[EM] Kevin--Your method and SDSC

[MIKE OSSIPOFF]

Kevin--

You said:

VFA is a simple method where the voter votes for one candidate
and against one candidate. The candidate with the most "for" votes
wins, unless that candidate has more than half of the "against"
votes, in which case the candidate with the second-most "for"
votes wins.

At least in the three-candidate case, I like this better than
Approval, since it satisfies Majority (for solid coalitions) and
"Majority Last Preference," as in, the last preference of a
majority can't win.

In VFA, a majority can make a candidate lose by voting against
him. It doesn't matter who any of them votes for.

So what do you say, Mike? Does VFA satisfy SDSC?

I reply:

When that majority vote against Y, to make Y lose, and if that's all they 
do,  then, for any Z, they're voting Z over Y.

Here's the short definition of voting for one candidate over another:

A voter votes X over Y if: If we count only his/her ballot, with all but X 
and Y deleted from it, X is the unique winner

[end of short definition of voting X over Y]

If we use that definition, and you vote against Y, but do nothing else, are 
you voting some Z over Y?

If we count only your ballot, with all but Y & Z deleted from it (but no one 
but Y was mentioned on your ballot), then Y is voted against by 100% of the 
voters. Y is disqualified.
Is Z the uniquie winner? Everyone but Y is un-disqualified, and they all 
have an equal number of votes for them: Zero. So they all win, and Z isn't 
the unique winner. So you haven't voted Z over Y, by that definition.

Well, when Richard proposed that defintion, which I'd considered and 
rejected previously, I agreed to it, because of its brevity. But there was 
always the possibililty that that definition could run into trouble, and so 
made it clear that I was keeping my own definition for use in such a 
situation:

A voter votes X over Y if it's possible to contrive a configuration of other 
voters' ballots such that, if we delete all but X and Y from the ballots, X 
is the unique winner. if and only if we count that voter's ballot.

[end of long definition of voting X over Y]

This is a situation in which the two definitions give different results 
about whether or not you're voting one particular candidate over another.

I adopted Richard's brief definition with reservations, and said that I was 
keeping my own definition in case Richard's briefer definition ran into 
trouble. This is such a time. When the two definitions give different 
results, it's clear from what I said then that I go with my definition.

When the short definition gives a different answer from that of my own 
definition, surely it's agreed that the short definition has run into 
trouble. I'm reluctantly going to have to drop the short definition, since 
it has now demonstrated that, even with a reasonably proposable method,  it 
can give a different result from that of my own definition. Applying  the 
long definition:

Suppose, for some Z, Y gets more votes for him than Z does, but Z gets more 
than anyone else  but Y.. But Y is one short of having the number of votes 
against him to disqualify him, and your ballot does nothing but cast a vote 
against Y.

That satisfies the long definition of voting Z over Y. So, if you do nothing 
other than vote against Y, then, for any Z, you're voting Z over Y.

You're voting everyone but Y over Y. And you aren't voting any of the others 
over eachother. Therefore, you're voting equal to eachother, everyone but Y, 
by my definition of voting one candidate equal to another.

Suppose that the majority who prefer X to Y consist of people who don't 
prefer any two candidates equally. Can they make Y lose without voting a 
less-liked candidate equal to a more-liked candidate? No. I've contrived a 
failure example. Your method fails SDSC.

Maybe Richard's definition could be brought into agreement with mine by 
requring instead that Y lose. Still, it seems safer to just use my own 
definition instead of waiting for the next possible situation in which the 
short definition might differ from it.

This is just a hasty first-glance answer. Unless I've made some error, your 
method doesn't comply with SDSC.

Mike Ossipoff

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