This reply to Kristofer's post is strengthened by the attachments it mentions and which Steve will also sent to you by email if you wish (stevebosworth@hotmail.com).

From: steve bosworth stevebosworth@hotmail.com
Sent: Sunday, May 29, 2016 7:54 PM
To: Kristofer Munsterhjelm
Subject: Re: Your 5th APR dialogue with Steve

To Kristofer,

Thank you for your response.  In one sense, I accept that if truncated ballots are not permitted, Condorcet will also support the winner with at least by 50% + 1, e.g. the MAM winner will be supported by at least 50%+1 of all the voters.  However, this ‘support’ will usually be composed mainly of indirect or ‘transitive’ support.  In contrast to IRVa, the percentage of voters who have expressly preferred the winner over each of the other candidates can easily be much less than 50%.  This is illustrated by the example discussed in the Endnote to the attached newly drafted Appendix 4 to my article (Super Equality for Each Citizen’s Vote in the Legislature’).

In that extreme example, ‘as a result of using IRV, candidate F is elected with a 61% majority and with an average intensity of preference of 9.62 out of 10.  In contrast, by counting the same 100 ballots using MAM, candidate E is elected with the explicit support of only 39% and with an average intensity of preference of 9.28.’
As a result, the attached appendix also claims that only ‘IRVa allows each citizen to guarantee:

1.  that her vote will continue quantitatively to count equally,
   

2.  that the winner will be elected by a majority of all the voters who have not deliberately cancelled the possibility of their otherwise wasted ‘default vote’ being counted until the majority winner is discovered, and
   

3.  that the winner will be supported more enthusiastically than any other candidate.’
   

This extract and the other passages that I see as being relevant to our discuss are printed green in the attachment.
Below, I have attempted to respond to each of the points made in your most recent email.

I understand that you may not have the time to answer until July.

Best regards,
Steve

From: Kristofer Munsterhjelm km_elmet@t-online.de
Sent: Friday, May 13, 2016 8:33 AM
To: steve bosworth
Subject: Re: Your APR dialogue with Steve, a short addition

On 04/29/2016 11:20 PM, steve bosworth wrote:

K:  If candidate X is the Condorcet winner, there always exists an
elimination order (i.e. "first eliminate Y, then eliminate Z, then
eliminate W") so that X ends up being supported by 50% + 1 at the end*.

Or, more generally: unless X is the Condorcet loser, there exists such
an order.
S:  Yes, but the winner at the top of this order may not have been expressly preferred by even a plurality of the voters (as illustrated by the attached Endnote).  This contrasts with the IRVa winner.
K:  However, this elimination order doesn't need to be (and often isn't) elimination in order of Plurality losers.
S:  I see no reason to disagree with this.
[….]
S: > … and for not allowing a voter’s different intensities of preference to

K:  Take a more indirect look. As far as I understand you, you mean that for
say,
1: A>B>C>D>E>F (vote a)
the A>F vote counts as much as in
1: A>F>B>C>D>E (vote b)
S: Yes.
K:  but you'd like it to count more when A and F are further apart because
that shows that the voter dislikes F more.
S: Yes.
K:  But in the Condorcet matrix,
one A>...>F vote counts the same as a direct A>F>... vote as far as the
contest between A and F is concerned, so the method doesn't seem to take
intensity into account.
S: Yes, and that is the problem.
K: However, in vote a, the voter says "I'm willing to get B, C, D, or E
before I get F".
S:  Yes.
K: What this does is increment the scores B>F, C>F, D>F,
E>F. So there is, in a sense, a difference of preference. By
supporting many candidates over F, the voter lessens the chance that F
will win. Suppose, for instance, that B and F were nearly tied. In vote
a, the voter does a lot more to raise B over F (to make B stronger than
F) then he does in vote b.
S: Yes, if I understand you, this ‘incrementation’ can cause each citizen’s vote to have a different value in the whole process.  If so, the principle of ‘one-citizen-one-vote’ is violated.
K:  So the voter's enthusiasm does have an effect by making more candidates
stronger before the candidate he dislikes becomes stronger. He doesn't
support A directly, but he supports a lot more of F's rivals.
S: Yes, the presence or absence of different numbers of these different ‘effects’ can result in giving some voters more weight in determining the order than other voters.  Again, this would seem to violate the principle of ‘one-citizen-one-vote’.
K:  The same argument goes for raising. When a voter goes from

