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on condorcet's paradox [Nov. 7th, 2007|10:36 am]
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%?


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.

[User Picture]From: hbergeronx
2007-11-07 09:23 pm (UTC)
I understand the math: what I don't know and don't trust is if San Bruno does. (seeing as he's the patron saint of the demonically posessed, and all :)

There were so many things wrong in how I perceived last night (lapis_lazuli summarized them pretty well) that I'm overwhelmed, and bereft of my usual logical placidity.
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