on condorcet's paradox 
[Nov. 7th, 200710:36 am]
matt

In an election where you pick two of three candidates (or write in), can the result of voting among the three candidates sum to ~100%?
http://sanmateodailynews.com/article/2007117sbcitycouncil
I suppose, each vote can be divided by 2 times the number of total voters (each voter votes twice) to give an ultimate "percentage". But, to my way of thinking, this means that Araujo's 25.3% return means that his name was on over 50% of the ballots, and he didn't win. Consider this:
of candidates A+B+C, pick two, if four people vote A+B,A+B,A+C,B+C: A=B=3/8= 38%. c=2/8=25%. And yet, fifty percent of people voted for C. If I detest the incumbents equally, by having to vote for two (and choosing randomly or the lesser of evils) I entrench incumbency.
Conversely, this means that if they didn't divide like that, and that's a percent of votes cast, that only 38% of people voted for the most popular candidate.
And, that's of people who actually showed up to vote.
How is this democracy a republic just?
update: as of 2pm, the vote is TIED for Measure F. Yes, your vote does matter, even when you're ready to damn it all. 