1: A>B>C>D
to
1: C>A>B>D

he helps C more by protecting C from A and B, i.e. by increasing A>C and
A>B.
S: I do not understand your phrase:  ‘i.e. by increasing A>C and A>B’.  This is because C>A>B>D favors C over A.
K: If that's not enough, consider Borda-elimination. It's like IRV, except
you count by Borda score rather than by first preferences when you
decide whom to eliminate. I imagine it'll be hard to say that
Borda-elimination fails to take enthusiasm into account, because
1: A>B>C>D>E>F
gives F 0 points and A 5 points in some given round, whereas
1: A>F>B>C>D>E
gives F 4 points and A 5 points.
Yet Borda-elimination passes Condorcet. It always elects the CW when it
exists. (Thus it also produces an elimination order where the Condorcet
winner ends up being supported by 50%+1)
S: Similar to Condorcet (MAM), and in contrast to IRVa, Borda splits each citizen’s one vote between the candidates.  Consequently, some citizens will be helping to defeat their most favored candidate by the lower scores they have given to the winning candidate.  Again, the winner may not be support by 50% +1 of the full value of each vote given by each citizen.  Nor is the average intensity of support given to the Borda winner likely to be as high as the winner discovered by IRVa.
S: > Lastly, unlike IRVawc and APR, each voter’s one vote in

K: IRV is nonmonotonic. That means that ranking someone higher may bump
them off the legislature, and whether that happens is unpredictable. How
can you then say that each voter's vote counts the same? It may
unpredictably count for a lot less - or more - in deciding whether
someone will get a seat.
S:  Using IRVa in the example in the attached Endnote, and when also used in your following example, each vote continues to count (directly or indirectly) as one until a majority winner is discovered.

K: * That is, unless there are truncated ballots, but then IRV doesn't give
a majority either.
S:  This is why I added ‘Asset Voting’ to IRV, to produce IRVa.  It seems to me that all ranking ballots should avoid truncation simply by interpreting each candidate not marked as having been rank equal bottom, except the candidates expressly ranked.
K:  E.g. single winner:

1: A
1: B
1: C
...
1: X
2: Z

Everyone but Z are eventually eliminated, but Z is supported by fewer
than 50%+1.
S:  Again, IRVa ballots would be made as easy as possible for citizens to use by counting each candidate not marked as equally ranked ‘bottom’ except for the candidates who had been expressly ranked.  Consequently, in this example, IRVa would firstly give candidates A, B, C & X the option either publically to elect Z by giving at least two of their ‘default votes’ to Z, or instead, to give all four of their ‘default votes’ to A, B, C or X .  If either two of these four candidates refused to give their ‘default votes’ to Z, or A, B, C & X found themselves unable to give their four ‘default votes’ to A, B, C or X, sequentially one of the four would be eliminated by lot and required publically to give his ‘default vote’ to one of the other candidates.  This elimination and vote transferring process would continue until one of the candidates had received four votes.

On 05/30/2016 04:53 PM, steve bosworth wrote:

What is the definition of "average intensity of preference"?

I tend to think that it's generally impossible to directly infer
enthusiasm from a ranked ballot, since we're dealing with ordinal data.
All you can say is whether "X likes Y more than X likes Z". You can't
know whether a ballot like

A>B>C>D>E>F

is

"I really like A, B, and C; I think D and E are so-so, and I loathe F"

or if it is

"I really like A, I think B is so-so, and I hate everybody else, but I'd
rather have C than D, D than E, E than F".

