[EM] Proportional preferential voting

Mon Sep 20 04:51:40 PDT 1999

At 13:00 99/09/20 , you wrote:
>By your own admission, though, it gets to be a little bit of a hassle by
>the 4th candidate- and finding general rules becomes more and more
[>important. ...

There are missing formula.
I quit researching into preferential voting field in 1996,
 and I am hoping to stop writing to this election-methods-list
 by Friday. Thanks to the experts, I believe I can set aside
 researches for perhaps another 15 years, and well, and why not have
 a thesis on the topic, if it is possible, in the meantime.

[>>...
>> STV: 3 candidates 2 winners:
>>    A  670
>>    AC  10
>>    B  320
>> Votes for A = 680
>> Quota = 333.3..
>> Transfer value = (680-333.33..)/?
>
>Obviously, in this case it would be pretty daft to distribute all of A's
>votes to C.

You mean rather than all of the "AC." votes perhaps.

>I like to come up with an instantaneous solution for the new quota by-
>q=(T-q-i)/n
>i=I(V-q)/V
>
See that "/V". It makes the boundaries to win-lose regions be
 curved. Who can think of a reasonable 'vote alteration rules' (rules
 that constrain the slopes of the win boundaries) that lead to
 curved surface boundaries.

>(Haven't given you the explicit solution, but you get my drift)
>
>where q is the quota, T the total number of valid votes, i the number of
>votes which fail to be effectively transferred, n the number of seats left
>to be filled including the one just filled by the candidate whose votes
>are being transferred, I the number of votes for that candidate that
>aren't going to any other candidate,  V the number of votes for that
>candidate. The value of a votes' transfer is then (V-q)/V

So V = 680 then. Then the transfer value is 10/680 = 0.0147.
In IFPP the AC votes would be transferred with a imagined transfer
 value of 1, at least in general, 

[>>..
>We all have our favourites on this list. Over time, this becomes
>really apparent- especially with respect to who gets involved in what
[>discussion. ...
>
>Arrow, etc. make it clear that the "strict reasonable criteria" are a
>threat amongst themselves. IIA and the sub-criteria of it which I
>make distinct are strict reasonable criteria used by theorists of voting
>and other theorists in social choice (e.g. market theory). The way to
>state IIA for deterministic models of choice is this-
>
>"The removal or addition of an option which has not been selected will not
>alter the outcome"

A 10 candidate election with 5 winners is held and the outcome
 was very nearly a tie between two candidates. The votes used real
 numbers, or the number of voters was very large.
Then, suppose a lot of preferences for candidate X were inserted
 into the voting papers. Suppose the number inserted were small
 enough so that even if they were 1st preferences, X would still
 lose.
Then a question is: is IIA consistent with the outcome changing.
As far as I know, enforcing invariance of the outcome is no
 problem (done in IFPP, although I don't recall checking (I quit
 writing equations of my theory ('IFPP') in Dec 1996)).

>for models of preference-
>
>"The removal or addition of an option should not alter the preference
>between any other pair of options"
>
>for probabilistic models of choice-
>
>"The removal or addition of an option which has not been selected will not
>reduce the chances of any outcome where that option is not chosen"

[>...
Does Condorcet (1 winner) satisfy 'Principle 1'
[>>...
>Where a Condorcet winner exists, principle 1 is satisfied as the
>preferences between the CW and other candidates are not altered by any
>swapping below (or above- as long as it's above-to-above or
>below-to-below it's alright) that candidate in anyone's preferences. As a
>result, the head-to-head considerations that lead to a Condorcet answer
>are unaltered and the CW remains a CW.

That isnot easy to understand, actually.

Condorcet satisfies the stronger 2nd version of principle (P1)
 if paradox/answer-undefined regions are excluded.

Proving Condorcet satisfied (P1)

(P1) is satisfied if an example showing the method fails can't
 be constructed.

Consider an example with any number of candidates. The rule does
 not hold if the winner changes from C to B if votes after (and
 perhaps including) any preference for B, are discarded.

An example can not be found.

There is no way to get the vector in the Cordorcet graph that points
 from B to C, to flip and point from C to B.

That is since all the voting papers fall into these four groups:

** no preference for B : these papers can't be altered
** preference for C is before B : can only reduce support for B
** preference for B is before C : C never gets any votes from this
              type of paper, and support for B can only be lowered
** preference for B and none for C : can only reduce support for B

[>...
>
>The "UQ" in my e-mail address stands for- University of Queensland
>(Brisbane)

It doesn't appear in the messages.

>> Voting in STV just requires a little extra thinking (or
>>  instructing) on the part of voters. It is not merely a
[>>...
>werks okae weyr aye cum frum.
That may not be true.

-----------------------------------------------------------------
Host: Craig Carey
            SPECIAL FEATURE: SOCIAL DECISION THEORY CORNER

Note: no references have been mentioned by me because these
 results are oringinal. I have 565 pages of maths equations
 [and much of it is dead ends, e.g. set theory formulations
 setting out what voting rule alterations applying to systems
 with large numbers of candidates, are nice.]

Is IFPP arbitrary?: who can fill in the missing steps below?.

--------------------

Solve V, a 3 candidate election system.
Find the answer set for the 1 winner case.
(2 winner ommitted).

V:
A   Zaa
AB  Zab
AC  Zac
B   b
BA  Zab
BC  Zbc
C   c
CA  Zca
CB  Zcb

(Step 1) Use (P1) or SPC:
    aV = aVA = (-bVA)(-cVA) = (-bVAB)(-cVAC)
[aV = (a wins V), VA is V fully truncated after A].

Now only need to solve for B, this

VBC:
A   Zaa
AB  Zab
AC  Zac
B   b
C   c

(Step 2) Require that nice rule that defines what truncating
 preferences means.
An example of it:  If the candidates are {A,B,C,D,E,F} then
 this vote

6 ABC { DEF

has an identical effect on the outcome to this:

2 ABCD
2 ABCE
2 ABCF

and also to this:

1 ABCDE
1 ABCDF
1 ABCED
1 ABCEF
1 ABCFD
1 ABCFE

That allows VBC to be rewritten as:

U:
AB Zab'
AC Zac'
B  b
C  c

Zab' = Zab + Zaa/2
Zac' = Zac + Zaa/2
a = Zab' + Zac'

Then normalise the system so that 1 = a+b+c, and plot
 the solution inside a triangle having at its vertices,
 the numbers: a, b, c.

(Step 3) Require this (the FPP, or proportionality
 requirement): When there are m winners and n candidates,
 and there also are no 2nd preferences, then the winners
 are the m candidates with the most votes (i.e. sort
 and select).

UA:
A  a
B  b
C  c

So:  aUA = (b<a)(c<a) = aU.
     bUAB = (bUAB)(-aUA) {otherwise regions have >1 winner}

(Steps 4...) The next step is to solve U.
 Define [Delta]bc = (b+Zab<c+Zac).
Cast shadows using alteration rules from the middle of
 the triangle. One rule woould prohibit a sharp corner
 near the exact centre.
... Maybe DEMOREP1 at aol.com can finish this.
The method for larger problems would include steps like
 the above.
References to work of others: I don't know of any others.
 (except for my reference to the 1998-89 paper of
 Mr Ron Holzman).

_____________________________________________________________
Mr G. A. Craig Carey
E-mail: research at ijs.co.nz
Auckland, Nth Island, New Zealand
Pages: Snooz Metasearch: http://www.ijs.co.nz/info/snooz.htm
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