[EM] Vermont IRV can elect a fanatical candidate

[Craig Carey]

At 03\03\03 22:04 -0700 Monday, Jan Kok wrote:
 >Recall that "Vermont IRV" immediately reduces the field to the two
 >candidates with the greatest number of first choices.
 >
 >If we define a fanatical candidate as one who is the first choice of a
 >minority, but last or second-last choice of a majority, in a race with at
 >least four candidates, then
 >
 >Vermont IRV can elect a fanatical candidate.  It will happen if the two
 >candidates with the most first choice votes are fanatical.
 >
 >I'd call that a disaster waiting to happen.
 >


Getting the wrong winners is what would be expected if the Alternative
Vote was used. That Vermont method is identical to the Alternative Vote
when there are 3 candidates.

I presume that a fairer method has a 1/3 quota in it and also a
fairer method favours the main opposition party that opposes the
Vermont Democrats party.

A way to proceed is this: find example pairs showing the AV method at
its worst and see how the other 2 methods compare. Since that is biased
against AV due to the lack of searching it ought be able to be way to
create opinions in favour of IRV if the alternatives are not better by
enough. This well known example could be a starting point.

The "irv-wrong-winners.htm" example:
 > >> >   ----------------------------
 > >> >     A       19999      80004
 > >> >     B           1          5
 > >> >     BA      19997      19997
 > >> >     CB      40002      40002
 > >> >     DBA     20001      20001
 > >> >   ----------------------------
 > >> > Total:     100000     160009
: >> > AV Winner:      A          B : IRV winners

Note: "A" won so by the IRV majority principle it had a CVD-50%.
Then (when 11 candidates) the election size doubled and now
candidate A loses with the final election having 3 parts out of 4,
all in favour of candidate A. The CVD majority could not have got
less principled when papers naming only candidate A were added.

The CVD brochure asserts that the "majority" principle is part of
US democracy.

However what was initially a CVD 50% majority, rose to hit the
CVD's final most thoughtful offer: the vote rose to hit the IRV
75% threshold, and then candidate became a loser. That is what
can happen when a rule that passes little but the lobbyists method,
a principle of the people.



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

The can be a whole class of methods of order K.
Suppose K = 3
They are designed like this:

Stage 1: Eliminate all but K and deem some candidates to be winners/losers

  * Eliminate (with full transferring) all candidates except for K
    candidates.

  * Using some formula/algorithm, deem some candidates to be losers
    and/or winners. They do not get eliminated yet, but just have
    their names noted.

  * When K = 3, what is needed (if the method is to be monotonic) is
    going to be close to this: all candidates (of the 3) with under 1/3
    of the remaining votes, are deemed to be losers. So one or two are
    deemed to be a loser.


Stage 2: Eliminate one more candidate

  * Now after that eliminating, there are new 1st preference counts.

  * One candidate is eliminated. It could be 3rd ranking under the
    new 1st-preference counts, or the 3rd ranking under the old
    counts, or using a quite different formula.

  * If K=3 and the 1/3 quota of the new counts with 3 candidates
    was used in the deeming of candidates to win/lose, then likely
    the original (pre-1st-stage) 1st preference counts would be
    ignored here too.

  * A different scheme is to use the average of the two sets of
    1st preference vectors. Other ideas are possible. Numerical
    testing (minimizing the maximum quantity of the public losing
    deserved power) might clarify that up, but 5 candidates might
    be needed and over 5 papers.


Stage 3: Find the winners from the remaining K-2 candidates

  * The winners are decided on using some method.
    If K = 3 then there are 2 candidates then the method is obvious.

  * The decision here has to give way to the decision of stage 1,
    where a candidate was deemed to be a winner or loser. The process
    of giving way gets more complex when K=4, since numbers that
    consider 3 candidates at a time are checked and terms like
    a+c<b+d become predominant in the algebra.

  * Stage 1 would do probably simply calculate a 1/3 quota to find a
    loser, but a well done K=4 solution will not simple calculate a
    a 1/4.

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