1 ABCBorda and plurality are exactly equal in the eyes of this example. No voter can break the tie in favor of their favorite; any voter can break the tie in favor of their second choice by swapping their first and second choice. In IRV, this won't work; it will just cause a tie between the voter's second and third choices. In Condorcet, it depends on whether you allow the winner to have pairwise ties - if not, the swap gains you nothing.
1 ACB
1 CAB
1 CBA
1 BCA
1 BAC
Is the BC is more manipulable than IRV or plurality vote, in this case? If so,
precisely how much more (in numbers)?
This example seems extremely contrived - this sort of symmetry is incredibly unlikely in any real election. How about something more familiar, like the weak centrist:
35% ABC
15% BAC
15% BCA
35% CBA
In IRV, the edge voters have some incentive to swap their first and second choice, so that their third choice doesn't win. In Borda, the centrist is strong enough that it takes collusion from both edge camps to elect anyone else.
So in this case, Borda definitely seems better than IRV. But in my mind, this is pretty irrelevant. Because a Borda election won't look like that - rather, it will look like this:
35% ABCDEFGHIJKL
15% BACDEFGHIJKL
15% BCADEFGHIJKL
35% CBADEFGHIJKL
And every voter will have to decide where to put the nine dummy candidates - should they go in front of or behind the second choice? And what if I can write in some candidates, just for good measure? It's a mess.
Don Davidson likes to argue that all the non-IRV methods are just fancy tricks designed to elect unpopular candidates. This argument is absurd when applied to Condorcet, or to Approval if the voters act somewhat rationally. But with the Borda count, where voters have every incentive to stash seemingly-irrelevant candidates in second place on their ballot, the risk of an unknown/unpopular candidate getting elected is very real.
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