In other words, the rankings themselves can't tell you whether

A>B>C>D

is

Washington > Lincoln > Bush > Khrushchev

or if it is

Lincoln > Khrushchev > Lenin > Stalin,

and how much weight you should put on first place compared with second
place is obviously different for these two scenarios, as far as
intensity is concerned.

The general approacah for dealing with intensity of ranking is to use
ratings instead, or use something like Majority Judgement where you're
permitted to skip ranks and each rank has a name ("Excellent", "Good", etc).

As for the MAM example, it seems that F wins in MAM as well.

If I'm not mistaken, your vote set is

38: F
23: G>F
25: E
9: M>N>E
5: K>P>E

candidates: E F G K M N P

The Condorcet matrix (row beats column) is

Option E F G K M N P
E 0 39 39 34 30 30 34
F 61 0 38 61 61 61 61
G 23 23 0 23 23 23 23
K 5 5 5 0 5 5 5
M 9 9 9 9 0 9 9
N 9 9 9 9 0 0 9
P 5 5 5 0 5 5 0

The wv matrix is

Option E F G K M N P
E 0 0 39 34 30 30 34
F 61 0 38 61 61 61 61
G 0 0 0 23 23 23 23
K 0 0 0 0 0 0 5
M 0 0 0 9 0 9 9
N 0 0 0 9 0 0 9
P 0 0 0 0 0 0 0

So the first few lock-ins go (equal strength defeats broken in
alphabetical order for simplicity):

1. F>E with 61 votes
2. F>K with 61 votes
3. F>M with 61 votes
4. F>P with 61 votes

Clearly, E can't possibly beat F since F is locked ahead of E on the
very first step. You must have calculated the Condorcet or wv matrix
incorrectly. In fact, regardless of whether you use wv or margins, F
becomes the CW: he beats everybody else pairwise.

If you'd like to check the calculation for other ballot sets,
http://www.ericgorr.net/condorcet/ will give both the Condorcet matrix
and the wv matrix when verbosity is set to "Tell me some things" or
"Tell me everything".

It'll also show what pairs MAM locks if you set it to "Tell me
everything", although it may not show them in order of defeat strength;
it locks large numbers of pairs at once when it knows that it can do so
without causing a cycle.

I'll further comment on a few others, as much as I have time to at least:

• As do all -a methods.

I'm not sure what that means.

Consider this example from rangevoting.org:

10: G > C > P > M
3: C > G > P > M
5: C > P > M > G
6: M > P > C > G
4: P > M > C > G

IRV elects M. But M only has 6 first preferences whereas C, the
Condorcet winner, has 8. Since C is the Condorcet winner, it's also
supported by a majority, so majority support can't be the reason C
doesn't win in IRV.
So if first preferences is the metric of enthusiasm, IRV fails.

And if it isn't, then it's possible to construct examples where:

- some candidate A has one more first preference vote than some other

candidate W,
- A has no second place preferences,
- W has some constant c number of second place preferences,
- A wins in IRV,
- W is the Condorcet winner,

meaning that if one first place vote is worth x second place votes, then
you can make IRV choose wrong no matter what x is.

E.g. for x=14:

4:      W>B>A>C>D
5:      A>W>B>C>D
5:      C>W>A>B>D
5:      D>W>A>B>C

W has 4 first preferences and A has 5. W has 15 second preferences (A
has none) and A has 14 third preferences.

So either enthusiasm is only based on first preferences, in which case
the first example shows that IRV gets it wrong, or it's based on first
and second preferences, in which case the second example would make
IRV get it wrong.

There's a continuum argument hiding in here, though I don't have time to
go into it in detail. Briefly: IRV is more Plurality-like than Condorcet
and more Condorcetian than Plurality. So any argument that favors IRV
has to show that the argument can't be taken to its logical conclusion
to favor either Condorcet or Plurality just as strongly.

Suppose we conduct a presidential election by Approval ("thumbs up" or
"like") voting, and there are 5 candidates. That is, each voter gets a
list of the 5 presidential candidates and is told to place a mark by
each candidate he likes. The candidate with the most likes/approvals wins.

