A faction could confidently bury A under B, creating an
> insincere defeat cycle with weakest defeat being the 60 percent C over A
> compared with the 75 percent B over C, and the 65 percent A over B.
>
> All of these standard Condorcet methods are built on the assumption that
> the smallest majority (60% C over A) is the one most likely to be "wrong".
>
> So they elect A, unwittingly rewarding the A faction for subverting the
> sincere CW.
>
> But how about Q&D/C?
>
> It would elect C because it is a ballot Condorcet method intolerant of the
> kind of subversion that rewarded the A faction under Schulze, etc. Since
> the subversion would just backfire, no faction would be dumb enough to try
> it.
>
> Let's look at how it would backfire were the A faction stupid enough to
> attempt the burial of C under B:
>
> The candidate with the lowest basic score would then be C, so B would win
> as the only candidate pairwise undefeated by C.
>
> So A's gambit would just change the winner from its second choice C to its
> last choice B.
>
> Do you think any major faction in Brooklyn would gamble on a sure loss
> like that?
>
> To be clear, what this example shows is that our Quick and Dirty/Clean
> method has Condorcet efficiency superior to that of Schulze, etc.
>
> And why is it more Condorcet efficient? Because it was designed on a
> realistic game theoretic principle, rather than a statistical error
> correcting technique designed mainly for filtering out inadvertent
> "mistakes" in judgment (or minority opinion).
>
> This example could be (and has been) multiplied indefinitely with the same
> pattern of results:
>
> Q&D/C is much more Condorcet efficient than Schulze et al in rough
> neighborhoods, as well as 100 percent efficient in easy neighborhoods (like
> any and every Condorcet method is).
>
> So here's the best advice for Condorcet method design: focus on
> frustrating cycle creators. Make sure that all attempts at cycle creation
> backfire automatically. Beyond that accommodate any (possible but
> vanishingly rare) sincere cycles by making sure the method always elects
> simply, monotonically, and clone indepently from the ballot Smith set, as
> does Q&D/C, a method that Pareto dominates all of the old Condorcet methods
> on these criteria.
>
> To put it bluntly, all of those older methods are pretty much passe ...
> uniformly dominated by Q&D/C when it comes to single winner elections for
> public office.
>
> I'm sorry to talk so bluntly, but subtle words don't seem to register in
> this context, especially among the self-styled movers and shakers [who
> probably won't even read them anyway🤔]
>
> On a technical note ... to make sure that Q&D/C is ISDA, use a version
> that immediately restricts to Smith, or at least counts the "basic scores"
> relative to the Smith Set candidates.
>
> Forest
>
>
> El vie., 7 de ene. de 2022 4:38 a. m., Kristofer Munsterhjelm <
> km_elmet@t-online.de> escribió:
>
>> On 07.01.2022 07:05, Forest Simmons wrote:
>> > Most designers of Condorcet methods asume that the gentlemanly thing to
>> > do is to give the votes a benefit of a doubt and assume that they must
>> > have voted sincerely but cycles are a result of errores of judgement.
>> >
>> > Because of these assumptions they attempt to filter out the erroneous
>> > preferences statistically
>> > .. the main heuristic is that larger majorities are less apt to hold
>> > erroneous opinions than smaller ones ... hence cycles are broken by
>> > annulling the defeats with the smallest majorities.
>>
>> There is probably a tradeoff between strategic resistance and honest
>> VSE. The more you want one, the less you get the other, and at the
>> extremes you completely disregard one or the other.
>>
>> So if we could construct methods to spec, the best approach would be to
>> somehow infer just how much strategy the method needs to resist, and
>> then maximize VSE in a suitable model (probably spatial) subject to this
>> constraint.
>>
>> But we don't really know how strong the barrier has to be against
>> strategy. As I've mentioned before, I think it differs based on culture:
>> IIRC Ireland saw much less vote management under STV than did New York.
>>
>> If we only have one shot, it's reasonable to err on the side of strategy
>> resistance. That's not to say that the honesty-favoring methods don't
>> have their place, though: Debian seems to do pretty well with Schulze,
>> for instance.
>>
>> As for methods like Plurality and (probably) IRV -- well, they're just
>> Pareto-dominated. You can find methods with better VSE and the same
>> level of strategic resistance, or methods that handle strategy better
>> while providing the same VSE.
>>
>> -km
>>
>
