[EM] 3 Choices

[Craig Carey]

At 01:07 12.Oct.99 , DEMOREP1 at aol.com wrote:
>For the benefit of newer EM folks, I mention again that apparently ALL 
>methods have problems with sincere versus insincere votes when there 3 or 
>more choices (especially due to having divided majorities).
>
>"Sincere Votes" example using YES votes and head to head number votes.
>One executive officer to be elected.
>25 AB
>26 BA
>49 C
>
>100 total
>
>Both A and B have YES majorities. B would beat A with all sincere votes.
>
>Presumably a pre-election poll showing that the A and B folks have a majority 
>might cause some of the C voters to insincerely vote for the lesser of the 
>perceived evils using number votes on election day.
>
>Thus, I mention again (and again) that an election method must work on the 
>votes cast on election day (and not replays with more or less voters or 
>choices).  Presumably some of the voters in a Droop Quota minority will vote 
>for the lesser of the perceived evils.



The Droop Quota...

      "[the] Droop Threshold is the number of ballots divided by
       (the number of seats plus 1), then adding 1"

  From: http://www.fairvotencal.org/gloss.html


Irrespective of whether the previous author was thinking of 1 winner or
 2, the following assumes one winner and three candidates.

I take a little time here to respond to the anonymous Mr DEMOREP1 at aol.com


Another place where the problem is desribed is at:

   http://www.ccrc.wustl.edu/~lorracks/projects/techreport/subsubsection3_4_1_1.html

There is no soltion to the problem, on the page.

The mathematics in this message tweaks the 3 candidate Alternative Vote
 method by applying a quota that sometimes eliminates both candidates
 that do not have most 1st preference votes.
Then it is shown that the quota needs to be 1/2 to prevent vote splitting,
 but it cannot be >1/3 otherwise all 3 canidates might fail the threshold.


Replace numbers with variables. Calls this V

  AB a
  BA b
  C. c

Assume a < b < c.

Have the supporters of C prefer A over B.

Supporters would consider voting like this:

  AB a
  BA b
  CA c

But in both IFPP & STV, the 2nd preference in the 'CA.' papers
 are ignored, unless (c<a)(c<b).

Supporters of C could vote strategically and give some votes to
 candidate A.

  A. x
  AB a
  BA b
  C. c-x

(0<=x<=c)

Suppose there is a quota, and candidates that do not reach the
 quota fail. Let the Quota be Q, Q = a constant times the total
 number of votes. Modify STV using a Quota. If the Quota is 1/3
 of the total number of votes then the method is IFPP, which
 doesn't exhibit any vote splitting problem in this example.

When should C not transfer votes?, and if transferring will stop
 B winning, how much should be transferred?:


CASE (b < Q). Then a<b<Q, and C wins, so there would not need to be
  vote splitting, by the supporters of C.

Consider larger & larger values of x:

CASE (Q < b).(a+x < Q). Candidate A fails to reach the quota,
  and so either B or C wins, so supporters of C do not need to
  divert their votes to A in this case when (x < Q-a)

CASE (Q < b).(Q < a+x).(a+x < b). Candidate A still loses because
  the methods under consideration are constrained to be the Alternative
  Vote method with a quota applied.

CASE (C wins) There is no need for diverting votes to A to cause B to
  lose, since C will win with x = 0.

CASE (Q < b).(b < a+x).(C loses). B now has the least votes by first
  preferences, and A has more than the quota. B loses and the winner
  is the winner of:

     A. a+x+b
     C. c-x

  after B was elimintated from:
     A. x
     AB a
     BA b
     C. c-x

  The {A,C} 2 candidate system shows that the less votes C's supporters
  give to A, the more likely C is to win. 
  So in this last case, x should equal b-a+e, where e is such that:
    0<e and e approx = 0
  In none of the other 4 cases was vote splitting by C's supporters
  appropriate.

     AB a
     BA b
     C. c
  (a < b < c)
  C loses if B would win an STV election and (Q<b).
  Eliminate A, and find: (C loses) = (c<a+b)(Q<b)

----------

Vote splitting can occur and be appropriate if H is true:
H = (Q<b).(b<a+x).[(c<a+b)(Q<b)].(x=b-a+e).(0<=x<=c)
H = (Q<b).(b<a+[b-a+e]).(c<a+b).(0<=b-a+e).(b-a+e<=c)
H = (Q<b).(c<a+b).(a<b).(b<a+c).(a<b<c)   { a<b<c }
H = (Q<b).(c<a+b).(a<b).(b<c)

If a+b+c = 1, then H can be visualised in a triangle.

Note that the quota must be <= (1/3 * (number of votes) otherwise
 it causes the method to not find winners sometimes (i.e. when
 a and b and c are all about equal to each other).

Let Q = (a+b+c)*f, 0<=f<=1/3

H = (f[a+b+c]<b).(2c<a+b+c).(a<b).(b<c)
H = (f<b).(c<1/2).(a<b).(b<c)

So H most surely is not false, i.e. it is not an empty region.
For it to be an empty region, f should be at least 1/2.
The area covered by the small triangle H, is minimised when f = 1/3.
Setting f equal to 1/3 causes the modified STV method to just
 be IFPP. So unless there is an error, IFPP is the what results from
 modifying 1 winner 3 candidate STV so as to minimise the vote
 splitting.
So there it is: subject to a doubtful constraint that the solution
 only be a tweaked STV (with 3 candidates and one winner), the most
 vote splitting resistant method is IFPP. Also another constraint
 is that the papers be only ('AB', 'BA', and 'C.').

This is a point inside the triangle H:

   AB 2
   BA 3
   CA 4

   A. 1.01
   AB 2
   BA 3
   CA 2.99

C has to give at >1 vote to from 'CA' to 'A.' to change the winner
 from B to A, in both STV & IFPP.

It seems to me that quality IFFP/STV-like methods that never
 sometimes make vote splitting reasonable, do not exist.
So this idea of DEMOREP is suspect just like IIA. I suppose no
 other STV like practical method will make it past a no-vote-splitting
 test.

[There may be errors in this message's derivations.]
I read Mr Catchpole's last reply to me: something about (P1) going
 failing, or something(?)]


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

A similar matter is when gifting votes to lesser candidates causes
 the major group giving the votes, to switch their candidate from a
 loser into a winner. That can't happen when (P1) is held. STV doesn't
 actually satisty (P1).


__________________________________________________________________
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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