Hello Raphael again! > If so, there is an odd number of winners in that case, which might > prove problematic.
For example:
2 abc
2 bca
3 cab
Starting with a uniform distribution, we get:
no. of
papers for
iteration a b c
0 1 1 1
1 7 6 8
2 43 38 66
3 217 274 538
4 727 2286 4190
5 ?
Algebraically, one finds that the following is an equilibrium:
no. of
papers for
iteration a b c
0 1 2 2
1 7 14 14
Jobst
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