1 Jun 2010 18:52

### Range PAV results

```> Also of note is that even though STV only has access to ordinal
> information, none of the cardinal methods manage to dominate it.

In STV the candidates represent particular voters, and get no additional credit
for pleasing some of the other voters.  In PAV the voters that get satisfaction
from other voters' representatives are considered already partially covered.

Consider, for example, the two winner election profile

10 A
40 A>B(99%)
45 C>A(99%)=B(99%)
5  C

If I am not mistaken, the set {A,C} would be the STV winner, but {A,B} would be
the PAV winner, whether greedy or exhaustive.

The STV set has perfect proportionality, but the PAV set gives double coverage
to 85 percent of the voters at the expense of 5 percent getting nothing.

Consider that (1) in a two winner election a faction of 5% has no guarantee of a
representative, (2) even so in this election if all of the C supporters got
their act together they could have gotten C into the junta, but
(3) they are better off as a whole taking the double coverage.

It seems to me that on the whole, {A,B} is a more representative set than {A,C}.

How does the automatic rating system of the simulation compare them?
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1 Jun 2010 21:12

### Tanking advantage of cycle proof conditions

```
Forest Simmons wrote (29 May 2010):

Here's a four slot method that takes advantage of the impossibility of beat
>cycles under certain conditions:.
>
>
>Use range style ballots with four levels: 0, 1, 2, and 3.
>
>(1) First eliminate all candidates that are pairwise defeated by a ratio greater
>than 3/1.
>
>
>(2) Then eliminate all of the candidates that are pairwise defeated by a ratio
>greater than 2/1 based on only those comparisons that involve an extreme rating,
>i.e. 3 beats a 0, 1, or 2, while 1, 2, or 3 beats a 0, but don't count a 2 as
>beating a 1, since neither 1 not 2 is an extreme rating on our four slot ballot.
>
>
>(3) Finally, eliminate all of the candidates that are pairwise defeated by any
>ratio greater than 1/1 on the basis of comparisons that involve a rating
>difference of at least two, i.e. 3 vs. 0 or 1, and  2 or 3 vs. 0, while
>considering 3 vs. 2,   2 vs. 1, and 1 vs. 0 to be too weak for this final
>elimination decision that is based on a mere 1/1 defeat ratio cutoff.
>
>The candidate that remains is the winner.
>
>
>
>If there is a pairwise tie in step three, use the middle two levels to resolve
```

1 Jun 2010 22:37

### Range PAV results

```Here's an example (two winner) that's even more to the point of the difference
between STV and PAV:

1 A(`100)
49 A>B(99)=D(99)>>C(0)
49 C>B(99)=D(99)>>A(0)
1 C(100)

I think that STV would elect A and C, while Range PAV would elect B and D.
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```
2 Jun 2010 00:00

### methods based on cycle proof conditions

```A couple of other possibilities for methods based on cycle proof conditions:

I.  BDR or "Bucklin Done Right:"

Use 4 levels, say, zero through three.  First eliminate all candidates defeated
pairwise with a defeat ratio of 3 to 1.  Then collapse the top two levels, and
eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
candidates are left, among these elect the one with the greatest number of
positive ratings.

II.  SSCPE or "Six Slot Cycle Proof Elimination"

Use six levels, zero through five.  First eliminate all candidates with a
pairwise defeat ratio of five to one.  Then allowing only those ballot
comparisons with strength 2 or greater (i.e. the preferred candidate is rated at
least two levels above the other), eliminate all candidates with a defeat ratio
of two to one.  Then allowing only comparisons with strength three or greater,
eliminate all candidates beaten with a defeat ratio of one to one, i.e. all
defeated candidates.  If there are two or more undefeated candidates, elect the
one with the greatest number of positive ratings.

III.  SPE or Strong Preference Elimination:

Use 2n levels.  First eliminate all of the candidates that are defeated when the
only ballot preferences counted are of strength n or greater, i.e. the rating of
the preferred one is at least n levels greater than the rating of the other.  If
there are two or more unbeaten candidates, collapse the bottom three levels to
zero, decrement the other levels by two, and decrement 2n to 2(n-1) and repeat
the process recursively.

