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matt

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on condorcet's paradox [Nov. 7th, 2007|10:36 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/2007-11-7-sb-city-council

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.
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Comments:
[User Picture]From: hbergeronx
2007-11-07 09:16 pm (UTC)
In principle, I could have voted for one, or Araujo + a write for Me, if (as with some election machines) they mandate two votes.

In some elections, if you only select one, the ballot is considered invalid and discarded. (I don't know the local politics well enough to know if any of the above apply, to my discredit, though I'm sufficiently motivated now to not make those kind of choices uniformed next time).

On the particular machine, it's tremendously difficult to write in. You must select each letter the way you select your initials in a videogame, by scrolling through all the letters.

The reason he didn't run a slate is that the opposition candidate is young (25) and inexperienced- the whole "no on F" campaign has a certain wacky quality which is what has had me following it these past two or so months. His command of math in many of the "no on F" campaigns clearly voiced that he hadnt thought out the mathematical consequences of how he ran his election.

I am not really contesting that those who won, won fairly. Over two thirds of people voted for either of the other candidates, which is fair and square in my book. I'm more concerned with the "how to make this better in the future" angle, and getting a little cheap bitching in as well fter what was a night of a comedy of errors. Ths was my first election after having moved out here where I had hoped I was ready to vote, and things were just so different than in NJ that I suffered culture shock.

What worries me is that whatever the circumstances, there's a situation where a small majority of people can (in an election where the rules prejudice the situation I describe in my post through subterfuge or ignorance or whatever) vote for a candidate and they are not elected, and what that means. I think there's a possibility where the winners can win fairly and at the same time, the loser can lose unfairly, and that's a concept I'm trying to wrap my head around.
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[User Picture]From: twoeleven
2007-11-08 12:03 am (UTC)
i guess i don't understand what you mean that the loss was unfair.
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[User Picture]From: hbergeronx
2007-11-08 12:16 am (UTC)
Each person gets two votes, but you can only vote for a person once. This means that even if half the people vote for you, you only get 25% of the votes, and you lose- you need 33%+1. By this line of resoning, it seems unfair: it seems like you need to have a supermajority (2/3) of people vote for you to win, which is a bigger hurdle than one on one.
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[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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