Craig Carey wrote:
> >1. Check to see if any candidates get more than 1/3 of the 1st
> >preferences. If not, the winner is the candidate with the most 1st
> >preferences (like FPP). Otherwise, go to step 2.
> >2. Eliminate all candidates with 1/3 or under of the 1st
> >preferences. These candidates are eliminated from ballots in the same
> >way candidates are eliminated in AV.
> >3. Hold an AV election between the remaining candidates.
> >
> >Is there any part of that procedure that you do not view as
> >sufficiently defined?
>
> Suppose there are 20,000 candidates. It is highly unlikely that any
> would get more than 1/3 of the vote. Very unlikely. Rather than the
> method becoming incomprehensibly adequate at recovering support
> at deep preference levels, it just pick the FPP winner, i.e. it
> ignores all preferences after the 1st.
So you do understand the method. BTW, what does "incomprehensibly
adequate" mean?
> That method has got to be worse than AV and AV1, and STV can get
> support that is 5 to 20 preferences deep contrubuted to a candidate.
I never said that AVQ was better than AV, or than I advocate AVQ.
However, it is true that AVQ passes P1, while AV does not.
You are right that AVQ would often be similar to FPP. However, there
is no reason to expect that the vote would be split 20,000 ways in AVQ
any more than it is in FPP.
> >> A method won't pass (P1) if all the surfaces do not meet properly and
> >> form a gentle piecewise-flat surface. Any method with an "if then else"
> >> in it, comes under more suspicion of failing (P1). Your AVQ has this
> >> feature: "if all are rejected at any stage then one will be picked".
> >
> >Are these assertions based on anything other than intuition? Isn't
> >it possible that P1 does not relate to a smooth graph? Isn't it
> >possible that a smooth graph is totally meaningless?
> >
>
> Inside the simplex of possible elections (with the papers corresponding
> to vertices), an "if-then-else" positions a flat that quite possibly
> that has win regions that do not match up and align properly. So part
> of the flat becomes part of the surface of the win region. It gets less
> obvious to argue futher.
No need to go further. It's already not obvious.
> Anyway, AVQ is already shown to be almost identical to FPP.
"Almost identical" seems like an exaggeration. AVQ says that a
candidate is entitled to a simulated run-off, as long as he/she gets
more than 1/3 of the vote. It is quite common to have candidates
split the vote, but not so much that both fall under 1/3. I suspect
most people in FPP countries, outside the US, can recall a few
elections where this has happened.
> >violates P1. Otherwise, you will have to face the possibility that
> >your intuition is incorrect, and that either AVQ gives a smooth graph
> >or P1 does not imply a smooth graph.
>
> What is that word "graph" mean: "a directed net", or "plot"?.
A plot, as you describe above when you talk about "regions that do
not match up and align properly."
> ...
> >> Ideally the 100th preference should if possible,
> >> have the same power and influence as a first preference, contrary
> >> to the nature of STV.
> >
> >Methods that consider candidates two-by-two (pairwise) have this
> >property. I'm going to describe such a method further down.
>
> Condercet has that other problem of not finding enough winners.
I didn't even say "Condorcet". I just said pairwise methods.
> >> Condorcet is a method that returns the wrong
> >> number of winners. As far as it goes (which isn't far) , it satisfies
> >> (P1).
> >
> >When I said Condorcet-type methods, I meant methods that meet the
> >Condorcet criterion. There are lots of suggested methods along these
> >lines, however, I am going to describe one of the better ones,
> >Tideman's method.
> >
>
> Now that AVQ was lost into history as a failed method ...
AVQ was designed to answer the question, "Are there any methods other
than FPP that pass P1?" It serves this purpose very well. I also
think it compares well with FPP and AV.
> >Compare every candidate to every other to get one-on-one (pairwise)
> >majority decisions. That is, if there are candidates X and Y, you
> >would find out how many people rank X over Y, and how many rank Y over
> >X. Then, the candidate with more votes is said to have a majority
> >over the other, with a margin of whatever the difference is.
> >
> >For example, if 30 people vote X over Y, and 20 vote Y over X, I will
> >say that X has a majority over Y with a margin of 10.
> >
>
> There are 5 cases. You actually ignored two cases:
> .......
> ...X...
> ...Y...
> ..X..Y..
> ..Y..X..
>
> I do not know what is to be done with papers marked (...X...).
I don't understand. What do you mean by a paper marked "(...X...)".
Has Y somehow been omitted from the paper? If the voter has left Y
unranked, then that is usually interpreted to mean that the voter has
implicitly ranked it under all the explicitly ranked candidates.
It is quite possible that the paper ranks X and Y as equal, either
explicitly (if this is allowed), or by leaving them both unranked (if
this is allowed). In that case, the vote is not counted either as
ranking X over Y, or Y over X, since it doesn't do either.
> Also, until you add the word "Condorcet" (or whatever), there is no
> summation of counts on papers.
Is this a complaint? You'll have to rephrase this, because I have no
idea what you mean.
> The word "rank" can only apply to
> a single paper (since it is defined for [3 of the 5 cases for] a
> single paper).
I also use the word "rank" to apply to the final result. As in, the
method ranks candidate A over candidate B. Often it is desirable to
get a complete ranking of the candidates by the method. This complete
ranking is essentially a bi-product of Tideman's method.