```

2 Jun 2010 02:44

### Re: methods based on cycle proof conditions

How about "Condorcet-compliant Range-2 with Runoffs": use 3 levels, 0 through 2 (these numbers matter). Find the largest number of approvals for any candidate. Any candidate with a range score higher than that number makes it into the runoff. This cannot eliminate the CW (or indeed, any member of the Smith set. To extend this property to more than two levels, you have to implicityly collapse to 2 levels first.) Then use a condorcet-compliant method for the runoff.

JQ

2010/6/1
A couple of other possibilities for methods based on cycle proof conditions:

I.  BDR or "Bucklin Done Right:"

Use 4 levels, say, zero through three.  First eliminate all candidates defeated
pairwise with a defeat ratio of 3 to 1.  Then collapse the top two levels, and
eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
candidates are left, among these elect the one with the greatest number of
positive ratings.

II.  SSCPE or "Six Slot Cycle Proof Elimination"

Use six levels, zero through five.  First eliminate all candidates with a
pairwise defeat ratio of five to one.  Then allowing only those ballot
comparisons with strength 2 or greater (i.e. the preferred candidate is rated at
least two levels above the other), eliminate all candidates with a defeat ratio
of two to one.  Then allowing only comparisons with strength three or greater,
eliminate all candidates beaten with a defeat ratio of one to one, i.e. all
defeated candidates.  If there are two or more undefeated candidates, elect the
one with the greatest number of positive ratings.

III.  SPE or Strong Preference Elimination:

Use 2n levels.  First eliminate all of the candidates that are defeated when the
only ballot preferences counted are of strength n or greater, i.e. the rating of
the preferred one is at least n levels greater than the rating of the other.  If
there are two or more unbeaten candidates, collapse the bottom three levels to
zero, decrement the other levels by two, and decrement 2n to 2(n-1) and repeat
the process recursively.

The idea of SPE is that the most important eliminations are done by strong
preferences, and weaker preferences are invoked only to break ties.  More than
one tie breaker step is needed only to ensure the technical compliance with
Pareto.  Random ballot could be used as a tie breaker just as well where ever
voters are not allergic to it.
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```
2 Jun 2010 23:27

### Re: Range PAV results

```fsimmons <at> pcc.edu wrote:
>> Also of note is that even though STV only has access to ordinal
>> information, none of the cardinal methods manage to dominate it.
>
> In STV the candidates represent particular voters, and get no additional credit
> for pleasing some of the other voters.  In PAV the voters that get satisfaction
> from other voters' representatives are considered already partially covered.
>
> Consider, for example, the two winner election profile
>
> 10 A
> 40 A>B(99%)
> 45 C>A(99%)=B(99%)
>  5  C
>
> If I am not mistaken, the set {A,C} would be the STV winner, but {A,B} would be
> the PAV winner, whether greedy or exhaustive.
>
> The STV set has perfect proportionality, but the PAV set gives double coverage
> to 85 percent of the voters at the expense of 5 percent getting nothing.
>
> Consider that (1) in a two winner election a faction of 5% has no guarantee of a
> representative, (2) even so in this election if all of the C supporters got
> their act together they could have gotten C into the junta, but
> (3) they are better off as a whole taking the double coverage.
>
> It seems to me that on the whole, {A,B} is a more representative set than {A,C}.
>
> How does the automatic rating system of the simulation compare them?

The automatic rating system can't determine the proportionality of an
arbitrary ballot set. The way the system works, it first generates a
model (in this case, every candidate and voter having n binary issues
which are randomly yes or no according to a predetermined fraction of
ayes for each), and then it derives the ballots based on that (voters
preferring candidates closer to them, rating them proportional to the
number of bits by which their opinions agree).

That said, I think my system would give a scenario like the above a
better regret score at the cost of (exact) proportionality, because the
majority (the 85%) get a better outcome with the 5% getting nothing.

If the scores are:

10: A(100%) > B = C (0%)
40: A(100%) > B(99%) > C (0%)
45: C(100%) > A = B (99%)
5: C(100%) > A = B (0%)

then A+B has a utility sum of 100*10 + 100*40 + 99*40 + 99*45 + 99*45 =
17870 whereas A+C's utility sum is 100*10 + 100*40 + 45*100 + 5*100 = 10000.

Thus, {A,B} would probably get a better regret score than {A,C}, but a
somewhat worse proportionality score.

-

Your example makes use of preferences that are very close to 100. Would
an ordinary (rank-ballot) multiwinner method (satisfying DPC etc) elect
{A,B} if the preferences that are very close to 100 were to be brought
to 100? I.e. if

10: A > B = C
40: A = B > C
45: (null)
5: C > A = B

would an ordinary method elect {A,B}? I guess what I'm asking is whether
Range PAV electing {A,B} is because of a property of Range PAV as
compared to other (reasonable) cardinal multiwinner methods, or of
cardinal multiwinner methods in comparison to ordinal ones.
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```
3 Jun 2010 20:54

### methods based on cycle proof conditions