Now suppose that voter A chooses to approve of two candidates, while
voter B chooses to approve of three. Does that violate one citizen one
vote? After all, voter A "spent" two points (gave points to two
candidates) while B "spent" three.

I'd say no, because each ballot has one state for each candidate: either
liked or unliked. Each ballot adds the same information to the voting
pool: whether the candidate chose to strengthen candidate X, candidate
Y, candidate Z, etc... So if there are 5 candidates, each ballot has 5
aye/nay votes, and each voter spends 5 such votes. The voter who liked
two candidates used 2 ayes and 3 nays, and the other voter used 3 ayes
and 2 nays.

Analogously, a Condorcet election could be conducted by asking the voter
"in which of these runoffs would you support the first candidate on the
ballot?". There are n^2 hypothetical runoffs for n candidates, so that
kind of ballot would have n^2 aye/nay votes.

In a three candidate election, someone who votes

A>B>C

would say aye to "would you support the first candidate on the ballot in
an A vs B runoff?", "in an A vs C runoff?" and "in a B vs C runoff?",
and nay to all the others. It's just that, to spare the voter from
having to fill out a bunch of tedious yes/no questions, the method
infers the answer to these aye/nay questions from a ranked ballot.

And if the Approval ballot doesn't violate one man one vote, then
neither does the Condorcet ballot. Yes, some voters increment more
runoffs than others (if there's equal rank and truncation), but deciding
to not increment someone's runoff is also a vote.

Either I meant that he helps C by protecting C from an increase of A>C
and from an increase of A>B, or that was a typo. In either case, what I
meant was that if someone raises C, the number of votes it takes to make
A beat C is also increased.

I could probably have made that clearer.

Monotonicity failures require multiple ballot sets. See
http://www.rangevoting.org/Monotone.html. My point is that the voter's
power is an unpredictable function of his ranking. If he moves a
candidate higher on his ranking, he may cause the candidate to lose and
vice versa. That's not something that says "equal power" to me.

Under no method (except possibly Random Pair, Random Dictator, and
combinations of them) are votes completely equal in power anyway. It
depends on what they vote and when they do it. Suppose you have a
majority election:

51: A
50: B

and two voters show up, and they're the last two voters of the election.
If they vote for B, they change the outcome (great power); if they vote
for A, nothing happens (not so much).

Yes, but we were initially talking about IRV without -a:

That's what I was responding to.

To Kristofer,

Thank you for your earlier than expected response and especially for putting me onto Majority Judgment (MJ).

MJ seems to be even a better system for electing a single-winner than my IRVa.  In different ways, each allows each citizen to express the intensity with which she supports or opposes each candidate and guarantees the election of a winner who is supported by an absolute majority.  However, in contrast to IRVa, MJ enables each citizen within the MJ winner’s majority to be a part of that majority purely as a result of her own judgments.  She would never have to rely on her IRVa first choice but eliminated candidate sequentially to transfer her ‘default vote’ to the remaining candidate he currently sees as the one mostly likely to represent him and her most faithfully, i.e. until a majority candidate is discovered.  Also, while both IRVa and MJ would seems to offer no practical opportunity for anyone to manipulate an election with many voters and candidates, theoretically MJ would seems to be even more resistant to strategic voting.

Do you agree?  Is MJ currently your own preferred method for electing a single-winner?

Your answers to these questions may help further to clarify any later discussion we may have both about IRVa and MAM.

From: Kristofer Munsterhjelm km_elmet@t-online.de

Sent: Wednesday, June 1, 2016 9:14 PM

To: steve bosworth; election-methods@lists.electorama.com

Subject: Re: [EM] Kristofer and Steve on IRVa vs. MAM

On 05/30/2016 04:53 PM, steve bosworth wrote:

K: What is the definition of "average intensity of preference"?

I tend to think that it's generally impossible to directly infer

enthusiasm from a ranked ballot, since we're dealing with ordinal data.

