Demorep wrote: >Head to Head (Condorcet) Table > >BR Browne BH Bush BN Buchanan G Gore N Nader > > BR BH BN G N > >BR xx 62 70 52 49 >BH 38 xx 98 49 49 >BN 30 2 xx 49 49 >G 48 51 51 xx 52 >N 51 51 51 48 xx
At last, someone has displayed this voting set in an fashion that doesn't make me go blind. So... Gore, Nader, and Browne are the members of the Smith set. There is a circular tie between them. Gore beats Nader 52-48, Nader beats Browne 51-49, and Browne beats Gore 52-48. Since the ballots are all fully expressed, Ranked Pairs and Schwartz Sequential Dropping will have the same effect; dropping the Nader-Browne victory. This gives the election to Browne. If I understand Borda-seeded bubble sort right, it too will have the same outcome. The Borda rankings (ignoring those outside the smith set) are 1) Browne, 2) Gore, 3) Nader. So Nader fails to advance on his turn, and Gore fails to advance on his turn. Browne then is left in first. If, on the other hand, say the cyclic ties went the same way, but due to Nader running a few clone candidates, the initial order was 1) Nader 2) Browne 3) Gore, Then Nader wins. Basically, whoever starts in first place in the cyclic tie wins, since whoever makes it up to challenge them must be the candidate that loses to them. So insofar as Borda seeded bubble sort uses Borda count to decide things, it has the same strategic problems as Borda. -Adam
