- ... [...] ... I am not challenging this assertion. Rather, what I have tried to state is that IF a Condorcet-type method is advocated, that Kemeny-Young, aliasMessage 1 of 57 , Aug 1, 2009View Source--- In RangeVoting@yahoogroups.com, "warren_d_smith31" <wds@...> wrote:
>I am not challenging this assertion. Rather, what I have tried to state is that IF a Condorcet-type method is advocated, that Kemeny-Young, alias VoteFair ranking, is a good one to try, in part because it is well-documented (and this is why my reading from Markus and others that it was _poorly_ documented caused a strong respose). By contrast, I cannot understand Markus' method as described in, say, Wikipedia. I would still maintain that range/score voting is more expressive (and thus, by _my_ criteria better) than any method based on ranking alone; nothing I have said in this thread is intended to contradict this statement of Warrens.
> I would however, claim that range voting, while NOT "always"
> producing the best possible result, does produce better quality ON
> AVERAGE, for either honest or strategic voters, than any rank-order
> method including Kemeny-Young.
> This claim can be and has been, precisely defined, and supported by
> both theorems and computer simulations.
- Huh! You SAID eliminate the canddt with the fewest top-rank votes . So I presented a case where this caused A to get eliminated first, while I would expect AMessage 57 of 57 , Aug 4, 2009View SourceHuh!
You SAID "eliminate the canddt with the fewest top-rank votes".
So I presented a case where this caused A to get eliminated first,
while I would expect A to win with true Condorcet counting (you had
said nothing about involving a pairwise table).
On Aug 4, 2009, at 8:15 PM, warren_d_smith31 wrote:
>>> --there is another IRV-like voting method, which *is* a Condorcet
>>> you eliminate the canddt with the fewest top-rank votes, until
>>> (among the still-surviving candidates) a beats-all winner exists.
>>> That seems to be a pretty good voting method.
>> Sounds like an IRV weakness. Try:
>> 6 Z>A
>> 5 X>A
>> 4 A
> --ok, in this situation, pairwise table is
> * A X Z
> A * 10 9
> X 5 * 7
> Z 6 8 *
> hence A is a beats-all winner. Therefore, zero eliminations occur
> since a beats-all winner already exists, and A is the winner.
> A variant of this method, was recommended by N.Tideman in his book.
> Chris Benham pointed out D.Woodall had already proposed the
> (seemingly better) simpler method
> I just described. This method also is one of those implemented in
> the "condorcet internet voting service" CIVS by Andrew Meyer.