## Massive 1988-2008 Exit Poll Discrepancies: A Probability Analysis

21 Jun

Massive 1988-2008 Exit Poll Discrepancies: A Probability Analysis

Richard Charnin
June 21, 2012
Update: July, 3, 2014

This post reviews probability calculation methods used to analyze the 1988-2008 unadjusted state and national exit polls.

The 1988-2008 Unadjusted State and National Exit Poll Spreadsheet Database consists of individual election worksheets. The tables contain unadjusted state exit poll samples, vote share, margin, votes cast and recorded, margin of error, win probability, electoral vote, region and the weighted aggregate of the unadjusted state exit poll shares. In addition, the workbook includes graphics, exit poll time lines, red-shift analysis, early vs. late voting statistics, net uncounted votes, 2006 and 2010 exit polls, etc.

The Democrats won the 1988-2008 unadjusted state exit polls by 52-42% and the recorded vote by just 48-46%, an 8% reduction in margin. In 274 state exit polls, there were approximately 375,000 respondents – a very large sample of six presidential elections. Like much other statistical analysis, the results are based on the Law of Large Numbers.
http://en.wikipedia.org/wiki/Law_of_large_numbers

Unadjusted exit polls tell us exactly how respondents said they voted. Exit polls are always adjusted to conform to the recorded vote count.The discrepancy is the difference between the exit poll and recorded vote margins.

US Count Votes did a comprehensive analysis of the 2004 exit poll discrepancies which disproved the exit pollster’s reluctant Bush responder hypothesis.

The True Vote Model estimates voter intent based on mortality, turnout and vote shares. It has confirmed the unadjusted exit polls to within 1%.

The polling margin of error (MoE) is a function of the number of respondents (n) and 2-party shares.
MoE = 1.3 * sqrt (p * (1-p)/n), where 1.3 is 30% cluster effect factor.

Example: Florida 2004 (2862 respondents)
Kerry led the exit poll by 50.8-48.0% (two-party 51.4-48.6)%.
But Bush won the recorded vote by 52.1-47.1%

The FL exit poll MoE = 2.38% = 1.3 * 1.96 * sqrt (.514*.486/2862)
Kerry’s 2-party vote (x) would be expected to fall within the following interval 95% of the time.
Mean – MoE < x < Mean+ MoE or
49% < x < 53.8%.

The probability that Kerry won FL is given by the Normal Distribution Function:
http://en.wikipedia.org/wiki/Normal_distribution

P = Normdist (Poll, .5, MoE/1.96, true)
P = 87.6% = Normdist (.514, .5, .0238/1.96, true)

Example: 2004 National Exit Poll

Kerry led the unadjusted poll (13660 respondents) by 51.7-47.0% (2-party: 52.3-47.7%). But Bush won the recorded vote by 50.7-48.3%.
The National exit poll margin of error was 1.1%.

Sample Kerry Bush Other
13,660 7,064 6,414 182
Share 51.7% 47.0% 1.3%

The probability P that Kerry won the popular vote is calculated as:
P = 99.97% = Normdist (.523, .5, .011/1.96, true)
Let’s be conservative and increase the margin of error to 2.0%.
P = 98.7% = Normdist (.523, .5, .02/1.96, true)

The National Exit Poll, with no change to the 13660 respondents) was forced to match the recorded vote. This is standard procedure for all exit polls. The NEP implied that 43% of the 2004 electorate (52.6 million) were returning Bush 2000 voters. But Bush only had 50.5 million recorded votes in 2000. Approximately 2.5 million died and another 1-2 million did not return in 2004. There could have no more than 46-47 million returning Bush voters. The 2004 NEP indicated an impossible 110% Bush 2000 voter turnout in 2004. The exit pollsters forced an impossible poll to match an impossible vote using impossible returning voter weights.

Example: 274 state presidential exit polls (1988-2008)
A total of 226 polls (82.4%) shifted from the poll to the vote in favor of the Republican. Only 48 shifted to the Democrat. Normally, as in coin-flipping, there should have been an equal shift to the Republican and the Democrat. What is the probability P of 226 polls red-shifting to the Republicans?

