At 10:44 PM 8/30/2005, Warren Smith wrote:
**Second, I am concerned about "Favorite Betrayal" and "2-party domination."
Earlier on EM I basically constructed a mathematical proof that all
Condorcet methods lead to 2-party domination. (It may be seen at
http://math.temple.edu/~wds/crv/CondStratPf .)
I'm concerned that Mr. Smith presents his paper as a "mathematical
proof" of what it does not mathematically prove. He purports to show
in the paper that the answer to the "Question" is yes.:
MATHEMATICAL MODEL: 3-candidate V-voter Condorcet elections
with random voters (all 3!=6 permutations=votes equally likely).
QUESTION: Is there a subset of identically-voting voters, who, by
changing their vote
to rank the two "perceived frontrunners" max and min ("betraying" their
true favorite "third party" candidate) can make their least-worst
frontrunner win
(whereas, their true favorite cannot be made to win no matter what they do)?
However, I must say I find it rather hard to follow his proof. The
model would produce, in a large election, would produce near ties, if
I am correct; that is, for N votes, the occurrence of each vote
pattern would approach N/6. I constructed a simple example to show a
near-tie by Condorcet, and to see what would happen if all
identically-voting voters changed their vote to "betray" their
favorite so as to shift the election.
101: A>B>C
100: A>C>B
100: B>A>C
100: B>C>A
100: C>A>B
100: C>B>A
pairwise
A:B 301:300
A:C 301:300
B:C 301:300
A is the Condorcet winner.
Can the B voters who prefer C to A change the result?
101: A>B>C
100: A>C>B
100: B>A>C
100: C>B>A (revised to "betray" B
100: C>A>B
100: C>B>A
Pairwise
A:B 301:300
A:C 301:300
B:C 201:400
It appears not.
Which makes the paper unintelligible to me. All voters who prefer
another candidate to A have already done their utmost to defeat A
under the original conditions, changing their vote does not increase
the pressure against A.
To shift the result, voters who actually preferred A would have to
reverse their choice.
But since I don't understand the math in the paper, perhaps indeed
there is something I've overlooked.
My proof was then attacked
because (a) it lived in a somewhat unrealistic "random elections" model
(all elections equally likely) - that objection actually did not bother me -
though it is a somewhat valid criticism (as it is of many unlikely
scenarios). However, that does not seem to be the major problem.
and (b) my proof had assumed that the Condorcet method employed
full-rank-orderings as votes ("equalities" not permitted, truncation of
preferences not permitted, and "margins" assumed to be used).
"margins"? As I understand basic Condorcet, simple rankings are used,
and the Condorcet winner is the candidate who beats all other
candidates in the isolated pairwise elections, as I showed in my
example. I'm not sure that truncations make a difference, they merely
complicate the model and examples. It was fine to assume no
truncations for the purposes of this examination, I'd think. Again,
I'm not sure what "Margins" means. In the example given, perhaps this
would refer to the "margin" by which C beats B; some methods might
consider that, and this example might be a good reason why this is a
bad idea, but that is not Condorcet, strictly speaking.
Looking deeper, I find that Mr. Smith was asked to construct an
example, and he claims that he has already done so. The example is
found at: http://math.temple.edu/~wds/crv/IncentToExagg.html
8 B>C>A
6 C>A>B
5 A>B>C
pairwise:
A:B 11:8
A:C 5:14
B:C 13:6
However, as he notes, this election has a "Condorcet cycle." That is,
there is no Condorcet winner. The variation between the votes is
quite large, so this election is only reasonably possible if the
number of votes cast is small (assuming his equal probability model,
which he did not state here.)
Can the 6 C>A>B voters manipulate the election to cause A to win?
Yes, they can, by changing their vote to A>C>B, which eliminates the
C win over A in the original vote and thus breaks the Condorcet
cycle. This indeed is a "betrayal" of their favorite, and a real
betrayal. Consider this: they would have to vote this way *in advance
of knowing that C was not going to win.* Under the model conditions,
C and A are equally likely to win the election. Since they prefer C,
why in the world would they betray their favorite?
It is only under the unlikely circumstance of a Condorcet cycle that
this kind of scenario becomes possible. And even then, it is
*extremely* unlikely that voters would actually do it. Further, what
is to stop the 8 B>C>A voters from doing the same thing? They have
the same opportunity; by ranking instead C>B>A, they would throw the
election to C. Likewise the A voters can throw the election to B.
But these strategies make no sense. Far from Q.E.D., these proofs
seem to be O.T.L. -- out to lunch.
Of course, I'm not a professional mathematician, and my math
education was years ago. Still, I have a habit of not assuming I'm an
idiot but waiting until it is clearly shown.... which does happen at times.
This
latter criticism b actually does bother me, and it was due to my
ignorance - I not having been exposed to EM was under the false impression
that these assumptions were all part an parcel of what it meant to be
a "Condorcet" method. Actually my proof can readily be extended even
to permit equality-rankings.
The margins are the problem, I suspect.
But anyhow - the essential point which was made to rebut me (by Adam Tarr)
was: Condorcet methods employing BOTH winning-votes AND equalities-permitted
can escape my 2-party-dominance proof and maybe can escape 2-party dominance.
This it seems to me is very important. It seems to me that from now
on, it is just
silly to talk about Condorcet methods unless it is both wv and =permitted,
because 2-party dominance is a failure state as far as I am concerned.
Mr. Smith, it seems, is far from showing the alleged 2-party effect
with Condorcet. His alleged proof shows, defectively I think, that
there is a "plurality-like" strategy possible. And from that he jumps
to the non sequitur that two-party dominance is inevitable with such a system.
These allegations of solid and irrefutable proof, asserted in one
place, never proven in the public arena, and then continually cited
as if Mr. Smith is his own irrefutable source, are beginning to get to me....
If you want to create advanced math for all time, by all means, stick
to your guns. But if you want to convince people *today*, you will
have to show them in ways that they can understand. They are not
going to lay down and play dead because you are able to cite a proof
that is written somewhere else. Most of them won't bother to look. If
your proof really is cogent and convincing, the few that do look will
report their opinions back, so it will no longer be just you making
the claim. And others will repeat the work or validate it. This is,
indeed, how *science* works. It is not a solitary enterprise, though
certainly there are solitary aspects. On the other hand, if the proof
is unintelligible, even if somehow it is true, if only people would
spend the serious time that it would take to truly "get" it, it is
definitely not for today. And not for Iowa '08.
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