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 ---- Election-methods mailing list - see http://electorama.com/em for list info
