Methods of elimination in quota preferential STV

[Craig Carey]

At 22:09 07.10.00 -0700 Saturday, Bart Ingles wrote:
 >
 >
 >DEMOREP1 at aol.com wrote:
 >>
 >> Mr. [Ingles] wrote in part-
 >>
 >> Voters simply vote for multiple candidates, as they would with approval
 >> voting, except that each choice gets an equal fraction of the vote (e.g.
 >> if you vote for five candidates, each gets 1/5 of your vote).  You then
 >> eliminate the weakest candidate and recount, so that the remaining
 >> candidates get a larger share (if one candidate was eliminated from your
 >> ballot, the remaining candidates now each get 1/4 of a vote).  Continue
 >> eliminating candidates in the same fashion until the required number
 >> remain.
 >> ----
 >> D- As with Approval Voting with single winners, the A=B=C notion is false 
in
 >> most cases.
 >
 >What is 'the A=B=C notion'?
 >

In my theory it is named P2: smudging votes into papers equally with
subsequent preferences permuted out (just the next preference or the
next two). It seems to get the right results.

If the example shows preferences for only 3 candidates, then there indeed
can be 20 candidates. I haven't a crisp proof that I am right, but
the dispute I have is that your assumption is less natural than to assume
that the method embeds into larger methods correctly. I.e. in the case
where additional papers (but not preferences) are added and each paper
has a weight of 0 votes.

There is a mathematical principle called (probably) freedom: variables are
unconstrained when not actually constrained. I don't add constraints.

Corollary:
The number of winners can exceed the total number of preferences on
all the papers, since it only need be under the number of candidates.
Of course there is a bit of a tie amongst winners problem. There is
no problem resulting from this with me since I mandate embedding in
methods with lots of additional 0 vote papers. I was unable to derive
a method until I did that. The other rule assessors pick and choose
using whimsy and error, and then completely miss major rules. I can't
explain Bart Ingles' reasoning for his position possibly adverse to
embedding (or having good rules).




 >
 >> However, it is simple (but probably requiring computerized elections due to
 >> the fractions).
 >>
 >> A possible minor problem with the end result-
 >>
 >> 3 member legislative body
 >>
 >> winners and their votes
 >>
 >> A       B       C
 >>
 >> N1     N1     N1  (1/3 votes)
 >> N2     N2           (1/2 votes)
 >>          N3     N3  (1/2 votes)
 >> N4              N4  (1/2 votes)

[Demorep's messages name an "AOL 3.0.1 for Mac sub 85" e-mail client.]

 >> N5                     (1 vote)
 >>         N6             (1 vote)
 >>                   N7   (1 vote)
 >>

Children's scribble?. When I look at that I see "N1" and "N2" and
"(1/3 votes)". It might be an attempt to define a method, but it can't
be because there is no tying the text up with ballot papers at one
end, and with a set of winners at the other. I wrote to Demorep
privately and he might as well present his methods better and describe
the ballot papers. I asked Demorep to describe the ballot papers in
the YES/NO method. That hasn't been done. Is this the YES/NO method I
asked for a clear description of. I asked for no numerical examples.


 >> Would each winner have a voting power in the legislative body equal to the
 >> votes he/she receives ???
 >
 >I'm not sure what is being shown in the above example.  Are A, B, and C
 >the candidates, and N1-7 the voters?
 >
 >If instead you meant to show a 7-way tie, then it would need to be
 >resolved in the same way as any other system.
 >
 >Bart



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

At 17:59 07.10.00 -0700 Saturday, Bart Ingles wrote:
 >
 >Craig Carey wrote:
 >
 >> To Mr Mike Ossipoff: what is the probability of this election example
 >> occurring?
 >>
 >>  >   99 ABCDEFGHIJ
 >>  >    1 J
 >
 >
 >I can't speak for Mike, nor do I know the probability of the above
 >example occurring, but if it does I can tell you the probability that J
 >would win (assuming the voters are rational):  zero.
 >

**  The probability of J winning in an Approval Vote seems to be able to
**  equal 1. That would occur in the 2 winner case. It would occur even in
**  an election where there were 26 candidates and 25 winners and the same
**  "99, 1" votes.


 >Of course, if the voters are irrational then it doesn't matter which
 >method you use.  Maybe the 99 are inmates in a mental institution, and J
 >is the director of the institution.


Can someone name a single council or city or public district that claims
to use a method that they name to be the "Approval Vote" ?. I am not asking
for any name of a region that uses a method their electoral officers would
name with a different name. I guess no place outside of USA can be named
easily. There has not been sufficient checking. I browsed to Mr Donald
Davison's site in the last days and there was a bit of suggestion that
the choice vote is not at all STV. If the CVD is tweaking up STV, then
my Politicians-and-Polytopes mailing list is a highly suitable place to
discuss such actions. Message posters here reject truncation resistance.
This is the list for methods that have check boxes on their ballot
papers. Does anybody want to write on proportional representation?.