[EM] Voting-System Choice for Polls (Just one more thing I want to say)

[Michael Ossipoff]

On Wed, Dec 21, 2016 at 3:59 PM, Markus Schulze <
markus.schulze at alumni.tu-berlin.de> wrote:

> Hallo,
>
> Michael Ossipoff wrote:
>
> > It couldn't get any more minimal & ideal than that.
> > MAM (unlike Beatpath, etc.) never unnecessarily
> > disregards a pairwise defeat (by electing someone
> > who has the defeat).
>
> The problem with Mike Ossipoff's argumentation is
> that he doesn't define the terms "necessarily" /
> "unnecessarily".
>

Actually, I did.

Shall I explain it again?

A pairwise defeat, by itself, says something. It says that the public have
voted that X is better than Y. They've voted to have X instead of Y.

But what if the X>Y defeat is in a cycle: X>Y>Z>X  ?

Say all 3 of those defeat are equal in strength.

 (I suggest that the measure of strength be chosen for the best
anti-strategy properties. But it isn't in dispute here--MAM & Beatpath use
the same measure of strength.)

Now, with a cycle of 3 equally-strong defeats, the electorate have said
*nothing* about which alternative would be best, or which one they want.

But say the defeats aren't equal:

We've got 3 defeats, which, taken together, say nothing. Each one those 3
defeats can be truly said to be making the other two meaningless--making
the other two defeats into defeats-in-a-cycle, as opposed to defeats
by which the electorate are saying something.

Of course, if we could disregard one of the defeats, then the other two
would be meaningful statements by the electorate.

Suppose that one defeat, Y>Z, is weaker than the other two.

Each defeat can be regarded as in opposition to the other two. Y>Z prevents
both X>Y & Y>Z from being meaningful statements by the electorate. And each
of X>Y & Z>A prevents Y>Z from being a meaningful statement by the
electorate. And X>Y & Y>Z can also be said, together, to prevent Y>Z from
being a meaningful statement by the electorate.

If they weren't there, it would be, wouldn't it.

So, from the above:

Y>Z nullifies X>Y & Z>A.

and

X>Y & Z>A nullify Y>Z

Which nullifies the other(s) more strongly?

That's right: X>Y & Z>X nullify Y>Z more strongly than Y>Z nullifies X>Y &
Z>A, because they're both stronger than Y>Z.

So Y>Z should be counted as the nullified defeat in that cycle.

Y>Z is nullified by being the weakest defeat in a cycle.

Here's the primary definition of the MAM count:

A defeat is affirmed if it isn't the weakest defeat in a cycle whose other
defeats are affirmed.

[end of definition]

What that's saying is that a defeat is nullified if it is the weakest
defeat in a cycle whose other defeats *are not *nullified by that
definition.

Yes, that's recursive. No, it isn't circular.

The strongest two defeats are automatically not nullified, because it's
impossible for either of them to be the weakest defeat in a cycle. So there
are un-nullified defeats (to which the above definition can refer, when
evaluating the other defeats.).

I'll repeat that:

A defeat is nullified if it's the weakest defeat in a cycle whose other
defeats aren't nullified.

It wouldn't make a whole lot of sense to say that a defeat is nullified (&
therefore disregsrded) because it's the weakest defeat in a cycle some of
whose other defeats, themselves, are nullified (and being disregarded).

But that's what Beatpath does.

Disregarding a publicly-voted preference is undemocratic. It shouldn't be
done unless necessary.

Markus wants to know how I define "necessary".

It's necessary to disregard a defeat if it's nullified by being the weakest
defeat in a cycle (obviously that would be a cycle whose other defeats
aren't themselves disregarded, nullified in that manner).

There, Markus, that's what I mean by "necessary". "Unnecessary" means "not
necessary". (for which, refer to the 1st sentence in this paragraph.)

By that obvious, natural & fair definition, Beatpath unnecessarilly
disregards defeats. Unnecessarily disregards preference statements made by
the public.

That's why, when MAM & Beatpath give different winners, MAM's winner
pairwise-beats Beatpath's winner in the vast majority of instances.



>
> In those instances, where Tideman's ranked pairs method
> satisfies this desideratum and the Schulze method
> violates this desideratum Mike Ossipoff will say that
> the Schulze method _unnecessarily_ violates this
> desideratum.
>

The "desideratum" that I said shouldn't be unnecessarily violated is the
honoring of a publicly-voted preference. I've told what, for that purpose,
"necessarily" means.  No, I didn't define "necessary" as "done by MAM".

You're all confused, & you've got it backwards. I told why it can be
accurately said that MAM, and not Beatpath, only disregards a defeat when
it's necessary (where "necessary" isn't defined according to which method
does it.)


>
> And in those instances, where Tideman's ranked pairs
> method violates this desideratum and the Schulze method
> satisfies this desideratum Mike Ossipoff will say that
> the ranked pairs method _necessarily_ violates this
> desideratum.
>

See above.

And, by the way, you're calling MAM  "Tideman's Ranked-Pairs method".

Incorrect. I don't advocate Tideman's Ranked-Pairs method.

MAM is different from Tideman's Ranked-Pairs method in at least two
important ways:

1. MAM uses winning-votes (wv) as its measure of defeat-strength.

2. MAM incorporates tiebreaking bylaws that give it desirable
criterion-compliances even with a very small electtorate. ...something that
can be important in polls.

...especially my EM polls :^)

Michael Ossipoff


>
> Markus Schulze
>
>
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