Jan Kok said: > I'm curious if anyone can mathematically justify such statements as > "Voting method A exhibits property P 'more often' than method B"?
Well, for methods that use strictly ranked ballots to pick among N candidates I would represent all possible electorates with an N! dimensional vector space. Each direction would correspond to the number of voters with a given (sincere, normally) preference order. I'd look at an N!-1 dimensional "slice" of that vector space given by the constraint that Sum(j=1,N!)V(j)=1 and V(j)>=0 where V(j) is the percentage of voters with preference order j. Within this region, I'd identify the areas in which property P is satisfied by a particular method and use standard integration to compute a volume, and divide that volume by the total volume of the region in question. I don't think that this method could be interpreted as giving a probabilistic answer to the question "how often" because it assumes that all possible scenarios are equally probable. Still, it allows transitive comparisons, i.e. if property P1 occurs more often than property P2, and P2 occurs more often than P3, then P1 occurs more often then P3. One easy application of this method is to figure out "how often" there will be a Condorcet winner. With N candidates we have N(N-1)/2 pairwise contests. Since each contest has 2 possible outcomes (ignoring ties) there are 2^(N(N-1)/2) possible breakdowns for the pairwise results. Each combination of pairwise results covers a fraction 2^(-N(N-1)/2) of the total electoral space. Now, say that we have a Condorcet winner. Among the other N-1 candidates there are (N-1)(N-2)/2 pairwise contests and 2^((N-1)(N-2)/2) possible ways that the pairwise contests among them could break down. Each of those outcomes covers a fraction 2^(-N(N-1)/2) of the total volume of electoral space (for a fixed number of voters). A little algebra shows that the region in which candidate A is the Condorcet winner covers a fraction 2^(1-N) of electoral space. Finally, since there are N possible Condorcet winners, the fraction of electoral space in which a Condorcet winner exists is N*2^(1-N). A quick sanity check indicates that for 2 candidates, the above formula says that 100% of electoral space has a Condorcet winner, as we'd expect. The interpretation is that as we add more candidates, to be a CW one must win more and more pairwise contests. This becomes harder and harder to do, so the fraction of electoral space in which a CW exists becomes smaller and smaller as we add more candidates. Alex ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
