The family of "regular" probabilistic (stochastic) electoralsystems

[Craig Carey]

Note: This is another message that finds nothing of great value.


>At 17:51 09.12.99 , DEMOREP1 at aol.com wrote:
>>s349436 at student.uq.edu.au wrote in part--
>>
>>Okay- "Regularity" is the name used earlier by Albert Langer (Craig might
>>recognise the name ;) ) to describe the probabilistic analogue of IIA. It
>>goes like this-
>>
>>The addition (removal) of a candidate does not, for any other candidate, 
>>increase (decrease) the probability of that other candidate winning.
>>---
>>D-
>>A simple example-
>>
>>2 A>B>C
>>1 C>B>A

I can't see how to show an error in the rule with that.
This is no better:

At 18:11 09.12.99 , Craig Carey wrote:
>Yes!. ...
>
>1 A
>1 B

Correction: My second to last post is likely to be rejected by others since
 I interpreted the removing of candidate A to be the removing of a paper.

Nevertheless, the rule is still quite useless. Contrary to what I wrote, it does
 not really tend to impose FPTP.

Let the papers be these:

a AB
b B
c C

Inside the triangle having vertices {a=1, b=1, c=1}, note that removing candidate
 A is the same as scribing points along lines parallel to the {AB,B} edge. The
 points are taken towards the vertex b=1. I had earlier incorrectly said the
 direction was parallel to the line b = c (i.e. b(1-k)=c(1+k)+ka) (which led me
 to my FPTP comments).


---------------------------------------------------------------------------
:>>   The removal of a candidate does not, for any other candidate, 
:>>   decrease the probability [i.e. fact] of that other candidate winning.
---------------------------------------------------------------------------

1. GEOMETRIC ARGUMENT Allowing REJECTION Of The 'Regularity' Rule:

The rule is implausible since it says:

(1) near (a,b,c)=(0,1/2,1/2) bottom edge, the {B},{C} boundary is b=1/2. If that
    were not so then B could change from a winner into a loser near that region.

    This is worse than FPTP and the opposed to what proportionaility would do,
    which is exapnd B's region out to (c<1/2). Mr Catchpole' rule reduces B's
    region.

And

(2) near (a,b,c)=(1/3,1/3,1/3) (the triangle centre), the {B},{C} boundary
    is b=1/3.  No problem near the centre of the triangle.

An unfixable problem is that the two lines, b=1/2 and b=1/3, have to be joined,
 and that can't be done without the rule being violated. So it cannot be a
 rule at all.

If the aim were to make the method proportional then B ought fill
 (a<c)(b<1/3)(c<1/2). However this rule tries to have B fill this:
 (1/3<b)(1/2<c). Hence it has nothing to do with proportionality.


2. NUMERIC ELECTION EXAMPLE TENDING TO SHOW THAT THE RULE IS WRONG:

(A numeric argument needs to be constructed so that peoples' preferred
 methods agree and also show the rule to be wrong.)

An earlier discussion was about something called "Catchy-IIA", which
 defined that the outcome was unaffected by the removal of a loser.

The rule:

   "The removal of a candidate B does not change A, if it wins, into
    a loser"

CASE 1: REMOVING A CANDIDATE DELETE PAPERS WITH A PREFERENCE FOR THE CANDIDATE

If removing a candidate means removing all papers mentioning the candidate
 then this example shows a problem (two AB papers are removed):

4 A
2 BA
5 C        IFPP/STV winner = A      (the IFPP 1/3 quota = 3.666)

4 A
5 C        Now candidate A loses.   The IFPP/STV winner = C


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

CASE 2: REMOVING A CANDIDATE ONLY DELETES PREFERENCES FOR THE CANDIDATE

This old example that failed the Catchy-IIA rule also fails this rule.

The candidate B is removed and 2 (BC) papers are changes in (C) papers.
Before that is done A wins, and after that is done, A loses

--------------------------------------------------------------------
At 03:43 09.10.99 , Craig Carey wrote:
>At 16:54 08.10.99 , David Catchpole wrote:
>...
>>IIA's not a general criterion -- the criterion which it applies to an
>>electoral system is that _where_ it may be satisfied, it should be.
...
>
>>> Here's an example which both STV & IFPP agree, and which the rule fails:
>>> 
>>> A. 2
>>> BC 2
>>> CA 1
--- ------ A wins prior to the removal of loser B

           (The IFPP quota = 1.666, so STV=IFPP here)
...
>:   A. 2
>:   C. 3
--- ------ A loses after B is removed

...
>>Just because IIA as a specific rule on an election schema may not be
>>satisfied does not mean that it is not a useful principle where it may be.
>>This has been the point of my past umpteen e-mails.
...
>How are you going to modify it?. ...

It never was modified.

I doubt cubes have a place in voting theory. If the 3 candidate either win
 or lose, then the cube has no interior, it is just 8 points. It hardly matters
 if it is a 

At 15:39 08.12.99 , David Catchpole wrote:
...
>looks like- It's the Saari cube! Geez, that makes it easier. So the

----
(Errata: there was a mistake around the "c=0.499" area in my 2nd to last
 message.)



Mr G. A. Craig Carey, research at ijs.co.nz
Auckland, New Zealand.
Snooz Metasearch: <http://www.ijs.co.nz/info/snooz.htm>