All you can say is whether "X likes Y more than X likes Z".

S: Yes and I accept that ‘likes more’ does not always amount to ‘enthusiasm’.  However, it is always a ‘preference’.

K: You can't know whether a ballot like

A>B>C>D>E>F

is

"I really like A, B, and C; I think D and E are so-so, and I loathe F"

S: No, if this is what the voter feels, this is better expressed as

A=B=C>D=E

And leave F to be counted equal as ranked bottom with any other candidate not ‘expressly’ preferred.

K: or if it is

"I really like A, I think B is so-so, and I hate everybody else, but I'd

rather have C than D, D than E, E than F".

S:  Again, if ‘hate’ means ‘I do not want to support their election in any way, I see these feelings as more clearly expressed as follows:

A>B

However, if, in this case, ‘hate’ includes degrees of dislike, and the fear that F might only be prevented from being elected by the election of C, D, or E; this would be best expressed as:

A>B>C>D>E

K: In other words, the rankings themselves can't tell you whether

A>B>C>D

is

Washington > Lincoln > Bush > Khrushchev

or if it is

Lincoln > Khrushchev > Lenin > Stalin,

S:  A given voter herself knows.

K: … and how much weight you should put on first place compared with second

place is obviously different for these two scenarios, as far as

intensity is concerned.

S:  Nevertheless, each voter must decide.

K: The general approach for dealing with intensity of ranking is to use

ratings instead, or use something like Majority Judgement where you're

permitted to skip ranks and each rank has a name ("Excellent", "Good", etc).

As for the MAM example, it seems that F wins in MAM as well.

If I'm not mistaken, your vote set is

38: F

23: G>F

25: E

9: M>N>E

5: K>P>E

candidates: E F G K M N P

The Condorcet matrix (row beats column) is

Option  E      F      G      K      M      N      P

E      0      39      39      34      30      30      34

F      61      0      38      61      61      61      61

G      23      23      0      23      23      23      23

K      5      5      5      0      5      5      5

M      9      9      9      9      0      9      9

N      9      9      9      9      0      0      9

P      5      5      5      0      5      5      0

The wv matrix is

Option  E      F      G      K      M      N      P

E      0      0      39      34      30      30      34

F      61      0      38      61      61      61      61

G      0      0      0      23      23      23      23

K      0      0      0      0      0      0      5

M      0      0      0      9      0      9      9

N      0      0      0      9      0      0      9

P      0      0      0      0      0      0      0

So the first few lock-ins go (equal strength defeats broken in

alphabetical order for simplicity):

1. F>E with 61 votes

2. F>K with 61 votes

3. F>M with 61 votes

4. F>P with 61 votes

Clearly, E can't possibly beat F since F is locked ahead of E on the

very first step. You must have calculated the Condorcet or wv matrix

incorrectly. In fact, regardless of whether you use wv or margins, F

becomes the CW: he beats everybody else pairwise.

If you'd like to check the calculation for other ballot sets,

http://www.ericgorr.net/condorcet/ will give both the Condorcet matrix

and the wv matrix when verbosity is set to "Tell me some things" or

"Tell me everything".

It'll also show what pairs MAM locks if you set it to "Tell me

everything", although it may not show them in order of defeat strength;

it locks large numbers of pairs at once when it knows that it can do so

without causing a cycle.

I'll further comment on a few others, as much as I have time to at least:

• As do all -a methods.

I'm not sure what that means.

Consider this example from rangevoting.org:

10:    G > C > P > M

3:      C > G > P > M

5:      C > P > M > G

6:      M > P > C > G

4:      P > M > C > G

IRV elects M. But M only has 6 first preferences whereas C, the

Condorcet winner, has 8. Since C is the Condorcet winner, it's also

supported by a majority, so majority support can't be the reason C

doesn't win in IRV.

So if first preferences is the metric of enthusiasm, IRV fails.

