[EM] Hybrid Beats-All/Approval v. Straight Approval

Forest Simmons fsimmons at pcc.edu
Tue Oct 30 12:48:01 PST 2001

On Mon, 29 Oct 2001, Forest Simmons wrote:

> On Mon, 29 Oct 2001, Bart Ingles wrote:
> > 
> > 
> > Forest Simmons wrote:
> > > 
> > > If voter X is almost sure that his ballot will make the difference between
> > > a hated (by X) Condorcet Winner and a Condorcet tie (to be settled by
> > > chance), voter X might be tempted to deliver up the election to chance
> > > even if that required him to vote his favorite last and his most despised
> > > first.
> > > 
> > > Of course, if all like minded voters gave in to the same temptation, the
> > > tie breaker odds of favorite to most despised would be adversely affected.
> > 
> > 
> > But if the aim is to sabotage the Condorcet candidate, the insincere
> > voters would be more likely to manipulate lower preferences, leaving the
> > first choice intact.  So the random tie breaker would be unaffected (I
> > think).
> > 
> > If the strategic aim is to help the Condorcet candidate using insincere
> > strategy, and the voters rank the likely Condorcet candidate equal or
> > higher than their sincere favorite, then this would of course affect the
> > random tiebreaker.  This may not be a concern, though, since the
> > motivation in this case is to prevent the tiebreaker from being used in
> > the first place.
> > 
> You're probably right. But all it would take is one simple example to lay
> all of the doubts to rest.

Here's a simple example (utilities in parenthesis next to candidate

49  A(100), B(50), C(0)
49  B(100), C(01), A(0)
02  C(100), A(99), B(0)

The two members of the C faction prefer to (insincerely) switch the order
of A and C (to make A into a CW) over the alternative of breaking the
Condorcet tie by random ballot in which despised B has a 49 percent chance
of winning. 

This example shows that random ballot completed Condorcet does not satisfy
the Favorite Betrayal Criterion.  This was a surprise for me because I
always thought that if a method satisfying the FBC were used as the
Condorcet tie breaker, then that combination would also satisfy the FBC.

Any method that doesn't satisfy the FBC can be manipulated by bogus polls,
so Bart was right (as usual).


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