Condorcet(x( ))

[Mike Ossipoff]

Lucien Saumur writes:
> 
> In an article, dfb at bbs.cruzio.com (Mike Ossipoff) writes:
> 
> >About the use of x = .5 instead of 0:
> >
> >True, if you rank 2 candidates together in last place, or if you
> >don't rank either of them, then you're voting against them in the maximum
> >way: You're voting everyone else over them. _That's_ the sense in which
> >you've indicated that you want a maximum count against them.
> >
> >What you haven't done is indicate that you meant to rank each of them
> >over the other. As I said, it would be counting something that the voter
> >hasn't said. Harmless in, say Copeland, but not so harmless in a method,
> >like Condorcet, that counts votes-against.
> >
> >Most assuredly there's something wrong with x = .5 in Condorcet for
> >last-ranked or unranked pairs of alterntaives: Condorcet would lose its
> >properties of getting rid of the lesser-of-2-evils problem, getting
> >rid of the need for defensive strategy, and protecting majority rule.
> 
>           I do not understand what you mean by
> vote-against.

Condorcet's method, as I've defined it, says that:

If no 1 candidate beats each one of the others, then the winner is
the candidate who has the fewest voters ranking over him someone who
beats him.

In other words, for each candidate, determine which candidate who beats
him is ranked over him by the most voters. The number of voters ranking
that other candidate over him is the measure of how beaten he is. The 
winner is the candidate least beaten by that measure.

Because, in each pairwise comparison in which X is beaten, Condorcet's
circular-tie-breaker counts only the votes for the other candidate over
X, I call that "votes-against".

> 
>           I understand Condorcet to compare each candidate
> to every other candidate. Whether you count the unranked
> candidates as having obtained .0 or .5 vote against each
> other does not change the result of the comparison.

Not if there's a candidate who beats each  one of the others. In
that case, when we're determining if A beats B, of course it
doesn't matter which way we treat a pair of unranked candidates.

But, then there's no 1 candidate who beats each one of the others,
that's when Condorcet uses its circular-tie-breaker, based on votes-
against. Say you've not ranked A or B. Say A beats B, so we want
to know, for Condorcet's circular-tie-breaker, how many voters have
voted A over B. If we count you as voting A over B &  B over A,
then you're counted as voting A over B even though you never really
voted A over B. This counts as the vote against B in its pairwise
comparison against A. This can make B lose when it otherwise wouldn't.


> 
>           Example: if 85 of 100 voters do not rank A and B
> while 10 voters prefer A to B and 5 voters prefer B to A,
> the result will be the same (i.e. A beats B) whether we
> consider that A beats B 10 to 5 or (10 + 42.5) to (5 +
> 42.5).


In the sense that A beats B, nothing is changed. But if no 1 candidate
beats each one of the others, so we're using Condorcet's circular-tie-
breaker, then it makes a difference. In your example, if we don't
invent preferences that the voter didn't vote, then B has 10 votes
against him in the A vs B comparison. But if we falsify preference
votes by counting them from people who didn't vote them, then B has
47.5 votes against him in the A vs B comparison. Big difference.


Mike

> 
> __________________________________________
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>           http://www.igs.net/~lsaumur/
> 
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> 


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