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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 08:35 AM
Original message
A Mathematical proof: Exit polls are accurate
Edited on Thu Mar-10-05 09:01 AM by TruthIsAll
How do we know exit polls are accurate?

Let's do simple math.
It's simple algebra- actually arithmetic

Let's just assume the loss of democratic votes was due to natural spoilage. This is DOCUMETED fact in EVERY election - millions of votes are lost due to punch card machine anomalies.

Let's disregard fraud for now.
KISS.

So we have the Exit poll identity:
Intended vote = recorded vote + lost vote (spoilage).
The equation: I= R + S

Now the Exit Poll is reflection of the INTENDED VOTE, NOT the RECORDED VOTE tally.

Therefore, the Exit poll can NEVER equal the recorded vote and must therefore be greater than the recorded vote. And that is why the Exit Poll is closer to the truth and favors the Democrats.

Let's now expand the analysis to assume a fraud factor.

Our equation becomes:
Intended Vote = Votes Recorded + Lost (spoilage) + Stolen (fraud).
or I = R + S + F

That brings the exit poll all the way home.

Exit Poll Deviation = I - R, where R = S + F

Notice that it's a Deviation, NOT an Error. The Recorded vote is in error - and is corrected to the Intended vote by adding back the spoiled votes and the stolen votes.

But Mitofsky does just the opposite. He adjusts the correct preliminary exit poll weights in deriving the final exit poll to match the recorded vote, which of course we have proven is in error to begin with.

Thus he surgically removes the natural spoilage and fraud factors from the equation as if they never existed.

Like cleaning up the blood from the crime scene.


Q.E.D.

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acmejack Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 08:44 AM
Response to Original message
1. You overlook one salient point...
Exit polls are only accurate in foreign countries. They are able to identify fraud off shore, but heck we got computers, so just relax...
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Marnieworld Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 08:44 AM
Response to Original message
2. Thank you Truth is All
I read your posts all of the time. Thank you so much for all of your tireless efforts. What has happened to elections in this country is insane and hard to believe. It's so frustrating knowing the truth is so different from what is believed out there. Your posts always reassure me that I'm not crazy-there was a stolen election-facts, and especially for you, MATH does not lie. The world is round dammit and the earth revolves around the sun. The sky is blue and Bush's "re-election" is based on theft and fraud. You prove it everyday and you have no idea the impact of your work. Thank you so much.


:toast:

Marnieworld
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Mnemosyne Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 09:13 AM
Response to Reply #2
3. I'll second that respectfully. nt
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eomer Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 09:24 AM
Response to Original message
4. Great post, as usual...
but there is a booboo in this line:
Exit Poll Deviation = I - R, where R = S + F

Maybe you meant to say:
Exit Poll Deviation = I - R, where I - R = S + F

If you can edit to fix it then I can edit to delete this post.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 12:19 PM
Response to Reply #4
7. I saw and fixed it just before reading your post. Thanks.
Edited on Thu Mar-10-05 12:24 PM by TruthIsAll
I'm at work and just noticed it - 3 hours later.

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slackmaster Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 09:40 AM
Response to Original message
5. One small issue with this analysis
Edited on Thu Mar-10-05 09:41 AM by slackmaster
You have assumed correctly that some votes that should get recorded always get lost one way or another.

But you have assumed a 0% loss rate for exit polling data; that the data collection process in exit polls is consistently flawless and free of bias. And of course you continue to assume that the weighting process applied to raw exit poll data is always done correctly and also free of bias.

A vote is not the same as a poll, but if you want to view it as one in which a 100% sample set is used, your assertions of systematic fraudulent manipulation of the recorded vote can be viewed as a form of researcher bias. You haven't accounted for researcher bias on the part of pollsters.

Of course the biggest problem of all is that exit poll results have no official value. Only the recorded vote counts.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 12:47 PM
Response to Reply #5
8. You are correct. There is always a small sampling error factor.
Edited on Thu Mar-10-05 12:53 PM by TruthIsAll
We can modify the equation:
I = R + S + F + e, where e = the true polling error, which is a non-biased positive or negative term. It applies equally to Bush or Kerry, so we can assume it is zero. We know that e (really the MoE) is very small for exit polls.

Exit Poll deviation from the Recorded Vote (based on Democratic loss) has averaged near 3% in the last 5 elections. Naturally spoiled votes (a declining factor as mechanical punched card machines are replaced by automated equipment) have accounted for approximately 1-2% of the total votes cast. The Fraud factor (BBV/optiscan) must therefore account for the remaining deviation.

THERE IS NO PAPER TRAIL TO MEASURE THE DEVIATIONS DUE TO PROGRAMMED TOUCHSCREENS WHICH DEFAULT TO BUSH - AS WELL AS THE CENTRAL COMPUTERS WHICH SUM THE VOTES.

