[EM] Proportional preferential voting

Thu Sep 16 10:03:09 PDT 1999

At 19:40 99/09/16 , you wrote:
>Craig... could you give another description of this? I don't quite catch
>on...
>
[>> >...
>> > ----------------
>> > |A.. |     |   |
>> > |AB. | Zab | a |
>> > |AC. | Zac |   |
>> > ---------------
>> > |B.. |     |   |
>> > |BC. | Zbc | b |
>> > |BA. | Zba |   |
>> > ----------------
>> > |C.. |     |   |
>> > |CA. | Zca | c |
>> > |CB. | Zcb |   |
>> > ----------------
>> > 

(A wins) = (b+c<2a)[(b+Zcb<a+Zca) or (2b<a+c)][(c+Zbc<a+Zba) or (2c<a+b)]
(B wins) = (c+a<2b)[(c+Zac<b+Zab) or (2c<b+a)][(a+Zca<b+Zcb) or (2a<b+c)]
(C wins) = (a+b<2c)[(a+Zba<c+Zbc) or (2a<c+b)][(b+Zab<c+Zac) or (2b<c+a)]

Sort the candidates by first preferences votes (i.e. c < b < a):

(A wins) = (b+Zcb < a+Zca) or (2b < a+c)
(B wins) = [(a+b+c)/3 < b] and (a+Zca < b+Zcb)
(C wins) = false

So this the method is a modified STV (1 winner 3 candidates case),
 but subject to a new rule that all candidates with less than one
 third of the 1st-preference votes, do not win.

---------------------------
The 2 Winners case

({B,C} win) = (b+c>2a)[(b+Zcb>a+Zca) or (2b>a+c)][(c+Zbc>a+Zba) or (2c>a+b)]
({C,A} win) = (c+a>2b)[(c+Zac>b+Zab) or (2c>b+a)][(a+Zca>b+Zcb) or (2a>b+c)]
({A,B} win) = (a+b>2c)[(a+Zba>c+Zbc) or (2a>c+b)][(b+Zab>c+Zac) or (2b>c+a)]

Sub-case: a < b < c

({B,C} win) = [(b+Zcb>a+Zca) or (2b>a+c)]
({C,A} win) = (c+a>2b)(a+Zca>b+Zcb)
({A,B} win) = false

(C win) = True
(B win) = (b+Zcb > a+Zca) or (b > (a+b+c)/3)
(A win) = ((a+b+c)/3 > b) and (a+Zca > b+Zcb)

[The STV 2 winner formula is too complex and yet wrong.
The transfer denominator is wrong in "the" method as far
 as I know. Maybe it reviewed every 100 years or so.

---------------------------
Typos that were fixed: (a+a<2b), (b+b<2c), {C,A}, were replaced
 with: (c+a<2b), (a+b<2c), {A,B}.

At 18:08 99/09/16 , you wrote:
>On Thu, 16 Sep 1999, Craig Carey wrote:
>> 
[>>...
>
>One can't remove paradoxes of voting (I'm talking here about paradoxes
>ala Condorcet) but admittedly other "artificial" problems with voting are 
>removable. It's odd that you say that removing some of these problems
>leads us towards FPP. In fact, I would say the opposite- as I understand
>it, your interpretation is the same as a writer on computer voting (her
>name eludes me- Donald knows, I think...) who neglected to acknowledge the
>fact that norms of voting extend to the _complete_ preferences of a voter,
>not just the marks they put down on paper (in her case, she said
>something as silly as "FPP is always independent of irrelevant
>alternatives"). So for the norm you give-

I don't know much about "norms of voting".
I retract the words "paradoxes .. can be removed".
Some subsequently attached comments can be paradoxical.
What are "norms of voting". The "_complete_" preferences
 of voters get separated from the quantities of bundles
 of papers. It is possible to make the 'loss' point with
 sample election examples.

This rule doesn't even hold for FPP:

>>  'permuting all preferences before a preference for a
>>   particular candidate, never makes a difference to
>>   that candidate's win/lose status'.
>
>which is covered by considerations of IIA, FPP is nasty (worst?)...
>because obviously, where there are preferences above a winner, how they 
>are arranged makes quite a significant impact on whether that candidate
>wins (A has 49 votes, B has 50 votes, C has 30 votes... if two other
>voters have voting preference A>C>B or they have C>A>B... obviously this 
>is for one-seat FPP but similar problems would arise for multi-seat
>methods).

That example, proves that, contrary to what I wrote, the FPP method
 is ruled out by that permuting rule in quotes above.

I'll name my method IFPP (Improved FPP or enhanced STV)

The two cases:

    A  49
    ACB 2
    B  50
    C  30
         FPP: A wins, STV: A wins, IFPP: A wins.

    A  49
    B  50
    C  30
    CAB 2

        FPP: B wins, STV: A wins.
        IFPP: A wins (since (49+50+30+2)/3 = 43.666.. < 49, else B)

Therefore these alterations occur in FPP:

[A wins (ACB)]  <--> [B wins (CAB)]
[B loses (ACB)] <--> [B wins (CAB)]

The last line disproves the "nice" assertion.
Probably a better example would be the (any) alteration
 that would reject STV (3 candidates, 1 winner)

-----------------
[>...
> 
>> I'd hope to see an example proving that the Condorcet method
>>  violates the "one man one vote" idea.
>
>Huh? (Sorry for abruptness...)

("man" means "vote", and "vote" means 'effect' or influence).

I presume that in the Condorcet (and many of its fixes), a
 single vote can have a little more effect than it is supposed
 to.

 An example: "In an election with 2,152,370 candidates and
 430,927 winners," politicians don't need to be told by
 their maths advisors that "some voting subgroups in the
 other 138 major known groups had more power than they were
 entitled to ... You know, we lost the vote over the voting
 system".

STV passes that test of "one man, one vote" flawlessly.

>> PS. Thinking numbers is a dead end if the method is based on
[>> >...
>Well, maths is more than numbers... Could you tell us what the paper is
>(it's pretty intuitive that a general approach is needed...)?

I lost my photocopy. I really don't who it was. Mr Ron Holzman
 had a paper called 'To vote or not to vote, what is the quota',
 Discrete Applied Mathematics, 2('88/89). It had something
 called the "participation axiom", and that was about the
 discarding of votes. (Lines through vertices).

G A Craig Carey
Auckland, New Zealand

Mr Craig Carey

E-mail: research at ijs.co.nz

Auckland, Nth Island, New Zealand
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