[EM] Plurality with Condorcet polling is effectively Condorcet. Condorcet for 2012!

[MIKE OSSIPOFF]

(I emphasize the obvious fact that some of the most useful and helpful participants in this mailing list have been people who reside
in countries other than the one that I reside in. The project that I'm proposing is polling for the U.S. presidential election of 2012)
 
I've briefly mentioned this idea in a previous post.
 
I suggest that all here who want a better voting system for the U.S., and who reside in the U.S., so that they can conduct polling here,
work together on the following project:
 
All of us, in our own counties &/or cities, conduct a public poll. We can do that polling in places such as public
plazas, etc.
 
The poll consists of a rank-ballot, of the candidates in the 2012 U.S. presidential election.
 
The reason why I suggest polling in person, in public places, is to avoid the ballot-stuffing possible in Internet voting, and
to, avoid possible selection biases on the Internet.
 
Of course, one should only do one day's session of polling, because, if one polls for several days, it will be difficult to
recognize people who have already voted.
 
Each poll-conductor, after his/her polling session, will post his/her rankings to the EM mailing list. (We'll know what the
names of the polling volulnteers are, so only their rankings will be counted).
 
When every volunteer has done his/her polling session, and all of their ballots have been posted at EM, anyone, including
me, can count the ballots to find a CW.
 
But, instead of just counting the raw ballots, I suggest weighting them according to the number of ballots in each local poll, and the
population of the U.S. region in which that particluar local poll was conducted.
 
Here's how I'll do that (unless someone has a better suggestion):
 
On a U.S. map (conic projection or locally-centered azimuthal equidistant), I'll draw a line between each pair of
neighboring polling cities. Then I'll draw the perpendicular bisector of that line.
 
The set of perpendicular bisectors, together, will form a set of irregular polygons. I'll refer to those irregular polygons as
"regions".
 
For each local poll, each of its ballots will be weighted by multiplying it by the the population of the region in which that local
poll is conducted, divided by the number of ballots in that local poll.
 
How to find the population of a region? 
 
Of course if a state is entirely in that region, then its population is simply added into the region's population.
 
What if a state is partly in that region?
 
Sum the population of the major cities in that region, and add that sum into the region's population.
 
Assume that the state's population outside the major cities is uniformly-distributed.
 
Determine the area of that state that is in the region.
 
Multiply that area by the state's population density (adjusted by subtracting the populations of the major cities
that have already been added in)
 
Add, into the region's population, the result of that multiplication.
 
How to determin the area of a state that's in the region?
 
Unless someone has a better suggestion, I'll do as follows:
 
I'll use the method of transects, using, as the numerical integration method, either Simpson's rule, or another
closed Newton-Cotes formula.
 
In the method of transects, a line is drawn across the area to be measured, more or less through the region's center. 
The area's width, measured perpendicular to that line, are measured at regular intervals along the line. My measuring
interval will be the millilmeter marks on a ruler. The Newton-Cotes forumulas, including Simpson's rule, use regularly-spaced interval-divisions,
such as those on a ruler. Such forumulas give an area estimate.
 
Of course, the area-measurements needn't be exact, because the overall project will involve approximating assumptions less accurate than
the area-determinations.
 
Anyway, thereby will be gotten an estimate of how the ballots should be weighted, to simulate a national vote.
 
Using the weighted ballots, we find the Condorcet winner. That candidate has a win. We announce that CW to
various small parties, alternative candidates, political organizations, and progressive media (or of course any media you want to announce it to).
 
These people and organizations can make the CW known around the country, if they want to. Looking it it from the point of view of
a Progressive, I point out that, if the CW is a Progressive candidate, then the everyone who prefers a Progressive to the
Democrats will know that they can probably safely vote for that Progressive CW, because s/he has a win, even in Plurality.
 
At least, to the extent that voters and candidates are distributed on a one-dimensional political spectrum, the CW can win in Plurality
if everyone to one side of hir votes for hir.
 
Mike Ossipoff
 
 
 
 
 
 
 
 
 
 
  		 	   		  
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