[EM] Re: 49 A, 24 B, 27 C>B example

Forest Simmons simmonfo at up.edu
Mon Dec 27 16:10:07 PST 2004


I'm sure that Kevin has given more thought to this example than I have, 
but it just occured to me that C cannot win under approval no matter where 
the third faction members place their approval cutoffs, whether above, 
below, or equal to B.  Since C can never win (under approval) the best 
strategy for the C faction members is to approve B.  For them this 
strategy majorizes or dominates their alternatives.  So the rational 
approval winner under perfect information is B.

It is interesting to me that we can have a cycle of three candidates in 
which not every candidate is an approval winner for some allocation of 
approval cutoffs.  However, I can prove that if there are no equal 
rankings (including truncations) then in a cycle of three candidates, each 
of them is an approval winner for some allocation of approval cutoffs.

In fact, it turns out that this is the case even if all voters are using 
above mean utility approval strategy based on the same winning 
probabilities.  [Of course, to make C win you have to use different 
probabilities than those that make A win, etc.]

If anybody wants to see this proof, I would be happy to post it to the 
EM list.

One lesson to learn from this is that the process of symmetric completion 
can change an approval dominated candidate like C into a contender.  In my 
view this is not desireable.

Remember in the spruce up process, we eliminate covered candidates because 
they are not serious contenders under sophisticated strategy.  For the 
same reason we should also eliminate approval dominated candidates like C.

I think this would be easiest to do after having eliminated the pairwise 
covered candidates and having collapsed the beat clone sets, and certainly 
before any symmetric completion step that Chris might want to throw in as 
part of Weighted Median Approval, for example.

If after these three reductions we are still left with a cycle of three 
candidates, then why not use a non-deterministic method to resolve the 
cycle?  I think that's our best hope for completely doing away with 
incentives for insincere order on the ballots.

Forest



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