Replying to Bruce on Definitions

Mike Ossipoff dfb at bbs.cruzio.com
Tue Apr 16 03:11:25 PDT 1996


Bruce Anderson writes:
> 
> On Apr 15,  5:41am, Bruce Anderson wrote:
> > I picture calculating winners according to (the now defined) pairwise methods 
> > by using the array:
> > r(i,j) = p(i,j) + xq(i,j), 
> > where either x = 0, or x = 1/2, or x = 1.  When it is important to distinguish 
> > among them, I would specifically refer to the Condorcet(0), Condorcet(1/2), or 
> > Condorcet(1) method.  I say that i is a Condorcet winner if it has the largest 
> > minimum over j of r(i,j).  I use x = 1/2; but Mike (and now, hopefully, 
> > everyone else on this list who DOES NOT explicitly DISTINGUISH among the 
> > possibilities) uses x = 1.  I think there should be no problem here as long as 
> > I keep it clear which is which whenever it makes a difference.

That's news to me, Bruce. Saying that I use x = 1 means that I count
voters who haven't ranked x over y or y over x. I believe I've repeatedly
made it clear that I don't count preferences that aren't voted. I count
only voters ranking x over y & vice-versa. You'd have me count someone
ranking i & j equal as if he/she had ranked i over j. 

Bruce, this is why I suggest that you're intentionally trying to
obfuscate the subject. It's so simple to say that I use the
votes against a candidate in his/her pairwise comparisons, that
the number of people ranking candidate i over candidate j is used
in determing candidate j's score. That I don't count equal rankings
of candidates i & j in any way. But you come up with this formula using
x, multiplying x by q(i,j), and then tell us that I use x = 1. From
what I've been saying all along, it should be clear that, in your
formula, I'd use x = 0, since I don't count in any way the rankings
that rank candidates i & j equal, or don't rank either one.

But that's one use of formulese, isn't it. That extra distance from
what is actually being said makes it possible to slip things like that
in more unobtrusively, since the contradiction between your version
of what I said & what I actually did say is less obvious when it's
written in that fashion.

Well, in case I haven't made it clear yet, the winner by Condorcet's
method, as I use the term, is the alternative over which fewest voters
have ranked the alternative that beats it that is ranked over it by the
most voters.

To write this out in longer form:

For each alternative, determine which alternative that beats it is
ranked over it by the most voters. The number of voters ranking that
other alternative over it is the measure of how beaten it is. The winner
is the alterntive least beaten by that measure.

Another wording that I've used:

The winner is the alternative leasts beaten by any alternative that
beats it, as measured by how many voters rank that other alternative
over it.

***

I hope at least some of these wordings are a lot clearer than the
formulese in recent postings. And no, it wasn't the formulese that
clarified to anyone what I mean by Condorcet's method.

> >-- End of excerpt from Bruce Anderson
> 
> All this is true.  In particular, Mike uses what I call x = 1 here as part of 
> calculating the winners according to his definition of the Condorcet voting 

As I said: No I don't. I've stated my definition in plain English. I
have clearly stated that I don't count preferences that aren't expressed,
and that I don't count ballots that don't rank candidate i over candidate
j or vice-versa. 


> method.  However, Mike (and I presume others) do not JUST use what I call the 

I wa> Condorcet(1) voting method.  Instead, what Mike calls the Condorcet voting 
> method is what I would call the Beats-all//Condorcet(1) voting method, where the 

No, Condorcet's method, as defined by Mr. Condorcet himself, starts
by looking for an alternative that beats each one of the others. That
initial search doesn't need a separate name, since it's part of
Condorcet's original proposal.

I have, however used the name "BeatsAll//Approval" for a method that
uses a 2nd balloting by Approval if no 1 alternative beats each one of
the others.

> Beats-all voting method is defined as follows.  If there is a candidate who 
> pairwise beats every other candidate, then Beats-call chooses that candidate as 
> its unique winner; otherwise (i.e., if no candidate pairwise beats every other 
> candidate), then Beats-call chooses every candidate on the ballot as being tied 
> as its winners.
> 

It's a disagreement. I'd like to add that if we have a rule to pick the
alternative that has fewest people ranking over it something that beats it,
then an alternative that beats each one of the others wins by that rule,
and so there's no need to specifically have a rule to elect an alternative
that beats each one of the others. Steve pointed that out, and that
leaving out the BeatsAll rulee would lend brevity. I included the BeatsAll
rule in my Condorcet definition for step-by-step clarity, but Steve is
right, and that rule isn't needed. But, either way, whether we include
it or not, it's still Condorcet's method, and it needn't be called
by a new name, BeatsAll//Condorcet.

> This statement is just a clarification, not (I think) a disagreement.  Still, it 
> is an important clarification because, in my notation, Beats-all//Condorcet(1) 
> is NOT the same as Condorcet(1) and, for that matter, Beats-all//Condorcet(0) is 
> NOT the same as Condorcet(0).  However, Beats-all//Condorcet(1/2) is the same as 

Note how many ways Bruce has radically changed the meaning of Condorcet's
method. 


> Condorcet(1/2).  Further, since Smith//Beats-all is the same as Smith, 

Why would BeatsAll be used after Smith??? Not that there's any need
to include BeatsAll in the name, since that function has always been
part of what Condorcet's method means.

I mean, if you were going to name a method called BeatsAll, as a separate
method, then it would be quite pointless to use it after Smith, since
an alternative that beats each one of the others would be the only
winner by the "Smith method" that you define--a method that picks
everything in the Smith set.


> Smith//Beats-all//Condorcet(1) is the same as Smith//Condorcet(1).
> 
> For whatever it's worth, note that Smith//Smith is the same as Smith, and that 
> Beats-all//Beats-all is the same as Beats-all; but that 
> Condorcet(1/2)//Condorcet(1/2) is not the same as Condorcet(1/2), and I strongly 
> suspect (but have not worked out an example) that Condorcet(1)//Condorcet(1) is 
> not the same as Condorcet(1).  Certainly, Copeland//Copeland is not the same as 
> Copeland.
> 
> Bruce

Bruce, the Single-Winner Committee, which uses EM for its discussion,
has several agenda items. We want to discuss standards for evaluating
& comparing single-winner methods. We intend to soon vote on those
standards. I suggest we add "candidate-counting" (how many alternatives
a particular alternative beats, & how many it is beaten by) as a standard
that you propose. Of course you could name every one of the academic
candidate-counting criteria as proposed standards if you so wish.

The other issue, one that you brought to SWC, with your posting
about Condorcet's method's alleged faults, is a comparison of the
faults & merits of Condorcet (as I define it, if you don't mind) and
Copeland. As you know, I've asked you several questions on that
subject--I've asked what you say in reply to some statements I've
made, and Copeland faults that Steve & I have mentioned. It's beginning
to look as if you don't intend to answer. That's fine, and I hope
everyone notices that. You apparently don't intend to defend the
method that you propose, which you've implied is better than plain
Condorcet, or even Smith//Condorcet. In that case, the Condorcet
vs Copeland issue is concluded, since every reply that's going to
be made has been made. Good, because now we can return to the
topic of voting on the order in which to use the standards for
rating the methods.


> 
> .-

Mike 


-- 



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