``` Forest Simmons wrote (1 June 2010):
<snip>

I.  BDR or "Bucklin Done Right:"
>
>Use 4 levels, say, zero through three.  First eliminate all candidates defeated
>pairwise with a defeat ratio of 3 to 1.  Then collapse the top two levels, and
>eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
>candidates are left, among these elect the one with the greatest number of
>positive ratings.
>
<snip>

This seems to be even more Approvalish than normal Bucklin.

65: A3, B2
35: B3, A0

(I assume that zero indicates least preferred)

Forest's "BDR" method elects A, failing Majority Favourite.

Chris Benham

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```
3 Jun 2010 22:28

### Re: methods based on cycle proof conditions

```At 02:54 PM 6/3/2010, Chris Benham wrote:
>  Forest Simmons wrote (1 June 2010):
><snip>
>
>I.  BDR or "Bucklin Done Right:"
> >
> >Use 4 levels, say, zero through three.  First eliminate all
> candidates defeated
> >pairwise with a defeat ratio of 3 to 1.  Then collapse the top two
> levels, and
> >eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
> >candidates are left, among these elect the one with the greatest number of
> >positive ratings.
> >
><snip>
>
>This seems to be even more Approvalish than normal Bucklin.
>
>65: A3, B2
>35: B3, A0
>
>(I assume that zero indicates least preferred)
>
>Forest's "BDR" method elects A, failing Majority Favourite.

As will any method which optimizes expressed utility, assuming that
these numbers are a rough expression of utility. Because they can be
some kind of artifact, I would suggest that a vote like this calls
for a runoff.

Note, in Bucklin, those approvals would be spread, and, of course, A
wins in the first round. In real elections, though, with numbers like
this, and no interfering candidate, if the 65 actually do rate B with
a 2, they are expressing that they don't much care, so the election
can legitimately be decided by the voters who do care. This is normal
democratic practice! If a runoff where held with these primary
numbers, I would expect low turnout and B wins by a landslide.

But if the A faction was somehow duped into voting like that, the
reverse will happen.

Now, who would use BDR with only two candidates? It's like using
Range with only two candidates. Why would you care about "majority
favorite" if you decide to use raw range. I wonder why the A faction
even bothered to vote with that pattern of utilities ("ratings").

For many, for years, to note that a method failed the Majority
criterion ("Majority favorite") was the same as saying that it was
totally stupid. But real, direct, human decision-making doesn't
satisfy the criterion unless people just want a fast decision and
don't care much. If they don't care very much, and somebody does care
and makes a big fuss, what happens?

From the answer you can then tell what kind of society it is, its
sense of coherence and unity, its ability to negotiate consensus and
thus natural operational efficiency, etc.

If I let you have your way when you care and I don't, then you let me
have my way when I care and you don't. Therefore utility maximization
systems will generally improve outcomes for *everyone*. The overall
game is not a zero-sum game. Single-winner elections appear to be
only when they are divorced from the context.

There is something related in comparing Approval and Range.

I found this odd effect, studying absolute expected utility in a
zero-knowledge Range 2 election voted "sincerely" in the presence of
a middle utility candidate. The individual expected utility for the
voter was the same for the approval-style vote vs the Range vote.
(And, of course, it was the same if you voted for the middle
candidate or not, the situation is symmetrical). But if the *method*
was approval, the expected utility was lower than if the method was
Range. Compared to Range, Approval was lowering everyone's expected utility!

So everyone votes Approval style, the expected utility of the outcome
must be lower than if at least one person votes Range style.
(Otherwise Range is the same as Approval, if nobody actually uses the
intermediate rating.)

Nobody has bothered to confirm or disconfirm this result. Warren
validated some of it, but not that part, and to really nail it down
required more math than I could easily do.

In order to insist on the Majority criterion, we lower the expected
utility for nearly everyone, certainly for most people, when we
consider many elections.

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```
4 Jun 2010 13:43

### methods based on cycle proof conditions

