[EM] Condorcet vs Approval

[Martin Harper]

Forest Simmons wrote:
>> I have a couple of suggested compromises starting with Condorcet and
>> moving towards Approval, but stopping short of ordinary Approval.

Looking at just the first for now...

>> I.  Condorcet.  I prefer the version of Condorcet that allows voters to
>> give a partial preference list, with all unlisted candidates considered to
>> fall below the last listed candidate. (These implicit preferences are
>> counted along with the explicitly marked preferences.)  If voters were
>> instructed to list (only) their APPROVED candidates in order of preference
>> (and leave their unapproved candidates off of the preference list), this
>> would be one small step in the direction of approval.

Why not allow them to (if they want to) make an entire ranking, and then 
draw a line between the approved and unapproved candidates?

1 Jake
2 Fred
------
3 Harold
4 Emily
5 Rob

Or (view in fixed width):

Name   1|2|3|4|5
-----------------
Emily     |  X
Fred     X|
Harold    |X
Jake   X  |
Rob       |    X

 
>> To put teeth into the instruction, voters should be informed that if there
>> is a Condorcet cycle of winners, then that tie will be resolved by the
>> approval scores ... all of the listed candidates will be considered
>> approved, and the non-listed candidates unapproved. 

I like it - was considering the same thing myself (Smith//Approval, I 
guess it'd be called). If there's only strategy in Condorcet methods if 
there's a tie, then it'd make sense to resolve the tie using methods 
which have sincere strategies, rather than ones which could encourage 
insincerity...

Because your Approval/Disapproval vote has to agree with your ranked 
vote, it becomes much harder to use insincere strategy: if there is a 
sincere A>B>C>A cycle, then A>B>C voters might be tempted to switch to 
voting B>A>C - but if doing so means that you can't approve just A, but 
would have to approve A and B - then you might decide against it - or at 
least be more reluctant.

In other words, insincere rankings reduce your available options for 
approval tie-breaking. That's probably good - any disincentive for lying 
has to be a bonus.

Strategy would seem to be quite complex for deciding where to draw the 
line, though... You're looking at the probability, given that there's a 
first place tie in approval ratings amongst the members of the Smith Set 
(which necessarilly means that there's a Condorcet Cycle), that said tie 
is between I and J.

In practice the key stratgical consideration could simply be to make 
sure that your approval vote actually counts - if A-F are candidates, 
and the Smith Set is A-C, then someone who approved A-C has effectively 
no voice in the tiebreak - so this would encourage voters to only 
approve of one or two candidates to make sure this never happened. Which 
makes the thing dangerously close to Smith//Plurality, sadly.

Would it be better to hold the approval tie-breaker as a seperate 
balloting to the condorcet vote, only if required (about 10% of the 
time, apparently). It'd be cheaper, certainly - though I'm wary of the 
effect that voters whose favourite didn't make it will have: you might 
find that they voted for the least competent candidate, so that their 
party have a better chance in the next election.

Would there be difficulties if a Condorcet Winner was less than 50% 
approved though? Or if they had less approval than another candidate? 
Perhaps best then not to look at the approvals at all unless they are 
needed...

>> Alternatively, before this drastic cycle breaker is applied, the
>> preferences among the approved could be discounted by half to see if that
>> less drastic action would resolve the cycle.

Hmm. I think that this gives an identical result to the more "drastic" 
method...

eg, I vote A>B>C, and approve A.
My B>C preference is going to be discounted by 1/2(call this preference 
"weak"), which means B will drop (slightly), and C will rise (slightly). 
If it is fully discounted (the drastic solution), than B will drop (a 
bit more), and C will rise (a bit more).

If the 50% discount resolves the tie, it will surely resolve it in 
favour of the candidate who would win the drastic resolution anyway?

eg2 (fixed width again)
                          discount of weak prefs.
compare  strong  weak    0%    50%   100%
A>B       +180   -80    +100   +140  +180
A>C       +50    -200   -150   -50   +50
B>C       +10    +100   +110   +60   +10

Here we have a cycle at 0% and 50%, and it is resolved at 100%.
Given that there has to be a cycle at 0%, and at 100% we know there 
cannot be, is there any way that the winner at 50% can be other than the 
winner at 100%?