Strange polls: Jill Stein at 1% and just 14% of respondents are Independents?

07 Aug

Richard Charnin
August 7, 2016

Richard Charnin

Matrix of Deceit: Forcing Pre-election and Exit Polls to Match Fraudulent Vote Counts
Proving Election Fraud: Phantom Voters, Uncounted Votes and the National Poll
Democratic Primary spread sheet

Strange polls: Jill Stein at 1% and just 14% of respondents are Independents?

According to the Ipsos/Reuters poll,  only 14% of respondents were Independents and Jill Stein had just 2% of Independents. These results are implausible.

The latest Gallup Party-ID survey indicates that 42% are Independents, 28% Democrats and 28% Republicans. The 2-party shares:  60% Independents, 40% Democrats.

Are we expected to believe that all of Sanders’ primary voters have gone to Clinton and Trump?

Ipsos Pct Stein Clinton Trump Johnson
Ind 14% 2% 46% 46% 6%
Dem 47% 1% 81% 18% 0%
Rep 39% 1% 5% 80% 14%
Total 100% 1.14% 46.31% 46.22% 6.33%

If Stein matched Sanders’  primary shares of Independents and Democrats, she could win a fair election.

Party-ID Gallup Survey Stein (est) Clinton (est) Trump (est) Johnson (est)
Ind 42% 45%  30% 10%  15%
Dem 29%  40% 50%  5% 5%
Rep 29% 5% 5% 80% 10%
Total 100% 31.95% 28.55% 28.85% 10.65%
Votes 129,106 41,249 36,860 37,247 13,750
Elect Vote 538 308 3 227 0

In the primaries (25 exit and 2 entrance polls) Bernie Sanders had  65% of Independents, but just 45.3% of the total vote. 

The 42I-28D-28R Gallup Party-ID survey equates to  60I-40D in the primaries. Using this split for the 27 adjusted exit polls, Clinton needed 83.4% of Democrats to match the recorded vote. The adjusted polls indicate that Sanders had 64.6% of Independents.

This is highly anomalous.


Exit Poll States Gallup Pct Sanders Clinton
IND 60.0% 64.6% 35.4%
Dem 40.0% 16.6% 83.4%
Recorded Match  100.0% 45.3% 54.7%
Recorded Vote 45.3% 54.7%

If  Sanders had 37% of Democrats, he would have had a total 53.6% share.

Exit Poll States Gallup Pct Sanders Clinton
IND 60.0% 64.6% 35.4%
Dem 40.0% 37.0% 63.0%
Est. True Vote 100.0% 53.6% 46.4%
Recorded 45.3% 54.7%

Jill Stein Polling Sensitivity analysis

Assuming Independents are 40% of the electorate, then for Jill Stein to have
5%(implausible), she needs 12% of Independents and 0% of Democrats and Republicans.
10%(conservative), she needs 17% of Independents and 5% of Democrats and Republicans.
20%(plausible), she needs 35% of Independents and 10% of Democrats and Republicans.
30%(optimistic), she needs 52% of Independents and 15% of Democrats and Republicans.


Sanders had  52% of Independents in the 11 RED states. Clinton needed an IMPLAUSIBLE 97% of Democrats to match the recorded vote.

Sanders had  an estimated 65% of Independents in the 40 BLUE/OTHER states. If he had 30% of Democrats, he would have had 51%.

 RED STATES Pct Sanders Clinton
IND 58.6% 52.0% 48.0%
 Req. to Match Dem 41.4% 3.0% 97.0%
Calc Match 100.0% 31.7% 68.3%
Recorded 31.7% 68.3%
IND 60.0% 65.0% 35.0%
Dem 40.0% 30.3% 69.8%
Calc Match 100.0% 51.1% 48.9%
Total Vote 51.1% 48.9%
 RED STATES 2-party Recorded 160
IND IND Sanders EV
AL 37.6% 57.6% 19.8% 9
AR 39.6% 57.5% 31.0% 6
FL 44.5% 59.3% 34.1% 29
GA 38.7% 55.7% 28.3% 16
LA 58.9% 73.4% 24.6% 8
MS 37.4% 55.5% 16.6% 6
NC 45.5% 58.0% 42.8% 15
SC 38.0% 55.2% 26.1% 9
TN 39.0% 58.5% 32.9% 11
TX 41.7% 58.8% 33.7% 38
VA 38.6% 55.0% 35.4% 13
avg 41.8% 58.6% 29.6%
Weighted Avg 42.0% 58.5% 31.7%

Posted by on August 7, 2016 in 2016 election, Uncategorized


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8 responses to “Strange polls: Jill Stein at 1% and just 14% of respondents are Independents?

