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 [EMAIL PROTECTED] Citizens for Approval Voting http://www.approvalvoting.org/ __________________________________ Do you Yahoo!? Yahoo! Mail - 50x more storage than other providers! http://promotions.yahoo.com/new_mail ---- Election-methods mailing list - see http://electorama.com/em for list info