And if it isn't, then it's possible to construct examples where:

    - some candidate A has one more first preference vote than some other

candidate W,

    - A has no second place preferences,

    - W has some constant c number of second place preferences,

    - A wins in IRV,

    - W is the Condorcet winner,

meaning that if one first place vote is worth x second place votes, then

you can make IRV choose wrong no matter what x is.

E.g. for x=14:

4:      W>B>A>C>D

5:      A>W>B>C>D

5:      C>W>A>B>D

5:      D>W>A>B>C

W has 4 first preferences and A has 5. W has 15 second preferences (A

has none) and A has 14 third preferences.

So either enthusiasm is only based on first preferences, in which case

the first example shows that IRV gets it wrong, or it's based on first

and second preferences, in which case the second example would make

IRV get it wrong.

There's a continuum argument hiding in here, though I don't have time to

go into it in detail. Briefly: IRV is more Plurality-like than Condorcet

and more Condorcetian than Plurality. So any argument that favors IRV

has to show that the argument can't be taken to its logical conclusion

to favor either Condorcet or Plurality just as strongly.

Suppose we conduct a presidential election by Approval ("thumbs up" or

"like") voting, and there are 5 candidates. That is, each voter gets a

list of the 5 presidential candidates and is told to place a mark by

each candidate he likes. The candidate with the most likes/approvals wins.

Now suppose that voter A chooses to approve of two candidates, while

voter B chooses to approve of three. Does that violate one citizen one

vote? After all, voter A "spent" two points (gave points to two

candidates) while B "spent" three.

I'd say no, because each ballot has one state for each candidate: either

liked or unliked. Each ballot adds the same information to the voting

pool: whether the candidate chose to strengthen candidate X, candidate

Y, candidate Z, etc... So if there are 5 candidates, each ballot has 5

aye/nay votes, and each voter spends 5 such votes. The voter who liked

two candidates used 2 ayes and 3 nays, and the other voter used 3 ayes

and 2 nays.

Analogously, a Condorcet election could be conducted by asking the voter

"in which of these runoffs would you support the first candidate on the

ballot?". There are n^2 hypothetical runoffs for n candidates, so that

kind of ballot would have n^2 aye/nay votes.

In a three candidate election, someone who votes

A>B>C

would say aye to "would you support the first candidate on the ballot in

an A vs B runoff?", "in an A vs C runoff?" and "in a B vs C runoff?",

and nay to all the others. It's just that, to spare the voter from

having to fill out a bunch of tedious yes/no questions, the method

infers the answer to these aye/nay questions from a ranked ballot.

And if the Approval ballot doesn't violate one man one vote, then

neither does the Condorcet ballot. Yes, some voters increment more

runoffs than others (if there's equal rank and truncation), but deciding

to not increment someone's runoff is also a vote.

Either I meant that he helps C by protecting C from an increase of A>C

and from an increase of A>B, or that was a typo. In either case, what I

meant was that if someone raises C, the number of votes it takes to make

A beat C is also increased.

I could probably have made that clearer.

Monotonicity failures require multiple ballot sets. See

http://www.rangevoting.org/Monotone.html. My point is that the voter's

power is an unpredictable function of his ranking. If he moves a

candidate higher on his ranking, he may cause the candidate to lose and

vice versa. That's not something that says "equal power" to me.

Under no method (except possibly Random Pair, Random Dictator, and

combinations of them) are votes completely equal in power anyway. It

depends on what they vote and when they do it. Suppose you have a

majority election:

51: A

50: B

and two voters show up, and they're the last two voters of the election.

If they vote for B, they change the outcome (great power); if they vote

for A, nothing happens (not so much).

Yes, but we were initially talking about IRV without -a:

That's what I was responding to.

On 06/11/2016 06:04 PM, steve bosworth wrote:

Before I answer that point, let me pose a question to you. Do you agree
that MAM elects F given the ballot set you presented in your previous post?

With that done, let's continue.

There are different scenarios and my best methods depend on the scenario.