> Your definition above seems to not contain the
> idea of adding counts of papers.
I really think you must be reading too much into this. Let's say I
have the following votes
40 A>B>C, meaning 40 ballots rank A over B over C
35 B>A>C
25 C>B>A
40 voters rank A over B. 35+25=60 rank B over A. So, I say that B
has a majority over A with a margin of 20.
> >> I regard it as vital that the method get the right number of winners.
> >> That simplifies derivations from rules.
> >
> >Just because LIIAC says that certain candidates should not effect the
> >result (and should not win), does not mean that all the remaining
> >candidates must be declared co-winners. In my example for Tideman's
> >method, only D was excluded from having an influence by LIIAC, but a
> >single winner was still chosen. Tideman passes LIIAC, but without any
> >indecision as a result.
> >
>
> LIIAC uses pairwise comparing. Why should A win and B lose if A
> pairwise beats B ?. It shouldn't I presume. Maybe Condorcet never
> had an answer either.
Here's my opinion on the matter, which I think coincides with that of
Condorcet.
If A pairwise beats B, then we have reason to believe that A is a
better candidate than B. All things being equal, our end result
should rank A over B. However, majorities are not always correct, and
sometimes different majorities, composed of different people, can
contradict each other, even in a single election.
25 A B C
40 B C A
35 C A B
Here, although A has a majority over B, which gives us reason to rank
A over B, it is also true that B has a majority over C, and C has a
majority over A. This means that we have to decide which majority
decisions are more likely to be true. Condorcet suggested that
majorities with greater margins should be held with more confidence
than narrow victories, so it makes sense to accept the two higher
majorities, thereby rejecting the lower one.
B->C 65-35=30
C->A 75-25=50
A->B 60-40=20, so this majority decision is over-ridden
The end result is B>C>A.
> Does this Tideman method ever elect the wrong number of winners?
Well, in a situation like the following
50 A B
50 B A
Any non-random method will view this as a tie, and I guess you could
call that electing the wrong number of winners. Is that what you
mean? Tideman isn't particularly prone to ties, but they are
possible.
One way to think about ties is, if someone was designated as
tie-breaker, would they have more or less power than a normal voter.
In Condorcet's method or Copeland's method, they clearly would have
more. In Tideman, AV, or FPP, they would have less.
> Here's another argument for STV and all similar methods (i.e. about
> STV) (any number of winners).
>
> * * *
> It seems to me that proportionality mops up degrees of freedom, and the
> only other important reality is what a voter can detect when in the
> voting booth. They don't understand that LIIAC is important. That is
> a sort of rule that a method designer might (falsely) allege important.
Just because voters don't understand a criterion doesn't mean it
isn't important. Voters have never heard of GITC, but they are
certainly affected by vote-splitting, whether they understand it or
not. Voters have never heard of Pareto, but if a candidate wins
despite the fact that another candidate is unanimously preferred to
it, we can see that something has gone wrong. LIIAC isn't the kind of
thing that makes for good pamphlets, or 20 second political
commercials, but if we are going to seriously discuss electoral
methods, we should be prepared to understand them better than the
public at large.
> They can select a single candidate, and do these two things: (a) alter
> the preferences before that candidate's preference, and (b) alter the
> preferences after.
>
> It is not possible to get the method invariant to permutations of the
> preceding preferences if they are all losers or(?) if they are all
> winners. That could be desirable; it is not obtainable.
There are three major kinds of strategic voting (as far as I know).
Burying- Lower a candidate (with respect to sincere placement) in
the hopes of defeating it. This is what SPC prevents.
Compromising- Raise a candidate in the hopes of electing it. This is
impossible to eradicate in a non-random method, but some methods are
more affected than others.
Push-over- Lower a candidate in the hopes of electing it. This is
what monotonicity prevents.
Compromising is the most familiar, as this is the only kind of
strategy that affects FPP.
For example,
25 A B C
35 B A C
40 C B A
C wins under FPP. However, by compromising, the 25 A B C voters
could vote 25 B A C, and elect B. This result is better from their
point of view.
Tideman (like most Condorcet criterion methods) suffers from burying
and compromising. AV suffers from compromising and push-over.
It may be tempting to say, we can never eliminate compromising, so
why bother with it. Let's eliminate burying instead. But by that
reasoning, FPP would be the least strategically affected method
possible, since it is only affected by compromising, where AV, for
example, is affected both by compromising and push-over. This is
obviously not the case, however. FPP is far more affected by strategy
than AV. This difference is caused by the degree to which
compromising affects the methods.
As a result, it seems at least possible that a method that allowed
burying could still be as resistant to strategy as a method like AV
that is not.
Of course, we should also question whether strategy should be our
only consideration, or whether a method might be less strategically
influenced, but still be rejected on the grounds that its decisions
are poor when people vote sincerely.
> However, one can get an invariance of the winner set on altering
> trailing preferences. So that (named "SPC" in this list) may as well
> be done. SPC is an easy consraint to meet, and a major method (STV)
> satisfies it.
First, you keep mentioning STV. Can't we restrict the discussion to
single-winner methods, like AV and Tideman? Second, I see no evidence
that SPC is easy to meet. It's certainly easy to find methods that
meet it, FPP comes to mind, but it seems to conflict with other
criteria, which may be more important. In general, I think it is wise
to be slow to accept a criterion as necessary.
---
Blake Cretney
