[EM] "More often" (was: IRV and Condorcet operating identically)

[Jan Kok]

----- Original Message -----
From: "Venzke Kevin" <stepjak at yahoo.fr>
To: <election-methods-list at eskimo.com>
Sent: Thursday, February 27, 2003 11:49 AM
Subject: Re: [EM] IRV and Condorcet operating identically


> --- Dave Ketchum <davek at clarityconnect.com> a écrit :
> > On Tue, 25 Feb 2003 09:23:41 +0100 (CET) Venzke
> > Kevin wrote:
> >
> > > I wonder if the only reason IRV has more apparent
> > > backing than approval or Condorcet is because it
> > would
> > > permit our present politicians to be elected even
> > more
> >
> > The above makes no sense, for IRV and Condorcet use
> > identical ballots and,
> > most of the time, award identical winners.
      ^^^^^^^^^^^^^^^^
...
> If you put voting systems in order of to what degree
              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> they preserve the problem of the election of the
> lesser of two evils, I would draw it like this:
> IRV - Plurality - Approval - Condorcet.

I'm curious if anyone can mathematically justify such statements as "Voting
method A exhibits property P 'more often' than method B"?

Say there is an infinite set of scenarios S, and say A and B each exhibit P
for subsets Sa and Sb of S.  If Sa is a proper subset of Sb, then it would
be reasonable to say that A exhibits P "less often" than does B.
(Unfortunately it doesn't give one a sense of "_how_much_ less often" P is
exhibited by A compared to B.)

But what if Sa and Sb "overlap", i.e. Sa intersect Sb is a proper subset of
both Sa and Sb, or Sa and Sb are both non-null but the intersection is null?
Then the only way I can see to compare "how often" A and B exhibit P is to
somehow assign probabilities to the scenarios, i.e. what percent of actual
elections are members of Sa and Sb?

Is there a generally accepted way of assigning such probabilities (which
would involve lots of assumptions about distribution of voter preferences,
candidate characteristics, voter strategy, etc., etc.)?

As a concrete example, can someone show that some Condorcet method fails
Favorite Betrayal "less often" than IRV?

Cheers,
- Jan

P.S.  I ran some crude simulations a few months ago with no strategy
(sincere voting) which showed that IRV and Condorcet SSD chose different
winners something like 30% of the time.


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