[EM] A new election method...
bmbuck at 14850.com
Thu Aug 9 11:55:11 PDT 2001
As a demonstration as to how easy it is to come up with a new election
method, here's a technique that popped into my head in reaction to Roy's
questions concerning elimination ballots and monotonicity. I'm not
claiming that this method is monotonic -- I'm not quite good enough to
prove it -- but this method does have some features which might appeal to
1. It is an elimination method.
2. It uses a ranked ballot.
2. It utilizes first-place information, although not in the same way as IRV
3. At any stage, it is acceptable to terminate the procedure if any one
candidate has a majority of first-place votes.
It is also a Condorcet method, in that if a Condorcet winner exists, that
winner will be elected.
Here it goes...
Based on a ranked ballot, list the candidates in order of descending
first-place votes (if there is a tie, compare the number of 2nd-place
votes, etc, to establish an order. If two candidates (out of N) have
exactly the same number of 1st place, 2nd place, ... Nth place votes, list
them in alphabetical order).
Use this listing as a seed for a single-elimination tournament, with a
candidate advancing in the tournament if he, compared head-to-head with his
opponent, is defeated.
Eliminate the "winner" of the single-elimination tournament, and
relist/eliminate, etc until a single candidate remains.
Example: Four candidates
Listed in 1st-place order:
50 B B
50 D D D
More people prefer C to B (120:80), more prefer A to D (105:95), and more
prefer B to D (120:80), so D is eliminated.
Listed in 1st-place order,
050 B B
030 A A A
More prefer C to B (120:80), More prefer B to A (170:30), so A is eliminated
Listed in 1st place order:
080 B B
More prefer C to B, so B is eliminated.
Is that method clear?
Since no candidate is eliminated unless another candidate defeats him
pairwise, the Condorcet Winner (if one exists) cannot be eliminated and
therefor must win.
How would someone go about examining this method for monotonicity?
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