[EM] Improved Generalised Bucklin (another example-related blunder fixed)

[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:

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