And we must not forget:
THE EXIT POLLS DO NOT INCLUDE THE LARGE NUMBER OF VOTERS WHO FOR VARIOUS REASONS (DISENFRANCHISEMENT, LONG LINES, ETC.) NEVER GET TO VOTE AND ARE THEREFORE NEVER COUNTED IN THE POLL.

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slackmaster Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 01:00 PM
Response to Reply #8
9. I would use caution in assuming the true polling error is non-biased
Every person has biases, therefore every process potentially reflects researcher bias.

Carry on with your good work, TIA.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 12:12 PM
Response to Original message
6.  Correction: Deviation = I- R = S+ F
I never get it exactly right the first time.

Deviation = Intended - Recorded = Spoiled + Fraud
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andym Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 05:08 PM
Response to Original message
10. Not a proof, but raises a legitimate point
Edited on Thu Mar-10-05 05:28 PM by andym
This is not a proof of exit polls accuracy.

You correctly define I as intended vote. Your initial equation is reasonable.
I = R + S + F

although I would add M for mistaken votes, where a voter might have unintentionally voted for the wrong candidate as happened in FL in 2000.
I (candidate) = R + S + F + M

But Exit polling data are not equivalent to intended votes, that is, I is not equal to exit polling data. However, what is true is that exit polling data attempts to measure I. Therefore in theory, exit polling data may be a more accurate estimation of the intention of the voters than the actual results . I think this is a valid point.

But this does not cover potential problems with exit-polls. That is the statistical model underlying the exit poll and the sampling methodology have to be accurate or exit polls will contain sampling errors (different from the MOE) that may make the exit-polls less accurate than the actual vote in determining the value of I.

If you read the NEP paper and the rebuttal by the group of academics in http://uscountvotes.org/ucvAnalysis/US/USCountVotes_Re_Mitofsky-Edison.pdf
you will see that the statistical model used by Mitofsky/NEP only broke down in the "within-precinct" error (WPE). Mitofsky claimed that this was likely due to the Reluctant Bush Responder. But the uscountvotes paper uses Mitofsky's own data to show that this hypothesis is unlikely.

Now this opens the door, but does not prove the possibility that I was significantly different from R due to the variables F (fraud), M (mistaken votes) and S (spoilage). The reason that it does not prove it, is that there are other possible systematic errors in the exit-polling that have to ruled out (everything from possible poor design of the questionnaires to effects of the actual poll-takers on the people filling out the form to inability to correctly obtain data for absentee voters). However, this does not detract from the fact that the door is open to investigate F, M and S and that some combination of these may have actually caused R to deviate significantly from I to cause the wrong man to have been elected President.

Now, the best possible followup work would be to identify precincts statistically where I may have deviated significantly from R and then conduct carefully scrutinized full hand recounts there.







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kster Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 06:15 PM
Response to Original message
11. Kick
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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Thu Mar-10-05 08:38 PM
Response to Original message
12. Your equation is a good idea. We could tweak it further.

We should add "P"(Uncounted Provisional ballots)to the equation.

In addition to Fraud, we should add "M" (Mis-punched/cast to the wrong candidate but not by fraud), if we are talking about the Intended vote for a Democratic candidate. This happens most commonly in heavily Democratic precincts.


Intended vote for a DEMOCRATIC Candidate = Votes Recorded + Lost (spoilage) + Stolen (fraud)+ Uncounted Provisionals + Mis-cast (not by fraud)
I= R + S + F + P + M


However, if we are talking about the TOTAL intended vote - not just the intended vote for a Democratic candidate - Fraud should not necessarily be added to the Recorded vote. Fraud includes, deliberately switching votes from one candidate to the other and can also involve adding phantom votes to the total count. The net result might be that Fraud should be SUBTRACTED from the TOTAL Recorded votes. Or, better yet, NET fraud should added,

And, of course, if we are talking about the TOTAL intended vote, "M" would not be a factor.

TOTAL Intended vote = Votes Recorded + Lost (spoilage) + NET fraud + Uncounted Provisionals
I= R + S + NF + P


(Note: I first posted this response on the "Check. Mate." thread.)








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davidgmills Donating Member (651 posts) Send PM | Profile | Ignore Thu Mar-10-05 08:52 PM
Response to Reply #12
13. Pretty soon we will be using all the letters in the alphabet
to account for all the hanky panky.

How about a "B" for the Blackwell factor and a "J" for the Jeb factor.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 08:53 PM
Response to Reply #12
14. Good tweak to enhance the linear model.
Edited on Thu Mar-10-05 09:21 PM by TruthIsAll
That was what I had in mind.

Something simple.
Something factual.

