[EM] Improved Generalised Bucklin

[Chris Benham]

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:
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.

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
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 (100 voters)
Round 1:          A:46    B:44   C:10
Round 2:          A:51    B:95   C:54
    A is eliminated, all ballots have been counted towards B or C, so  B 
 is first finalist.
Reverse IGB Elimination
Round 1:         A:49    B: 5    C:46
Round 2:         A:54    B:56   C:90
     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, all 5 candidates in the Smith set.
IGB
Round 1:    A:0       B:31     C:23     D:36     E:10
Round 2:    A:66     B:65     C:34     D:36     E:10
  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:    A:66    B:76 ,  so  B is the first finalist.
Reverse IGB Elimination
Round 1:    A:0       B:25      C:0      D:64    E:11
  Eliminate D.
Round 1:    A:0       B:35      C:31    E:34
           2:    A:23     B:35      C:41    E:90
  Eliminate E.
Round 1:   A:34      B:35      C:31
           2:   A:65      B:69      C:41
     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, in this example IRV and Schulze elect D, despite the fact 
that D is ranked last on 65% of the ballots. Nanson and Baldwin elect B, 
while Ranked Pairs, Borda and Black elect A.
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 interesting 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).
Because of the relative popularity and ok track record of IRV, the idea 
of marrying two methods together via a runoff might be politically 
advantageous. Just add a clause to a favoured rank-ballot method that 
says that the IRV winner is also computed and if that produces a 
different winner, then the two runoff. Who could object to that? 

Chris Benham