[EM] Re: Proposal

[Rob LeGrand]

Jobst Heitzig wrote:
> But see the following example (which is the reverse of an example I gave
> on June 13): 9 voters, 4 options A,B,C,D. Sincere preferences:
> 1 B<C<D<A
> 1 C<D<B<A
> 1 D<B<C<A
> 2 B<A<C<D
> 2 C<A<D<B
> 2 D<A<B<C
>
> A is the CW but will only receive approval from the first 3 voters in
> each iteration, while B,C,D will always receive approval from 5 voters
> (1 of the first three and 2 of the last six), hence B,C,D end up having
> probability 1/3 each while A has nothing... Am I right?

Well, if all voters use an optimal approval strategy, this is something of
a knife-edge case.  If the electorate were

100:A>D>C>B
100:A>B>D>C
100:A>C>B>D
199:D>C>A>B
200:B>D>A>C
200:C>B>A>D

then the approval votes at equilibrium would be

100:A>>D>C>B
100:A>>B>D>C
100:A>>C>B>D
199:D>C>A>>B
200:B>D>>A>C
200:C>B>>A>D

and A wins.  (199 can be thought of as 200 minus epsilon.)  Reducing the
other two 200s in turn give similar equilibria, both of which elect A, the
Condorcet winner.

=====
Rob LeGrand, psephologist
rob at approvalvoting.org
Citizens for Approval Voting
http://www.approvalvoting.org/


		
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