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Otter Bytes
Posted by Faun Otter in Election Reform
Wed Feb 03rd 2010, 04:59 PM
I have obtained the information that finally explains how the Senate election scam in Mass was pulled off.
The quoted "turnout" percentages are the sum of COUNTED votes divided by the number of registered voters times 100 (to make it a percentage) Therefore the "turnout" actually includes the cast votes were discarded without being counted.

How did the optical scanners that are now all owned by the Republicans perform compared with hand counting?
No surprises, they flipped the Coakley and Brown totals. The figures are eerie for their obviousness. I didn't round the figures, I worked from the raw data. Those trailing zeroes are real.

Count method Brown % Coakley % Libertarian % C/O to B/McC ratio (1) "Turnout" (2)
___________________________________________

Republican owned counting machines 52.00 47.02 0.98 52.76 58.80

Hand counted 47.00 51.78 1.22 59.31 60.31

Independently owned counting machines (3) 49.32 50.01 0.67 61.61 70.00

Levers 65.08 33.59 1.33 44.64 50.00


(1) If the Coakley votes are expressed as a proportion of Obama's votes, and Brown's as a proportion of McCain's votes, and the ratio of these figures is calculated, it should be roughly constant since there is no reason for this ratio to change. I also calculated the relationships between Brown's votes on each equipment type and the votes for Barr/Baldwin, and the relationship between Coakley's votes and McKinney/Nader. The variation in these confirms that the Coakley/Brown figures were flipped on the Republican machines, and that Coakley's votes were undercounted on the lever machines.

(2) 100% - turnout % = the combined none voting number (reasonably constant unless suppressed) plus the uncounted votes cast.

(3) Used in only in Milton - acts as cross check by being very similar to the hand count result.

The lever machines were shimmed in favor of Brown by discarding about 8% of the votes cast for Coakley. The person that rigged those machines forgot that the third party candidate total would increase as a percentage of the total count under a three candidate scenario. Can we cross check this? Yes - the alleged turnout (which actually measures turnout minus the uncounted votes that were cast) is below the predicted percentages. The lever figures are easily detected as false because the person who 'fixed' them went way over the top on some machines. See Southbridge for example - 34% turnout (That shows how many uncounted Coakley votes were trimmed) and Coakley only getting 42% while Obama got 65% of the vote in 2008? These figures, combined with the third party 'surge' to almost 3% is an outlier that discloses the magnitude of skimming Coakley's votes.
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Faun Otter
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156 posts
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Otter Bytes are selections from my web posts at DU, Bartcop and Buzzflash since 2000. My background is in statistics and running a Fortune 500 company's Market Research department. I have recently retired to a solar powered farmhouse hidden in the dismal swamps of the Carolina colonies.
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