tie-breaking

[Mike Ossipoff]

Tie-breakers, & other small-election issues, can be an involved subject,
but I agree that it's something that we should face, since we could
encounter ties in future committee elections. Also, if we post a later
messasge to ER about recommendations for committee elections, ties have
to be covered.

Topics in this letter:

1. Improving on Bruce's Plurality.ext
2. Bruce's suggestion to "take it from the top" with the tie or
   the latest reduced tie
3. My "Beat-Something" tie-breaker

***

1. Improving on Bruce's Plurality.ext:

As I understand it, Plurality.ext says that, with the overall method
M//Plurality.ext, if M returns a tie, then Plurality is tried, so that
each ranking gives a vote to its 1st choice. If there's still a tie,
and that tie is not resoved by "taking it from the top" (starting over
with M, with the alternatives in the reduced tie), then each ranking 
additionally gives a vote to its 2nd choice. And so on.

Aside from the taking-it-from-the-top aspect, here's why that method
can be improved: Say the tie is between my 1st & 2nd choices. Do I
really want to then give a vote to my 2nd choice? Hell no. So I'd
modify Plurality.ext as follows:

Stepwise Plurality:

Each ranking gives a vote to its 1st choice.

If there's still a tie, then each ranking that hasn't yet given
a vote to an alternative that's in the current tie then gives a
vote to its next choice that's in the current tie.

Exception: If every ranking has given a vote to something in the current
tie, then every ranking gives a vote to its next choice in the current
tie.

This is continued until 1 alternative has more votes than any other
alternative.

***

Incidentally, if the rule is added to award the election to the
1st alternative to acquire a vote total equal to half of the number
of voters, then Stepwise Plurality becomes a better main method than
Bucklin, and I consider it the best easily hand-counted method. The
best method for large elections where a computer isn't being used, and
where it would be prohibitively time-consuming or expensive to count
a Condorcet election.

I believe that that version of Stepwise Plurality is the same as
the Iterated Plurality that Steve implied when he discussed Iterated
methods some time ago.

***

If simplicity is the goal, and having an exception is felt to violate
simplicity, &/or if familiarity is desired, then, instead of Stepwise]
Plurality, one could propose Plurality Elimination:

Plurality-Elimination:

Repeatedly eliminate from the rankings the alternative occupying
highest position in fewest rankings. Repeat till 1 alternative occupies
highest position in more rankings than does any other alternative.

Or just tell people "It's like MPV except that the winner is the 1st
alternative to get a Plurality".

I prefer Stepwise Plurality, but Plurality-Elimination may be quicker
& easier to define & explain.

***

2. Bruce's suggestion to take it from the top, with the latest
reduced tie:
 
I understand the rationale for that: If the 1st method is considered
better than the tie-breakers, then, if one of the tie-breakers has
reduced the tie, then why not start from the beginning again, with that
reduced tie.

BUT: What the 1st method, the main method is best for is being the 1st
method initially. That doesn't mean it's necessarily better during the
tie-breaking process. Bruce's suggestion assumes that the 1st method
is the best tie-breaker, just because it's the best main method. I
disagree.

I'm not saying it's real important; tie-breakers aren't a crucial
conroversial issue. But not taking it from the top with the reduced
tie greatly simplifies the definition & explanation of the procedure,
it seems to me. 

For example, say the method is Smith//Condorcet//Stepwise-Plurality.

Well, the purpose of Condorcet is to avoid the lesser-of-2-evils 
problem & protect majority rule. But Condorcet does that, merely
by being used as the 1st method (or the 2nd one, after Smith), the
1st time. 

Once it goes to a tie-breaker, after Condorcet has returned a tie, 
Condorcet has already done it's job, avoided a lesser-of-2-evils
problem or a violation of majority rule. There's no need to send
reduced ties back to Condorcet. 

So that's why I claim that it isn't at all necessary to take it from
the top with a tie or reduced tie.

In fact, since Condorcet has already done its job, and since favoriteness
has some appeal, it could be argued that it's better to continue with
Stepwise-Plurity (or Plurality.ext, if that's being used).

***

3. My "Beat-Something" tie-breaker.

I like Stepwise-Plurality as a tie-breaker, because, after lesser-of-
2-evils & majority rule have been dealt with, favoriteness is appealing.

But there's another tie-breaker I'd use as the 1st tie-breaker, just
before Stepwise-Plurality:

Beat-Something:

If there are more than 1 unbeaten alternative, and some, but not
all, of those beat something, then the winner should be chosen from
among those unbeaten alternatives that beat something.

That could be said more concisely:

If there are unbeaten alternatives that beat something, then the
winner should be chosen from among those.

***

So I'd do the election using methods in this order, going to the
next method in the event of a tie:

Smith//Condorcet
Beat-Something
Stepwise-Plurality

***

Of course, actually, if we really want to do it the most aesthetic
way, we'd use Schwartz//Condorcet in a small committee election, instead
of Smith//Condorcet.

Though Smith//Condorcet provides compliance with all of the same criteria
in a small election as in a large one, there could still be situations
where Schwartz is more aesthetic than Smith in a small election..

The Schwartz set has 2 equivalent definitions:

Definition 1:

An "unbeaten set" is a set of alternatives none of which is beaten by
anything outside the set.

A "small unbeaten set" is an unbeaten set that doesn't contain a
smaller unbeaten set.

The "Schwartz set" is the set of alternatives that are in
small unbeaten sets. In set terminology, it could be said to be
the "union" of the small unbeaten sets.


Definitions 2:

There is a "beat path" from A to Z if A beats B beats C...
beats Z (where there can be any number of intermediate alternatives
between A & Z).

If there is a beat path from A to B, but not from B to A, then B
is disqualified.

The Schwartz set is the sets of un-disqualified alternatives.

***

Returning to tie-breakers, of course it would be possible, and 
maybe desirable, in a small committee, to list a relatively
long list of tie-breakers, rather than just 2 of them.
But that's another topic.

***

I'm going to post Steve's report tomorrow morning, with the
modifications he mentions, to ER.

Also, I'm going to at that time reply to the other posting,
with a bad-example for the relative of Condorcet that looks
at votes-against in victories as well as defeats.

***

Mike












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