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James,<br>
<blockquote type="cite">
<pre>><i>the ratings should be scaled and maximised among the members of the
</i>><i>Schwartz set,
</i>><i>between steps 2 and 3.
</i>
Sorry, Chris, what exactly do you mean by "scaled and maximized"? Is that
like raising the highest candidate to 100 and the lowest to 0?</pre>
</blockquote>
CB: I should have written "maximised and scaled". Yes, the highest-rated
Schwartz-set member is changed to the maximum,<br>
the lowest-rated member is changed to the minimum, and the middle-ranked
are changed so as to keep their relative spots in the <br>
range between those two extremes. For example if there are three Schwartz-set
members then (assuming a rating scale of 0-100)<br>
100, 99, 98 becomes 100, 50, 0; and 0, 1, 4 becomes 0, 25, 100; and
80, 20, 2 becomes 100, 23.684, 0.<br>
<br>
<pre>26: Bush 100 > Dean 10 > Kerry 0
22: Bush 100 > Kerry 10 > Dean 0
26: Dean 100 > Kerry 90 > Bush 0
1: Dean 100 > Bush 50 > Kerry 0
21: Kerry 100 > Dean 90 > Bush 0
4: Kerry 100 > Bush 50 > Dean 0</pre>
In your example, the voters have thoughtfully done this themselves.<br>
If this is not done, then I think there would remain some incentive to exaggerate
ratings among the likelt top-tier contenders.<br>
<br>
You wrote (Wed.Jun.9):<br>
<blockquote type="cite">
<pre>Weighted pairwise comparison method
Ballots:
1. Ranked ballot. Equal rankings are allowed.
2. Ratings ballot. e.g. 0-100, whole numbers only. Equal ratings allowed.
Note: You can give two candidates equal ratings while still giving them
unequal rankings. However, if you give one candidate a higher rating than
another, then you must also give the higher-rated candidate a higher
ranking.
Tally:
1. Pairwise tally, using the ranked ballots only. Elect the Condorcet
winner if one exists.
If no Condorcet winner exists:
2. Determine the direction of the defeats using the ranked ballots for a
pairwise comparison tally.
3. Determine the strength of the defeats by finding the weighted magnitude
as follows. We’ll say that the particular defeat we’re considering is
candidate A beating candidate B. For each voter who ranks A over B, and
*only* for voters who rank A over B, subtract their rating of B from their
rating of A, to get the marginal utility. The sum of these winning
marginal utilities is the total weighted magnitude of the defeat.
4. Now that the directions of the pairwise defeats have been determined
(in step 2) and the strength of the defeats have been determined (in step
3), you can choose from a variety of Condorcet completion methods to
determine the winner. Beatpath and ranked pairs are my preferred choices.
<end of definition></pre>
</blockquote>
CB: On reflection, better still would be to derive simple approval scores
from the ballots by each ballot approving the <br>
Schwartz-set member they rate above average (of the members), and half-approving
those they rate exactly average.<br>
Then in step 3, the "strength of the defeats" are the scores derived thus
: on those ballots that rank the pairwise-winner <br>
above the pairwise- loser, simple approvals for the winner minus simple approvals
for the loser.<br>
(Another good alternative would be to simply eliminate the candidate with
the lowest approval score, and then start again.)<br>
<br>
One advantage of this version is that it should get rid of any need or demand
for separate ranked-ballots. Voters would <br>
have no disincentive to rate favoured candidates slightly differently, so
the rankings can be simply inferred from the ratings.<br>
<br>
In your example, all the candidates are in the Schwartz set. Converting the
ballots to simple Approval ballots, we get:<br>
48: Bush<br>
26: Dean, Kerry<br>
01: Dean>.5Bush<br>
21: Kerry, Dean<br>
04: Kerry>.5Bush<br>
Approval scores- Bush:50.5 Dean:48 Kerry:51 <br>
<br>
(From here I think all reasonable approaches elect Kerry. He has the highest
approval score, and if we eliminate the candidate<br>
with the lowest approval score (Dean), then he pairwise beats Bush.)<br>
Proceeding with the more complicated version . Bush pairwise beats Dean.
On those ballots that rank Bush above Dean, Bush's<br>
approval score is 50 and Deans is 0, so the "approval magnitude" of Dean's
defeat is 50.<br>
Dean pairwise beats Kerry. On those ballots that rank Dean above Kerry, Dean's
approval score is 27 and Kerry's is 26 so the<br>
"approval magnitude" of Kerry's defeat is 1.<br>
Kerry pairwise beats Bush. On those ballots that rank Kerry above Bush, Kerry's
approval score is 51 and Bush's approval <br>
score is 2, so the "approval magnitude" of Bush's defeat is 49.<br>
So now we have:<br>
Bush>Dean 50<br>
Kerry>Bush 49<br>
Dean>Kerry 1<br>
<br>
Kerry wins. MinMax drops the Kerry defeat. Ranked Pairs locks Bush>Dean
and then Kerry>Bush and then skips Dean>Kerry<br>
to give a final order Kerry>Bush>Dean.<br>
<br>
So that looks pretty good. What do you think? <br>
<br>
My old idea of completing Condorcet by compressing ranks also picks Kerry
here, so I'm interested in seeing how the two methods compare in future
examples.<br>
<br>
Chris Benham<br>
<br>
<br>
<blockquote type="cite">
<pre>"Condorcet completed by Compressing Ranks:
1.CR ballots, voters give each candidate a score out of 100 ( or any
other round number that is much greater than the number of candidates.)
2. Inferring the rankings from the ratings, elect the CW is there is
one. If not, then eliminate the non-members of the Schwartz set.
3.Those ballots which show as many preference-levels among the remaining
candidates as there are remaining candidates (N) are now converted to
ballots showing (N-1) preferences by disregarding the smallest gap in
the scores of adjacently ranked remaining candidates. If there is a tie
for "smallest gap" , then disregard the tied gap between the highest
adjacent preferences among the remaining candidates.
4. Inferring rankings from these ballots (amended so that none show more
than N-1 preference-levels), elect the CW if there is one. If not ,
eliminate non-members of the Schwartz set, repeat step 3 and so on.
An equivalent version which is easier to hand-count but less natural and
more awkward to vote is to have the voters rank the candidates and also
rank the the gaps in utility between the adjacently-ranked candidates,
from largest to smallest."
</pre>
</blockquote>
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