It's like the accounting identity:
Total Assets = Total Liabilities + Equity

Total Assets = Intended Democratic Vote
Total Liabilities = Net Democratic Votes Stolen + Spoiled

Total Equity = Intended Democratic Vote - (Net Votes Stolen+Spoiled)

The Dems are always losing equity in the intended vote.
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kster Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 09:40 PM
Response to Original message
15. PEANUTS !! GET YOUR PEANUTS HERE !!
I'm sorry for interrupting this important 2004 election result and discussion thread, it was on top so I thought it was a 2 day long "Bev Harris,John Kerry was mean to me thread". Kerry on
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 10:16 PM
Response to Original message
16. This is so obviously wrong its funny
Edited on Thu Mar-10-05 10:45 PM by Nederland
This equation:

I = R + S + F

assumes that R and F are separate numbers--they are not. Any successful fraud that occurs will show up as part of the recorded vote totals.

Think of it this way. Why are you adding the fraudulent votes to the other two variables in order to determine the intended vote? Since when are fraudulent votes part of the intended vote? Shouldn't you be subtracting the fraudulent votes from the recorded vote in order to determine the intended vote?

Also, your equations neglect the minor fact that there is more than one candidate. :eyes: The reality is that intended vote needs to be calculated separately for each candidate thusly:

I Kerry = (R Kerry - F Kerry) + S Kerry

I Bush = (R Bush - F Bush) + S Bush

In these equations in order to get the intended vote number we take the recorded vote for a candidate, subtract the votes that were fraudulently recorded for that candidate, and then add in the votes that were supposed to be for that candidate but spoiled.

When you do the equations correctly, it quickly becomes obvious that they prove nothing. They prove nothing because you cannot know the F and S numbers for either candidate. All you really know is R (the recorded vote).
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 10:47 PM
Response to Reply #16
17. Nederland, not quite. You need some help in arithmetic.
Edited on Thu Mar-10-05 11:07 PM by TruthIsAll
Intended Dem Vote = Recorded Dem votes + Dem votes lost to Spoilage + Dem votes lost to Fraud.

I = R + S + F

We know I. Check the Preliminary (not the Final!) Exit Poll.
We know R. It's given.
We know S. It's documented.

So we can calculate F = I - R - S

Like figuring your taxes.


Nice try, though.
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 12:21 AM
Response to Reply #17
22. Question
I claim you don't know what 'I' is.

R is around 120 million. That much we agree on.

Show me how you think you can calculate 'I' and I'll explain to you why you are wrong.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 08:12 AM
Response to Reply #22
25. Use preliminary exit poll and recorded vote percentages...
Edited on Fri Mar-11-05 08:14 AM by TruthIsAll
I = prelim. Kerry exit poll = 50.7%
S= NET spoiled votes lost to Kerry = 1.0%
R= recorded vote = final exit poll = 48.2%
F = votes lost to fraud

I= S + R + F
so
F= I - S - R = 50.7 -1.0 - 48.2 = 1.5%

For Bush, votes lost (gained) by fraud = 1.8%
R = I + S + F
R = 50.7
S = 1.0 (Net spoiled votes lost to Kerry = gained by Bush)
I = 47.9 (from prelim. exit poll)

F= R - I - S
F= 50.7 - 47.9 - 1.0 = 1.8%

So which is it? Is F = 1.5% or 1.8% ?
Depends on which deviation from the final exit poll you choose - Kerry or Bush.
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 08:55 AM
Response to Reply #25
26. Now I've got you
The title of your thread is "A Mathematical proof: Exit polls are accurate"

Please note that in the above post, you are using exit poll data and assuming that the data is accurate. Therefore you are assuming the very thing that you are trying to prove. Hardly logical, wouldn't you say?


QED
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 09:20 AM
Response to Reply #26
28. THE FACTS SAY OTHERWISE.
Edited on Fri Mar-11-05 09:59 AM by TruthIsAll
It is obvious that the PRELIMINARY Exit poll is MUCH MORE accurate than the FINAL Exit poll which MATCHES THE RECORDED, FRAUDULENT VOTE. I HAVE PROVED IT.

OF COURSE, THE PRELIMINARY EXIT POLL IS NOT 100% ACCURATE. NO POLL IS. YOU ARE BUILDING A STRAWMAN BY PLAYING ON WORDS..

TWO FACTS:
1) We KNOW that spoiled votes ARE NOT counted in the recorded vote (or in the FINAL exit poll, since it is matched to the recorded vote)

2) We KNOW that fraudulent votes (switched touchscreen votes + mistabulations) ARE included in the recorded vote (and in the FINAL exit poll).

THEREFORE THE FINAL EXIT POLL AND THE RECORDED VOTE MUST BOTH BE INCORRECT BY THE AMOUNTS IN (1) AND (2) BY DEFINITION.

