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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: twoeleven
2007-11-08 01:41 am (UTC)
i don't think so. both the first and second place candidates will be seated. the fraction of the vote the second place candidate needs depends on what fraction of the vote the first place candidate gets. consider the extreme case: everybody casts one of their votes for one candidate, who therefore comes in first. one of the other candidates can come in second w/ half + 1 votes.

the 2/3 limit applies as the candidates become more evenly matched, and i think it works perfectly in that case. the candidate w/ the most support wins (w/ as few as 2/3 + 1 of the voters) followed by a candidate w/ 2/3 support (in the limit).

the only reason "half the voters voted for a candidate and yet he lost" is significant is that we're used to thinking of one man, one vote systems. different rules apply for other systems.
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