On Wed, 23 Feb 2005 22:37:01 +0100 (CET) Kevin Venzke wrote:
Dave,
--- Dave Ketchum <[EMAIL PROTECTED]> a �crit :
Counting votes:
(wv) seems the appropriate choice. If two voters rank a pair of candidates (a=b) as equal, then (a>b) and (b>a) should each get one count.
This paragraph is confusing because in WV, a=b is not counted that way.
Ok, so lets call it wvx while we debate whether my idea makes sense.
He is ranking them as being tied, as in x>a=b>y. He likes each better than y, but not as well as x, but a exactly as well as b. Does not matter, except:
I want the array public, and think the counts will be more reasonable as above.
In resolving cycles I think this makes a difference, and WANT the above count to avoid pushing voters away from voting a=b.
Your method *would* push voters away from voting a=b, since the whole point of equal ranking is to *not* strengthen either candidate's win over the other.
A disconnect in thought. Try: 5 a>b 2 b>a 6 a=b By not counting the = we have 5:2 With counting them we have 8:5
The > and < determined that a gets 3 more votes than b - to me, strength of a's win over b.
I have the = making a have a total of 8 votes, useful (I think) in deciding how to untangle cycles, should that be needed.
Kevin Venzke
-- [EMAIL PROTECTED] people.clarityconnect.com/webpages3/davek Dave Ketchum 108 Halstead Ave, Owego, NY 13827-1708 607-687-5026 Do to no one what you would not want done to you. If you want peace, work for justice.
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