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<h1>[EM] Condorcet and Participation </h1>
<b>Markus Schulze </b> <a
href="mailto:markus.schulze%40alumni.tu-berlin.de"
title="[EM] Condorcet and Participation">markus.schulze@alumni.tu-berlin.de
</a><br>
<i>Sun Oct 5 02:48:02 2003</i>
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<pre>Dear participants,
this is Moulin's proof that participation and Condorcet
are incompatible.
Situation 1:
3 ADBC
3 ADCB
4 BCAD
5 DBCA
Situation 2:
Suppose candidate B is elected with positive probability
in situation 1. When we add 6 BDAC voters then candidate B
must be elected with positive probability according to
participation and candidate D must be elected with
certainty according to Condorcet.
Situation 3:
Suppose candidate C is elected with positive probability
in situation 1. When we add 8 CBAD voters then candidate C
must be elected with positive probability according to
participation and candidate B must be elected with
certainty according to Condorcet.
Situation 4:
Suppose candidate D is elected with positive probability
in situation 1. When we add 4 DABC voters then candidate D
must be elected with positive probability according to
participation and candidate A must be elected with
certainty according to Condorcet.
Situation 5:
Because of the considerations in Situation 2-4 we get
to the conclusion that candidate A must be elected with
certainty in situation 1. When we add 4 CABD voters then
candidate B and candidate D must be elected each with
zero probability according to participation.
Situation 6:
Suppose candidate A is elected with positive probability
in situation 5. When we add 6 ACBD voters then candidate A
must be elected with positive probability according to
participation and candidate C must be elected with
certainty according to Condorcet.
Situation 7:
Suppose candidate C is elected with positive probability
in situation 5. When we add 4 CBAD voters then candidate C
must be elected with positive probability according to
participation and candidate B must be elected with
certainty according to Condorcet.
Markus Schulze
</pre>
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