[EM] Proof Borda Count best in the case of fully ranked preference ballots

[Rob LeGrand]

Steve wrote:
> What do you think of circular triplets, such as:
>
> 	A>B>C
> 	B>C>A
> 	C>A>B,
>
> and reversals, such as:
>
> 	A>B>C
> 	C>B>A.
>
> If that is all the information that we have to go on (when ordinal preference
> ballots are used, it is ), shouldn't either of these profiles cancel out
> completely and yield a tie?

Well, yes, when they're taken alone, and any reasonable method would yield a
tie in these two elections, although IRV wouldn't in the second.  But I don't
think that's what Saari is talking about; he thinks adding symmetric groups of
votes should never change an election's winner.  Adding the second profile will
never change the winner of a Condorcet election since it would in effect add 1
to each cell of the pairwise matrix.  But say two of the first profile were
added to the election

5:A>C>B
4:C>B>A

to give

2:A>B>C
5:A>C>B
2:B>C>A
2:C>A>B
4:C>B>A

The Condorcet winner changes from A to C.  I don't see a paradox here.  After
all, the votes you're adding may be circular, but they're not entirely
symmetric.  The added votes in this case have 4 for C over A and only 2 for A
over C.  I can see how you might intuitively want circular triplets to cancel
completely, but is that really a good reason to reject Condorcet methods? 
Besides, the best Condorcet methods are just as monotonic as Borda.

Obviously, if you agree with Saari's criteria, then Borda is for you.  I'm not
attacking his mathematics, just his criteria.  He don't seem to find it
important that almost everyone in a well-informed Borda election will benefit
from voting insincerely, usually *very* insincerely.  Borda works fine in
zero-information elections, but how common is it in the real world for no voter
to have any clue as to how anyone else will vote?

Have you read Blake's arguments against Borda?  What do you think of them?

=====
Rob LeGrand
honky98 at aggies.org
http://www.aggies.org/honky98/

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