Election Method fans,
By "Generalised Bucklin" I mean the MCA-like method that allows voters to fully rank the candidates, including giving equal preferences. I believe it may also have been called Median Ratings. If any candidate receives a majority of first preference votes, then the candidate with the largest tally wins. If not, then second-from-the-top preference-level votes are added to the tallies, and if then a candidate has a majority, the candidate with the largest tally wins; and so on.
Its cheif merits are that it complies with Weak Favourite Betrayal Criterion, Generalised Monotonicity (i.e. is monotonic and, apparently unlike Condorcet, passes Participation), Clone Independence, and also voted Majority Favourite and voted Mutual Majority.
Unfortuneatly, in common with Approval and limited-rank MCA, it is still vulnerable to what I call the "bogey candidate" effect. Imagine this scenario: A country holds an election for President and there are 4 candidates: Far Left, Left, Centre, Far Right. Shortly before the election an accurate poll of the voters' sincere preferences is published. 5% FL>L>C 44% L>(FL=C) 7% L>C>FL 40% C>(FR=L) 6% FR>C>L Left is the sincere favourite of 51%. Understandibly most of the voters expect L to win easily. The FR supporters all see the writing on the wall, so to maximise the slim chance of a Left defeat they all give equal first preference to Centre. But some of Left's supporters, nervous of the extremes and especially fearful of a FR victory, decide to "take insurance" and also vote Centre as equal first. There could be several plausible reasons besides plain stupidity why they might do that. They might be an ethnic minority which largely doesn't understand the national language. They might not have seen the poll or they might not believe it.The FR could be racists who have this ethnic minority in it's sights. Maybe in the part of the country where these ethnic minority sincere L supporters live, the FR has a big profile. So these are the votes:
Example 1.
05: FL>L
44: L
07: ( L= C) >FL
40: C
06: (FR = C)
Adding up the first preference votes, Centre defeats Left with a "bigger majority", 53% to 51%. Left is not the voted outright majority favourite, and there is no tight mutual majority, but notice that Left is the pairwise beats-all Condorcet Winner.
L>C 49-46, L>FR 56-6, L>FL 49-5.
My big improvement is to add the rule:
"When any candidate has a majority, eliminate all those who do not. Any ballots showing a preference among the remaining candidates which have not been counted towards any of their tallies, shall be counted toward the remaining candidate/s for whom the ballot shows highest or equal highest preference."
In the above scenario, the initial tallies would be L: 51%, C 53%, FR 6%, FL 5%. Right and Far Left would be eliminated, and the 5% FL>L ballots express a preference between L and C and have not yet been counted as part of either's tally, so they are added to the tally of the remaining candidate they show a preference over the other, which is Left. The 6% making up the FR tally has already been counted towards the C tally. So the final result of the election is
Left: 56% defeats Centre 53%.
Can anyone show that this Improved Generalised Bucklin, with its greatly enhanced Condorcet-efficiency and reduced incentive to insincerely up-rank, does not still pass Generalised Monotonicity and other important criteria I mentioned earlier?
This idea leads me to seriously propose this more complicated version as a very good single-winner ranked-ballot method:
Voters rank the candidates, equal preferences ok, truncation ok. First compute the IGB (Improved Generalised Bucklin) winner.
(If at the conclusion of a round, exactly two candidates are tied with tallies greater than half, then the two candidates runoff. If the runoff is tied, then the candidate who was ahead of the other last round wins. If more than two are tied, then it will be the candidate who was ahead of the other tied candidates last round. If the tied candidates reduce to two, then the two runoff, etc.) This candidate has qualified for the final runoff, and may do so again.
Now, counting truncated ballots as showing equal last preferences, compute the Reverse IGB "winner" (i.e. reverse the preferences and then count the votes as before), and eliminate that candidate. Now, proceeding each time as if previously eliminated candidates never existed, repeat until one candidate remains. This candidate is the other "finalist". If both finalists are the same candidate, then that candidate is elected. If not, then they runoff and the winner of the runoff is elected.
Example 2.
A recent example from James Green-Armytage (Sun. 17-8-03).
46: A>B>C
44: B>C>A
05: C>A>B
05: C>B>A
100 voters, all candidates in the Smith set.
According to James, his 44 BCA voters are insincerely order-reversing (trying a Burial strategy) against A, and their sincere preferences are BAC, which makes A the sincere pairwise beats-all Condorcet winner.
IGB sub-election to determine first finalist. Round 1: A46 B44 C10
Round 2: A51 B95 C54
All candidates have a majority, so no cadidates to "eliminate".
B wins IGB sub-election and so is the first finalist. Reverse IGB Elimination
Round 1: A49 B5 C46
Round 2: A54 B56 C90
Eliminate C, A wins RIGBE election and so is other finalist.
The two finalists runoff: A>B 51-49, so elect A!
Example 3.
31: B>A>E>C>D
23: C>B>A>E>D
25: D>A>C>E>B
11: D>C>B>A>E
10: E>A>C>B>D
100 voters, the Smith set is ABC.
IGB sub-election to determine first finalist Round 1: A0 B31 C23 D36 E10
Round 2: A66 B65 C34 D36 E10
Eliminate C D E. All ballots have contributed to the totals of A and/or B, except the 11 DCB ballots, so 11 votes are added to B's tally. Round 3: A66 B76 , so B is the first finalist.
Reverse IGB Elimination sub-election to determine second finalist
Round 1: A0 B25 C0 D64 E11
Eliminate D.
Round 1: A0 B35 C31 E34
Round 2: A23 B35 C41 E90
Eliminate E.
Round 1: A34 B35 C31
Round 2: A65 B69 C41
Those 41 C votes are all on ballots that have contributed to A or B,
so Eliminate B, and then A>C 66-34 and so A is the other finalist.
The two finalists runoff: B>A 64-36 , so elect B.
Curiously, according to the on-line ranked-ballot calculator at
http://www.onr.com/user/honky98/rbvote/calc.html
this method is completely alone in electing B. Ranked Pairs, Schulze,Borda,
Nanson etc. all elect A, while IRV picks C.
Example 3(a)
31: B>A>E>D>C
23: C>B>A>E>D
25: D>A>C>E>B
11: D>C>B>A>E
10: E>A>C>B>D
100 voters, all candidates in the Smith set.
This is the same as the previous example,except that the last two preferences
on the top line have been reversed.
The counting and the result to determine the first finalist is the same:
IGB picks B as the first finalist.
Reversed IGB Elimination
Round 1: A0 B25 C31 D33 E11
Round 2: A11 B35 C31 D64 E59 ABC are "eliminated" from reverse mini-election, and all the ballots have
been counted towards the tallies of at least one of the remaining candidates,
so the one with the largest tally (D) "wins".
Eliminate D.
Round 1: A0 B35 C31 E34
Round 2: A34 B35 C41 E59
Eliminate E.
Round 1: A34 B35 C31
Round 2: A65 B67 C66
Eliminate B. A>C 66-34 (no longer reversing) and so Eliminate C leaving A as second finalist.
The two finalists runoff: B>A 65-35, so B is elected.
So here I have described three related methods (1) the relatively simple Improved Generalised Bucklin (or should I keep that name for the full version?), which fails James G-A's insincere down-ranking example, (2) Reverse IGB Elimination which might be ok but at the moment doesn't appeal to me
as an independent method, and (3) IGB-RIGBE Runoff, the method I am presently most
enthusiastic/optimistic about (but maybe needs a smoother name).
Chris Benham
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