Saari and Cyclic Ambiguities

[Adam Tarr]

>5: ABCDE
>6: BCDEA
>7: CDEAB
>8: DEABC
>9: EABCD

The only ranked method discussed here that does NOT hand this election to E 
is IRV.

Borda count reduces this to

1: BCDEA
2: CDEAB
3: DEABC
4: EABCD

And the Borda scores are 20, 15, 15, 17, 30, for A, B, C, D, and E, 
respectively.  E wins by a huge margin.

In Condorcet voting, the Smith/Schwartz set is every candidate, there are 
no ties, and every voter expressed a full ballot.  So all the Condorcet 
methods are pretty much the same.  Ranked pairs overturns D>E and C>E (in 
that order).  SSD drops C>E, B>D, D>A, A>C, E>B, and D>E, (in that 
order).  Both results leave E unbeaten.

In IRV, A is eliminated first, then C, then E (by a count of 11B 15C 
9E).  E gets screwed by having the wrong votes transfer.  B wins the 
election, even though B finished dead last in Borda Count and only beats 
one candidate pairwise.

If the Approval cutoff is consistent, then Approval elects E as 
well.  Plurality of course elects E as well.  So really, only IRV can screw 
this one up.  It's not tremendously illustrative (sorry).

-Adam