[EM] Hey guys, look at this...

[Kristofer Munsterhjelm]

On 2/20/23 01:21, Kevin Venzke wrote:

>> I'd wondered whether Robert didn't have any intellectual commitent to the criterion, but
>> had used it in argument against IRV and therefore found his options limited.
> 
> I don't quite follow that since IRV satisfies Condorcet Loser.

I think the point is this: Robert says that if X beats Y pairwise, then 
Y shouldn't be elected. This requires the Condorcet winner criterion to 
be satisfied, because otherwise X = the CW, Y = anybody else violates it.

But suppose that a method didn't pass Condorcet loser. Then X = someone 
who's not the Condorcet loser, Y = the Condorcet loser would violate the 
criterion.

So this leads to a problem when proposing Condorcet,Plurality because if 
a Condorcet loser is elected, then it seems to go against the spirit of 
the criterion. About the only response I can think of is that if there's 
a cycle, it's impossible to satisfy the criterion (because everyone has 
someone else beating him), thus the criterion is suspended entirely for 
such elections.

The hard mode interpretation would be that the constraint implies Smith. 
But the problem then is that there are so few ways to make a method 
that's both Smith and simple to express in natural language. There are 
two relatively easy approaches, each with drawbacks: Copeland -- or 
elimination (e.g. Benham); and with elimination you lose monotonicity 
and (usually) summability[1], whereas Copeland seems to in some sense 
always be vulnerable to strategy, no matter what you combine it with. 
That's why it's hard mode :-)

-km

[1] Some uncommon exceptions include Nanson, Baldwin, and Raynaud. 
Funnily, it seems that combining Copeland and Borda to Copeland//Borda 
produces something both summable and monotone.