> On Wed, Oct 20, 2004 at 02:13:23AM -0400, Robert A. Book wrote: > > I think what you want is the Banzhaf Power Index, developed by Banzhaf > > (surprise!) in the 1960s. I forwarded your post to a friend of mine > > who's done some work on this, and discovered he's giving a talk on > > this very topic on Friday at the GWU math department. > > > > His summary, with a link to a more detailed web page, is below. > > I read his web page quickly, but did not find it particularly relevant. > The Banzhaf Power Index is apparently about figuring out how much power > each voter has in a block voting system, where everyone is not equal in > the sense that your vote has a different probability of influencing the > election depending on where you live. But he starts off by assuming that > every voter has an independent .5 probability of voting for each > candidate. That makes the analysis useless for the purpose of computing > the expected utility of voting, because it ignores all of the relevant > information that we actually have about the likely outcome of the > election, such as the IEM market data. >
Funny you should mention that -- I thought of the same thing after I posted Mark's message to the list, and I think I have a solution. The Banzhaf model could be tweaked to take this into account by weighting each coalition by (p^i)*((1-p)^(n-i) where p is the probability of a given voter being for candidate #1, n being the number of voters, and i being the number voting for candidate #1 in the particular coalition. This is the binonmal distribution with parameters (n,p), which for large n (and p not too close to 0 or 1) (or more precisely, when np(1-p)>5) can be approximated very well by a normal distribution with mean np and variance np(1-p). The result should be that the power of an individual vote should drop as p gets farther from 0.5, in either direction. That is, an individual vote in Nevada is more likely to be decisive than one in Utah even though both have 5 electoral votes -- because in Utah Bush is ahead by 37 points (in some poll anyway) and in Nevada Kerry is ahead by 1 point (again, in some poll). So a vote in Utah will not be decisive at all, but a vote in Nevada might be. I would suspect that a vote in Florida (25 electoral votes, 1 point difference) is more powerful than a vote in California (55 electoral votes, Kerry leading by 8 points). I think that in real elections, a model that takes current polls into account would be more useful -- it would tell candidates where to campaign. They should campaign where the poll-adjusted Banzhaf index is higher. Who knows, maybe they have figured this out already.... --Robert
