[EM] Later no harm, Condorcet, and randomization

[Kevin Venzke]

Dear Jobst,

 --- Jobst Heitzig <heitzig-j at web.de> a écrit : 
> Dear Kevin!
> 
> you wrote:
> > By the way, I think Later-no-harm is very important, in order to coax
> >  information out of the voters, and avoid de facto Clone-Winner
> > failures.
> 
> I agree with the first, that's why I try to find a compromise between
> LNH and Condorcet.
> 
> But what do you mean by "de facto Clone-Winner failures"?

Suppose we're using a method that satisfies Clone-Winner:

51 A
49 B

A wins. Now replace A with two clones, so

25 A1>A2
26 A2>A1
49 B

A1 or A2 will win, but only assuming this is how A voters really vote
after the cloning operation. In real life I suspect B has a very good
chance of winning.

> > I'm still digesting your suggestion. I believe Later-no-harm is
> > normally stricter than this, though: No candidate ranked above the
> > new preference is allowed to suffer a decrease in the probability of
> > being elected, even if that probability moves upwards in the voter's
> > ranking.
> 
> Yes, Woodall's definition for the non-deterministic case is different. I
> was only saying that for deterministic situations both agree!

I don't think this can be true, since a criterion for non-deterministic
cases can still be applied in deterministic circumstances. Revising
Woodall's criterion to be deterministic would just say that no candidate
above the new preference can turn into a loser.

> > But I don't think loosening this criterion would help much, since
> > normally if adding a preference can affect other candidates' odds of
> > election, you don't know whether these candidates are ranked above or
> > below the new preference.
> 
> Sorry, I don't understand this one... Anyway, I think my loosening of
> LNH is indeed of help: It is consistent with Condorcet, whereas LNH is not.

What I meant was, if you deal only with the pairwise matrix, it's not likely
that you can make a method which can tell which candidates can or can't
be harmed due to LNHarm.

I still have some difficulty understanding the sense in which "CW else
Random Candidate" satisfies a weakened LNHarm. I'll read it again.

> > A Condorcet winner always makes it into the CDTT set, incidentally. I
> >  think that's good enough for me.
> 
> If I understand CDTT right, it is equal to the Smith set, at least when
> no pairwise ties occurr, am I right? So what you suggested a few posts
> earlier was to use "Random Ballot on the Smith Set", right? Well, I
> think that would be more randomization than needed.

No, the CDTT can be defined two ways, unless I'm mistaken (and if I am
mistaken, I hope some list member will correct me!):

The set of all candidates which have majority-strength beatpaths to any
candidate possessing a majority-strength beatpath to them. (In other words,
candidate A is in the set unless there is some candidate B with a majority-
strength beatpath to A, while A doesn't have such a beatpath back.)

This is the set I intend to refer to (and implemented in my simulation), so
if the following definition by Woodall isn't the same, I will have to find
a new name for this set.

Woodall defines the CDTT as the union of all minimal nonempty sets where
no candidate in each set is dominated by any candidate outside the set.
(X "dominates" Y if more than half of the voters rank X above Y.)

The CDTT could be larger or smaller than the Smith set. One reason I suggest
Random Ballot is that, except for Woodall's DSC method, I don't know of any
other monotonic, clone-independent, LNHarm-satisfying method.

> I would prefer to
> restrict the choice to uncovered candidates, by using one of the Chain
> Climbing methods from yesterday, for example. What do you think about them?

I have to admit I haven't put any thought into uncovered candidates or their
significance.

I have a dumb question, though. You have a ranking (generated randomly, or
considering approval, etc.) and an "empty chain." I assume the first
candidate in the ranking goes into first spot in the chain. But what if
this is the CW? Then you won't be able to add any more candidates.

Kevin Venzke



	

	
		
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