The Binomial distribution function:
http://en.wikipedia.org/wiki/Binomial_distribution

Unfortunately, the spreadsheet Binomial function cannot calculate the probability; the inputs are too large. We need to break the problem into four equal pieces: 56 of 68 exit polls red-shift with probability p.

p = Binomdist (56, 68, .5, false)
P = p*p*p*p (equivalent to P = Binomdist (224, 272, .5, false))
P = 3.7E-31
P = 1 in 2.7 million trillion trillion trillion

Note E-31 is scientific notation for 31 places to the right of the decimal point. For instance, E-3 represents .001 or 1/1000

Example: The MoE was exceeded in 135 of 274 state exit polls
Only 14 would normally be expected to since there is a 5% probability that the exit poll margin of error would be exceeded in an election. Of the 135 polls, 131 moved in favor of the Republicans (only 7 would be expected).

The Poisson function is used for analyzing a series of events (like in queuing systems) in which each event has a very low probability of occurrence.
http://en.wikipedia.org/wiki/Poisson_distribution

The probability P that 131 out of 274 would favor the Republican is:
P = E-116 = Poisson (131, .025*274, false)
The probability is ZERO. There are 106 places to the right of the decimal.
P = .0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 0000000000 000001
P = 1 in trillion trillion trillion trillion trillion trillion trillion trillion trillion.

For each presidential election, the following table summarizes a) the number of state elections which there was a Republican red-shift from the exit poll to the vote, b) the number (n) of states in which the margin of error was exceeded in favor of the Republican, c) the probability that n states would red-shift beyond the MoE to the Republican, d) the Democratic unadjusted aggregate state exit poll share, e) the Democratic recorded share, f) the differential between the exit poll and recorded vote.

The Ultimate Smoking Gun that proves Systemic Election Fraud:

Year Exit% Vote% Diff
1988 50.3 45.7 4.6 Dukakis may very well have won a close election.
1992 47.6 43.0 4.6 Clinton won in a landslide, much bigger than recorded.
1996 52.6 49.3 3.3 Clinton won in a landslide, much bigger than recorded.
2000 50.8 48.4 2.4 Gore won by 5-7 million True votes.
2004 51.1 48.3 2.8 Kerry won a 10 million True vote landslide.
2008 58.0 52.9 5.1 Obama won a 23 million True vote landslide.

Total 51.8 47.9 3.9

Read this excellent article by Bob Fitrakis in The Free Press.

Polling is not an exact science. But the 274 state exit polls and 6 national exit polls over the 1988-2008 presidential elections confirm beyond any doubt that the massive discrepancies must be due to election fraud.

The exit pollsters are funded by the National Election Pool. The pollsters always seem to get it “wrong” in the US, and then have to make impossible adjustments to get it “right”. Why do the exit pollsters always get it “wrong” in the US but get it “right” in the Ukraine and Republic of Georgia?

Any standard statistical/probability analysis applied to publicly available data results in only one reasonable conclusion: election fraud is pervasive and systemic.

Take the Election Fraud Quiz.

Election Model Forecast; Post-election True Vote Model

2004 (2-party vote shares)
Model: Kerry 51.8%, 337 EV (snapshot)
State exit poll aggregate: 51.7%, 337 EV
Recorded Vote: 48.3%, 255 EV
True Vote Model: 53.6%, 364 EV

2008
Model: Obama 53.1%, 365.3 EV (simulation mean);
Recorded: 52.9%, 365 EV
State exit poll aggregate: 58.0%, 420 EV
True Vote Model: 58.0%, 420 EV

2012 (2-party state exit poll aggregate shares)
Model: Obama 51.6%, 332 EV (Snapshot)
Recorded : 51.6%, 332 EV
True Vote 55.2%, 380 EV

Posted by on June 21, 2012 in True Vote Models, Uncategorized

### 6 responses to “Massive 1988-2008 Exit Poll Discrepancies: A Probability Analysis”

1. June 22, 2012 at 6:27 pm

I’d like to see the stats on paper ballot elections and/or districts (as a control).

Richard Charnin's Blog

JFK Conspiracy and Systemic Election Fraud Analysis