  1. CarlAntoine

    August 7, 2016 at 12:40 pm

    Reblogged this on carlantoine and commented:
    #BernieSanders #FeelTheBern #JillStein #JillNotHill {#Clinton #Trump} #MSMbias #DemExit #GreenParty Polls bias

  2. George

    August 8, 2016 at 11:53 am

    On page 15 of the ‘Ipsos Poll conducted by Reuters’ re ‘All Adult Americans Identity’, Independents are given as 12%. I assume that the answerers gave this information, but cannot find out any more details about how the pollsters came up with the figure.

    Download at:

    • George

      August 8, 2016 at 12:04 pm

      Correction; my message should’ve read ‘Ipsos Poll conducted for (not ‘by’) Reuters’.

  3. sara

    August 10, 2016 at 5:37 pm

    don’t worry they can’t show how well Jill Stein is doing. They don’t want people to suspect anything when they rig this election too.

  4. Jim Lane

    August 12, 2016 at 12:10 am

    A more recent Ipsos/Reuters poll — — puts Stein at 3%. This is in line with what all the other pollsters are reporting (the Wikipedia summary at,_2016#Four-way_race is useful). It looks like Stein really is at about 3%, with reported values being scattered around that number. The alternative is to believe that every major pollster is part of a conspiracy to report grossly falsified results.

    • Richard Charnin

      August 12, 2016 at 12:54 am

      What was the percentage of respondents identified as Independents?
      Anywhere near 42%?

      • Jim Lane

        August 12, 2016 at 2:30 am

        I was trying to understand the 14% independents in the Ipsos/Reuters poll you discuss. OK, I didn’t try VERY hard, but I took a quick cruise through the site and didn’t notice an explanation of that aspect of the methodology.

        Unconstrained by bothersome facts, therefore, I indulged in some creative speculation. There are voters who self-identify as Democrats or Republicans or members of some other party. You can just call everyone else an independent. Among those so-called “independents”, however, many can be categorized as Democratic-leaning or Republican-leaning. Note that the Gallup poll you cite did ask that follow-up question of self-described independents. The result is that some of the “independents” confessed to a lean. Specifically, the 42% figure for independents comes from Gallup’s July 13-17 poll, but that poll also reveals this: Democrats plus leaners plus Republicans plus leaners came to 86%, leaving, ta-da, precisely14% for the independents. On that basis, I’ll hypothesize that the Ipsos/Reuters poll was using the narrower definition of “independent”, i.e., excluding each party’s leaners. I grumpily note in passing that it would be nice if the website said so, so that we knew for sure.

        Speaking of independents, you write: “The 42I-28D-28D Gallup Party-ID survey equates to  60I-40D in the primaries.” Obviously, you meant 42I-28D-28R, but my substantive question is about 60I-40D. This seems to imply that 60% of those voting in the Democratic primaries would be independents. If my reading is correct, though, that assumes (a) that independents were allowed to vote in all the primaries (not true, because several states have closed primaries), and (b) that in states with open primaries, every single one of the independents chose a Democratic ballot rather than a Republican ballot (extremely unlikely, given the media attention on Trump that would motivate people to vote for or against him). Or am I misunderstanding how you arrived at 60I-40D?

  5. Richard Charnin

    August 12, 2016 at 12:59 pm

    Jim, Yes, you are right. Many primaries were closed. But I am interested in calculating how Bernie would have done if all voted. The 60I/40D split is what one would expect in an open election. That to me is the definition of the True Vote.

    And don’t forget the rampant stripping of voter rolls and flipping of votes.

    Yes. there is a conspiracy of the MSM to have pre-election and exit pollsters match their projections/analysis to their desired recorded votes. The National Election Pool through Edison Research always adjusts pristine exit polls to match bogus recorded votes. It is absolute mathematical proof beyond any doubt.

    Just like there is a 100% probability that the MSM does not want Jill Stein to participate in the debates.


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