Scenario 1 is pure utilitarian voting, which I consider to be a very
unlikely setting for practical political elections. Here a voter knows
how much more benefit candidate A will give than candidate B, and
everybody has a common numerical (ratio) scale for evaluating the
candidates. An example of such a scenario would be a computer AI using
many different subprograms to determine which strategy is the best based
on what it currently knows; or, in a voting situation, all the voters
being perfect capitalists for whom log(dollars earned) is a good measure
of utility. Here Range is best, although if strategy is involved, I
would prefer something closer to DSV like SARVO-range to take that into
account. There may be even better ones that depend on making assumptions
about the statistical distribution the voters' votes come from, but I
haven't investigated that yet.

In scenario 2, the voters grade the candidates, but not on a curve. When
the voters grade the candidates (rank them, but are able to skip ranks),
and their grading of the candidates that do stand is independent of
which candidates stand, then we have this scenario, and MJ is best.
Most noticeably, it circumvents Arrow's impossibility theorem in this
situation. "Grading is independent of which candidates stand" means that
voter X will rank candidates A and B at Excellent and Good whether or
not candidate C is in the running. That is, even if candidate C is
completely awful, it won't affect X's decision of what grade to give A
and B. This property may hold or not hold - it depends on the voters'
behavior. Based on their exit poll data in France, Balinski and Laraki
argue that this property is preserved for real elections.
However, it's not hard to imagine a setting where pretty much every
candidate is good, so the voters spread out their ranks to make it
easier to distinguish them. Thus, if someone awful were to join, the
voters would compress their rankings of the good candidates so as to
make the scale large enough to show the system just how awful the awful
candidate is. E.g.
without the awful candidate:
- Agreeable Capitalist is Excellent
- Agreeable Socialist is Good
And with:
- Agreeable Capitalist and Socialist are both Excellent
- Horrible Authoritarian is Poor

Scenario 3 involves grading on a curve or ranking without skipping. I
think that Condorcet is best in that situation. Since introducing a new
candidate will affect all ranks, every method is subject to Arrow's
impossibility theorem and no method can pass IIA. However, we might get
close.I prefer a method that is:
- Condorcet (a candidate who wins every hypothetical runoff wins outright)
- Smith/Schwartz and similar extensions (Condorcet for sets)
- Independence from clones
- Monotonicity and reversal symmetry (to limit how bizarrely the method
acts given honest votes).
- Tougher independence criteria if possible (IPDA, ISDA).

Note that if there is a Condorcet winner, then the method obeys IIA. You
can remove any candidate (except the winner, obviously) and the outcome
will stay the same.

As long as the Condorcet method is "advanced" (pretty much cloneproof,
Smith and monotone), I don't have a problem switching from one to the
other. Different methods pick winners according to different objectives,
but they don't matter as much in my opinion. For instance, Schulze
passes a minimax objection property and I think Schulze argued that it
would make fewer voters unhappy on average than would not passing that
criterion. On the other hand, MAM is simple, passes another resistance
to complaints criterion, and passes local IIA.
Perhaps River would be my "ultimate" Condorcet method (though it doesn't
provide a full ranking) but I'm not sure. It is simpler than MAM, even
more Smith-like, and passes independence from Pareto-dominated
alternatives, which makes it very hard to dump loser candidates into the
system and have them affect the outcome.

If I recall correctly, Range advocates say that either we're close
enough to scenario 1 that Range is still the best, or that Range
degrades gracefully when ew move into scenarios 2 and 3, so we can still
use it. They would point at the simplicity and the real world use of
Range (e.g. star ratings on web sites). On the other hand, a problem
with Range is that there's an incentive to rate either maximum or
minimum, and again IIRC, so many users ended up doing so that YouTube
switched from star ratings to thumbs up/down rating.

If we are in scenario 2 or close to it, then MJ is a good choice. We
might also be between scenarios 2 and 3. If we are, we could possibly
still argue for MJ in this way: since it's so resistant to strategy and
to noise in general, if a few voters alter their grade scale based on
the candidates, that's no problem as long as most voters act like in
scenario 2. So if that argument is correct, we have to go pretty far
towards scenario 3 before MJ is no longer a good method.

