Beyond Reasonable Doubt: The New Hampshire primary was rigged
Once again, as in 2004 and 2006, the average of the final pre-election polls matched the unadjusted exit poll.
Once again, as in 2004 and 2006, the probability that so many polls would exceed the MoE is close to ZERO.
Once again, as always, we will be inundated with tortured explanations of why the polls were "wrong".
Once again, as always, the mantra that the polls were wrong implies that fraud could not have occurred.
Once again, as always, the CNN Final Exit poll was forced to match an implausible final vote count.
Once again, as always, there will not be a random, robust, credible ballot recount.
Once again, as always, there will be a hue and cry to eliminate exit polling altogether.
Once again, as always, the uninformed masses will believe whatever the media tells them.
Once again, as always, there will be a variety of bogus rationalizations to “explain” the astounding discrepancies.
Once again, as always, the media myth: Clinton’s emotional display won her a huge, last-minute female sympathy vote.
Once again, as always, they can hardly wait for November to be fleeced again.
In 2004 they produced another bin Laden tape the weekend before the 2004 election.
In 2004 the media fell for the Rove myth of a massive fundamental Christian voter turnout.
But the fact is that the Democratic GOTV effort overwhelmed that of the GOP.
Once again, it’s DÉJÀ VU all over again.
ALL 20 FINAL pre-election polls (3-4% MoE) had Obama winning by an average of 8% over HRC.
And the early (unadjusted) exit poll had Obama winning: by 8%.
http://www.bradblog.com/?p=5535
The Zogby polling trend was to Obama:
http://www.zogby.com/news/ReadNews.dbm?ID=1417
Date Pollster Sample Mix MoE BO HRC JE
106 StratVision 600 9.7% 4.0% 38 29 19
106 USA/Gallup 778 12.6% 3.5% 41 28 19
106 CBS News 323 5.2% 5.5% 35 28 19
106 Marist 636 10.3% 3.9% 36 28 22
106 CNN 599 9.7% 4.0% 39 30 16
107 Rasmussen 1774 28.7% 2.3% 37 30 19
107 Zogby 862 14.0% 3.3% 42 29 17
107 AmerResGrp 600 9.7% 4.0% 40 31 20
Weighted Avg 6172 100% 1.25% 38.6 29.3 18.8
Recorded Vote 36.9 39.5 17.1
Difference -1.7 10.2 -1.7
Z-score -2.6 16.0 -2.7
Coincidence or ….
Analysts at the Election Defense Alliance (EDA) have confirmed that based on the official results on the New Hampshire Secretary of state web site, there is a remarkable relationship between Obama and Clinton votes, when you compare votes tabulated by op-scan v. votes tabulated by hand: The percentages are exactly opposite and match to within .0001%.
Optical Scan
Clinton 91,717 52.9507%
Obama 81,495 47.0493%
Total 173,212
Hand Counted
Clinton 20,889 47.0494%
Obama 23,509 52.9506%
Total 44,398
Optical Scan 173,212 79.60%
Hand counted 44,398 20.40
Total 217,610
Note that the percentages are mirror images and match
to within .0001% (six decimal places).
Given that Obama won 52.9506% of the hand-counted votes, what is the
probability Clinton would win 52.9506% of the 173,212 optical scan
votes?
These are real votes, not samples, so we can derive an
estimate of the probability of voting machine fraud without considering a
statistical margin of error. We KNOW exactly WHAT happened. We don't know WHY
or HOW. But we can calculate a fair estimate of the probability that the result
was just a coincidence or was due to a miscounting of the votes.
One might be tempted to say that the probability is 1 in 173,212 since there were exactly 173,212 joint optical scan ballots. But that would be unrealistic. We need to consider a plausible range of outcomes. Let's assume that Clinton could expect somewhere between 45%-55% of the 173,212 votes. That is a plausible 10% range of 17,321 possible outcomes, from 77,945-95,267.
Plausible range of outcomes
Maximum 55% 95,267
Minimum 45% 77,945
Difference 10% 17,321
Given the range of 17,321 possible outcomes, what is the
probability HRC would get exactly 91,717 votes due to chance alone?
The approximate probability that the anomaly was due to chance and just a
coincidence: 0.0058% = 1/17,321
The approximate probability that the votes were manipulated: 99.9942%
Now we will try a different approach: calculate the probability based on the exit poll discrepancy.
Given that Obama led the poll at 8pm by 39-36%, what was the probability that HRC would win the official vote by at least 3% (39-36%)?
Assume that the exit poll margin of error was 1.5%.
Exit Poll Final Vote
Obama 39% 36%
Clinton 36% 39%
Use the Excel normal distribution function to calculate the probability that the discrepancy was due to chance:
Probability = normdist (.39, .36, .015/1.96, true)
The probability that the 6% discrepancy was due to chance = 0.0044% (1 in 22,577)
The probability that the votes were miscounted = 99.9956%
Sensitivity Analysis
MoE |
Prob. 1 in |
1.0% |
485,887,839 |
1.1% |
22,127,320 |
1.3% |
327,843 |
1.5% |
22,577 |
1.7% |
3,686 |
1.9% |
1,015 |
2.1% |
391 |
2.3% |
189 |
2.5% |
107 |
2.7% |
68 |
2.9% |
47 |
3.1% |
35 |