[EM] Later no harm, Condorcet, and randomization

[Jobst Heitzig]

Dear Kevin!

you wrote:
> 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 see, very good point! I haven't thought about that before...


>> 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.

Well, what does Woodalls LNH demand when the method always elects some
candidate deterministically, using randomness only for rare ties? It
demands that when some later preference is added, the winner cannot
change to a less-preferred candidate.
  And what does LNPMDO demand when the method always elects some
candidate deterministically, using randomness only for rare ties? It
demands that when some later preference is added, the winner cannot
change to a less-preferred candidate.
  That's exaclty the same, they only differ for non-deterministic methods.

> 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 don't suggest to use some pairwise matrix only. Most of the randomized
methods I suggested involve drawing a random ballot also.

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

I'll try to explain again: (i) When A is the CW, then adding a later
preference to some ballot that already contains A leaves A as the CW, so
nothing changes. (ii) When there is no CW then all candidates have a
positive probability of winning, including my last choice. So when I add
some further preferences to my ballot, then either nothing changes since
there is still no CW, or I produce a CW which will then get elected with
certainty. Since that CW cannot be my last choice, I will have the
positive effect that before the change my last choice was a possible
winner, but after the change it is not. That's what LNPMDO demands.

> 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.)

Ah, OK, I understand the difference now. The Smith set can be defined
similarly, but using ordinary beatpaths instead of "majority strength
beatpaths", by which you mean a beatpath in which every beat is
supported by more than half of all voters, right? Then it will make a
difference when equal rankings occurr, OK.

> 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.

Do you consider LNH more important than Condorcet-efficiency?

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

I think the significance is only that they possess beatpaths of length
one or two to all other candidates and so majority complaints can be
rebutted most easily.

> 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.

There's no such thing as a "dumb" question I guess. What you say is
completely correct. When the CW happens to be first in the sort order,
no further candidate would be added to the chain and the CW would thus
win like s/he should.

Don't think of the sort order as a ranking, it is only the order in
which I look at the candidates! When there is a CW, no matter where in
the order s/he comes, s/he will win. Only when there is no CW, there are
three or more candidates (=the Banks set) which could be the winner,
depending on the sort order. The significance of the sort order is then
that the later candidates have a better chance of winning since they
need only beat the candidates in the chain, while the first candidate in
the sort order has to beat all other candidates to win. In other words,
TACC elects the least approved candidate only when s/he is the CW, which
is a kind of "intuitive loser" criterion, isn't it?

Yours, Jobst