However, questions of which scenario we're in would ultimately have to
be answered by actual data about real elections. Dynamics also
complicate the picture: if a system provides relatively good results
under honesty and doesn't reward strategy too much, the voters might
become more honest over time simply because they don't get the payoff
for strategy. But it'd have to be found out by trying the method;
mathematics can't answer that. If voters were completely Homo Economicus
rational, they wouldn't bother voting in the first place, after all.

I see IRV to pretty clearly be a scenario 3 method. No surprise then
that I value Condorcet methods above it. For scenario 2, I value MJ over
it. For one, IRV fails IIA even where MJ passes it. I would also rather
have MJ than IRV in a scenario 3 setting, because I consider Bucklin
better than IRV, and scenario-3 MJ is basically Bucklin.

Finally, you can hedge your bets by using two methods and then holding a
runoff between the two. E.g. Range and MJ, or MJ and Condorcet. The
voters would use the most expressive ballot: rating candidates in the
former case, or assigning grades/valuations in the latter. But Balinski
and Laraki also suggest that the context of the ballot alters the
voters' behavior: a rating ballot may encourage the voters to normalize
whereas a grade ballot might not. If so, you'd have to give half the
population rating ballots and half grade ballots (at random), and that's
a little too complex.

I hope the above explains my thoughts about MJ and Condorcet well
enough, and that it'll help you reply to the rest of my previous post.

It does seem I didn't explain the problem with inferring preference
strength well enough, however. What I meant was, suppose voters have
internal preference intensities or utilities so that 100% is a saint and
0% is the devil. Then a ranked ballot can't distinguish between:

A (99%) > B (98%) > C (97%) > D (3%)
and
A (99%) > B (50%) > C (48%) > D (3%)

(Suppose for the sake of the argument that the voter wants to make all
his preferences clear, so he doesn't equal rank. After all, 99% is
better than 98%, and the voter might want to make that clear to the method.)

My argument is that trying to say anything about the utilities of the
candidates (the preference strength or intensity) based only on the
preference is flawed, because it tries to extract more information from
a ballot than that ballot actually does contain. The only thing that
ranking A ahead of B tells you is that the voter prefers A to B. That
holds whether A is in 99th place and B is in 100th, or A is in first
place and B is in second. Thus, the only thing the method can infer from
a candidate being ranked first is that the voter prefers that candidate
to everybody else, but there's not a substantial difference between
voter X ranking A first and B second, and voter X ranking A second and B
third, except that in the latter case, whoever he put first is someone
he prefers to both A and B.

Applying that argument to IRV: IRV focuses excessively on the first
preference by only taking first preferences into account when deciding
who to eliminate, and that doing so tries to extract more information
than the ballot supports. It is similar to assuming that A>B>B>C>D is
much more likely to be a

A (99%) > B (50%) > C(48%) > D(3%)

ballot than a

A (99%) > B (98%) > C(48%) > D(3%)

one. But we can't say that from rankings alone.

Actually, it's stronger than that. Because IRV uses the same logic after
eliminating each candidate, it's liable to misinterpret a

A (99%) > B (50%) > C(48%) > D(3%)

ballot as well, because if A gets eliminated, then B and C are very
close in utility even though there's no way IRV can know that. To be
even more simple about it, IRV's logic assumes that there are a few good
candidates and a whole lot of fringe ones. When the number of good
candidates increases, IRV runs into trouble, and the Burlington election
is a very good example of just that happening.

You might argue that the same applies to Condorcet, but in reverse.
Because there's no utility (scenario-1) information in a ranked ballot,
Condorcet can't distinguish a 99-98-48-3 ballot from a 99-50-48-3 ballot
either. That's true, but Condorcet doesn't try to infer anything about
the ranked ballot beyond what I said it can tell the method: if voter X
ranks A ahead of B, all that means is that he prefers A to B. All you
can tell from a ranking is relative preference, and that's all that the
Condorcet criterion needs.