50:AB 1:BA 49:B

[Mike Ossipoff]

In Don's example, with rankings:

50: A,B
 1: B,A
49: B

...If this is intended to show that Condorcet's method, or other
pairwise ("head-to-head") methods have a problem, then it's 
as desperate try. For 1 thing, as I've pointed out so many times,
in replly to examples like this, _any_ method is equally likely
to return ties in small committee elections. (The exception to
that is Copeland's method, which will readily return ties in
any size election).

The fact that you may have some other method that doesn't tie
when pairwise methods do is irrelevant. Different situations
can tie different methods. Condorcet's method, in that example,
registers a pairwise tie, & 2 unbeaten alternatives. So, in a
small committee election (which of course is the only kind where
that could happen), the count would go to its tie-braakers.

I've suggested Simpson-Kramer as a 1st tie-breaker. It counts
votes-against without regard to the distinction between defeats
& victories. But it still returns a tie here, since both A &
B have 50 votes-against in their worst (& only) defeat.

I don't remember the exact order of my tie-breaker proposals,
but it was an order that I was sure I liked at the time. I
posted a proposed list of tie-breakers.

Maybe I had BeatSomething after Simpson-Kramer. It too returns
a a tie in that example. But my final tie-breaker, Stepwise-
Plurality doesn't return a tie in that example; it elects B.
A & B start by getting 50 votes each. Since there's a tie,
and since everyone has given a vote to a member of the tie, then
everyone gives a vote to their next choice, which means that
B gets 50 additional votes, but A just gets 1 additional vote.

In that example, Bruce's Pluralitly.ext would also elect B.

***

So then, Don, the tie-breaker lists that we've already proposed
for small committee elections woudl break that tie in your
example.

***

In a public election, in the tremendously improbable event that
a result like yours happened, the tie would probably be solved
by drawing lots, as per existing provisions, which I don't
suggest changing in our sw reform proposals.

***

As I said before, tie-breaker lists for small committee elections
can become an involved topic, with only subjective individual
merit for the various tie-breaker list proposals I really doesn't
matter much how ties are solved, when Condorcet returns a tie,
because Condorcet has done its job even when it returns a tie,
in terms of the standards for which it was chosen.

***

Though I don't remember now what my tie-breaker list proposal
was, for small committee elections, it might have gone like this:

Smith//Condorcet (or Schwartz//Condorcet unless the Smith set exceeds
   some pre-specified size)

Simpson-Kramer

BeatSomething

Stepwise-Plurality

***

Since Smith//Condorcet is our recommendation for public elections,
and since it retains its properties in small elections too,
it seems that Smith//Condorcet would be the understood natural
choice for votes in EM. Though any proposals for different methods
are undesirable, because such proposals would delay voting, I'd
ideally suggest substituting Schwartz//Condorcet for Smith//Condorcet
if there aren't more than 5 alternatives in the Smith set, because
the aesthetics that suggest Smith can be extended in small
elections to suggesting Schwartz. The reason for the limit on
Smith set size is that I don't know of an efficient Schwartz
algorithm of if one is possible, or at exactly what size the
Schwartz determination would start getting unreasonably time-
consuming.

As I said before, though, in practice I wouldn't suggest changing
from Smith//Condorcet, because it's much easier to use the method
we've already chosen than to deal with proposals for other
variations. And there's no genuine merit difference between
Smith//Condorcet & Schwartz//Condorcet.

And, as I said, I wouldn't make an argument about tie-breakers
either, for the same reason--they don't really matter. Having
proposed my list, I'd also quickly agree to simply using
Plurality.ext

Mike




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