Third Candidates
Steve Eppley
seppley at alumni.caltech.edu
Fri Nov 22 11:37:46 PST 1996
DEMOREP1 wrote:
>Assume A > Z (somewhere between 99 to 1 or 51 to 49) (i.e. having a
>majority winner). Candidate M is added.
>
>3 possibilities (ignoring ties) ---
>1. A > M, A continues to win
>2. M > A and M > Z, M wins
>Is this possibility more likely when A beats Z badly (such as A 70
>to Z 30) ?
If M>A and A>Z, then I think it's more likely that M>Z also, and the
odds would increase as A's majority over Z increases. But it's not
always true.
>3. M > A but Z > M, circular tie.
>Is this possibility more likely when A barely beats Z (such as A 52
>to Z 48)?
Again speaking only probabilistically, if A barely beats Z the odds
of Z>M are higher than if A trounces Z.
>The general case remains with A being a majority and Z being a minority.
>Adding one or more candidates either does or does not divide the
>majority.
I don't know what you mean by the "general" case.
Whether or not adding M divides the majority depends on whether
the method rewards misrepresentation of preferences or penalizes
nonmisrepresentation. We want a method which won't divide the
majority if there is a sincere majority, no matter what other
choices are on the ballot. If the method fails this, it has a spoiler
problem which will keep candidates off the ballot, inflate the false
mandate of the winner, and suppress the voters' true preferences.
Example:
46:Z
54:A
A wins.
46:Z
20:A
34:MA <-- sincere preference for M over A
Which wins?
Which method will be used to tally these? If it's Condorcet's method,
A still wins, since there are 54 votes in the A vs Z pairing either
with or without M on the ballot.
But if it's Instant Runoff or Copeland//Plurality, Z wins. If the
method ignores the majority, the potential candidates or voters will
have to compensate. That's the source of the two-party system.
---Steve (Steve Eppley seppley at alumni.caltech.edu)
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