[EM] Improved Generalised Bucklin (example corrected)
Chris Benham
chrisbenham at bigpond.com
Thu Aug 21 09:34:07 PDT 2003
-------- Original Message --------
Subject: [EM] Improved Generalised Bucklin
Date: Thu, 21 Aug 2003 04:05:33 +0930
From: Chris Benham <chrisbenham at bigpond.com>
To: election-methods-electorama.com at electorama.com
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, the Smith set is ABC.
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
Round 2: A:23 B:35 C:41 E:90
Eliminate E.
Round 1: A:34 B:35 C:31
Round 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, 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: A:0 B:25 C:31 D:33 E:11
Round 2: A:11 B:35 C:31 D:64 E:59
ABC are mini-eliminated, 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 "wins". Eliminate D.
Round 1: A:0 B:35 C:31 E:34
Round 2: A:34 B:35 C:41 E:59
Eliminate E.
Round 1: A:34 B:35 C:31
Round 2: A:65 B:67 C:66
Eliminate B.
A>C 66-34, so A is the 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).
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
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