At 04:37 PM 11/11/00 -0800, Alex Robson wrote: >Alex Taborrok wrote: >> Returning to a more Armchairish point note that there has been an >> awful lot of discussion about the "will of the people." But we know >> from Arrow's theorem that the very notion of "the" will of the people is >> incoherent. > >I agree that the notion of "the" will of the people is incoherent, >but does Arrow's theorem really say this? Doesn't it just say that under >certain assumptions a non-dictatorial procedure does not exist? Arrow's theorem requires three or more candidates. But the question I have is how many candidates do we have here? This, like most non-1992 elections, boils down to two players, GOP and Dem. But there are outside players: Nader especially, but Browne and Buchanan to a lesser extent. Arrow's theorem wouldn't apply to a 1v1 election, which, in essense, is what we have. An in-state plurality translates into a winner-take-all sum (NB and ME not withstanding) -- electoral votes. Each player needs a majority of those votes, making a third party vote an exercise in futility. In other words, Arrow's theorem barely applies, as it bends "the will of the people" to a simple binary. But that gives us a problem. Nader supporters are clearly more likely to prefer Gore over Bush; Buchanan (and for the most part Browne and Phillips) supporters would likely prefer Bush. These peoples' preferences are not included in the electoral college, but are pretty clear to even the largest skeptic. If we are to attempt to determine the will of the people, why not include these people's preference? Just because they didn't? Sounds like a good reason to me. Again, it boils down to a third-party vote being a non-vote; a vote to abstain. Again we have two players -- Arrow's theorem does not apply. This actually boils down to a pretty good defense for the Electoral College, but that's another story. Dan Lewis www.WhatTheHeck.com