```

I.  BDR or "Bucklin Done Right:"
>
>Use 4 levels, say, zero through three.  First eliminate all candidates defeated
>pairwise with a defeat ratio of 3 to 1.  Then collapse the top two levels, and
>eliminate all candidates that suffer a defeat ratio of 2 to 1.  If any
>candidates are left, among these elect the one with the greatest number of
>positive ratings.
>
>
><snip>

This seems to be even more Approvalish than normal Bucklin.

65: A3, B2
35: B3, A0

(I assume that zero indicates least preferred)

Forest's "BDR" method elects A, failing Majority Favourite.

In response to the above, Abd Lomax-Smith wrote (3 June 2010):

<snip>

Now, who would use BDR with only two candidates? It's like using
>Range with only two candidates. Why would you care about "majority
>favorite" if you decide to use raw range. I wonder why the A faction
>even bothered to vote with that pattern of utilities ("ratings").
><snip>

I was content to simply prove that the method simply fails Majority Favourite, but to appease
Abd  here is a similar example with three candidates:

60: A3, B1, C0
35: B3, A0, C1
05: C3, A2, C0

A is the big majority favourite and the big voted  "raw range" winner, and yet  B wins.

Chris Benham

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```
6 Jun 2010 21:09

### Re: Preliminary Range PAV results

```Jameson Quinn wrote:
> Can you add a range-STV method? This would reweight ballots for elected
> candidate X by:
>
> max(0,(N-D(rN/R))/N)
>
> where N is num voters, D is droop quota, r is ballot's range score for
> X, and R is electorate's range total for X. This should be simple to
> code, a small variant on greedy PAV with a different reweighting formula.
>
> This is a more STV-like method, in that it elects candidates in order of
> some pseudo-"strong to weak", not in a proportional order as does greedy
> PAV.

If N = total number of voters, then (unless there's a bug somewhere) we get:

PA_Linear_Range_STV                                 0.29813 0.00031

If N = the number of voters voting according to the ballot being
reweighted, then the result is:

PA_Linear_Range_STV                                 0.20928 0.01991

As a comparison:

PA_Maj[Cardinal-20]                                 0.31348 0
PA_STV                                              0.11902 0.10007
PA_DHwL(L-R_offense/1/(wv),_0.925)                  0.20924 0.02172

The first is Range, and the second is ordinary STV. The third is the
closest ranked ballot method at this level of proportionality.

If you suspect a bug, I could give examples where "N = size of entire
electorate" gives a disproportional result (according to my program). "N
= number voting this way" seems more well behaved, but if you're
interested, I could also give an example where it differs from STV
and/or Birational.
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```

Gmane