# 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)

```