Finally, we KNOW that the INTENDED vote (captured by the PRELIMINARY exit poll) DOES NOT INCLUDE THE ERRORS IN (1) AND (2) ABOVE.

CONCLUSION:
THE PRELIMINARY EXIT POLL = INTENDED VOTE IS AN ACCURATE REFLECTION OF HOW THE VOTERS REALLY VOTED.

YOU CAN PLAY ON THE WORD "ACCURATE" ALL YOU WANT.
BUT YOU CANNOT DENY THE CONCLUSION.

ONE + ONE STILL EQUALS TWO IN MY WORLD, IF NOT IN YOURS.
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 10:33 AM
Response to Reply #28
30. You are in denial
You are assuming the very thing you are claiming to have "mathematical" proof for. That is illogical and no number of sentences written in all caps can change that. Give it up TIA, this post was a lost cause from the very beginning.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 10:38 AM
Response to Reply #30
31. A LOST CAUSE ONLY TO YOU.
.
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 10:42 AM
Response to Reply #31
32. LOL
Edited on Fri Mar-11-05 10:44 AM by Nederland
This is funny. I'm honestly not sure if you simply don't get it or if you are just too proud to admit you screwed up. Hopefully for you it is the latter. I'd rather be proud than stupid.
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TruthIsAll Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 11:50 AM
Response to Reply #30
33. TRY THIS ONE FOR SIZE...
Edited on Fri Mar-11-05 12:01 PM by TruthIsAll
FINAL EXIT POLL (DEM) = RECORDED VOTE:
RECORDED VOTE = INTENDED VOTE - NET VOTES LOST TO SPOILAGE - NET VOTES LOST TO FRAUD:
R = I - S - F

PRELIMINARY EXIT POLL (DEM) = INTENDED VOTE:
INTENDED VOTE = RECORDED VOTE + NET VOTES LOST TO SPOILAGE + NET VOTES LOST TO FRAUD

I = R + S + F

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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 04:21 PM
Response to Reply #33
34. Same problem
I = R + S + F

R is known
S is known
I is unknown
F is unknown

You are therefore left with an unsolvable equation because you need (at least) two equations to resolve two unknowns.



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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Thu Mar-10-05 11:04 PM
Response to Reply #16
18. From your expanded message, found when hitting "reply"...
"I Kerry = (R Kerry - F Kerry) + S Kerry
I Bush = (R Bush - F Bush) + S Bush"


I don' think so!


The INTENDED vote for Kerry should have the Fraud (i.e., votes switched to Bush) SUBTRACTED FROM IT in order to calculate the Recorded vote for Kerry. Thus, your equation should be

I Kerry - F Kerry = R Kerry + S Kerry

thus I Kerry = (R Kerry + S Kerry) + F Kerry


However, I do agree with your statement that Fraud (i.e., votes switched from one candidate to another) should not be included in the calculation of the single combined total for all candidates.
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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Thu Mar-10-05 11:17 PM
Response to Reply #18
19. Biased statement
You are assuming at the outset that fraud occurred in Bush's favor. Since that is precisely what you are trying to prove, you cannot simply assume it.
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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Thu Mar-10-05 11:29 PM
Response to Reply #19
20. I 'm not trying to PROVE that fraud occurred. However, here is a neutral-
statement of the candidates' equations:

"F Kerry" is fraud involving deliberate switching of Kerry votes to other candidates.

"F Bush" is fraud involving deliberate switching of Bush votes to other candidates.


I Kerry - F Kerry = R Kerry + S Kerry

thus I Kerry = (R Kerry + S Kerry) + F Kerry


I Bush - F Bush = R Bush + S Bush

thus I Bush = (R Bush + S Bush) + F Bush


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Nederland Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 12:08 AM
Response to Reply #20
21. Question
If you aren't trying to prove fraud occurred, what are you trying to prove?
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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Fri Mar-11-05 05:57 AM
Response to Reply #21
23. Yes, ultimately, we do hope to prove that fraud occurred. However...
the equations can not prove fraud. The equations are merely a description of the relationship of the Intended vote to the Recorded vote.

Personally, I find an expanded version of the equations - including (M)is-cast votes and uncounted (P)rovisional votes - useful at the precinct level. The equations illustrate the ideal precincts to check for fraud: exit poll precincts with a significant disparity between exit poll data (which reflects intended vote) and recorded vote, with minimal spoilage, minimal uncounted provisionals, and minimal mis-cast votes.

The actual PROOF of fraud will require full hand recounts of some of those precincts.
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eomer Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 07:16 AM
Response to Reply #23
24. F can be a negative number.
In the equation:
I = R + S + F
F could be a negative number in some particular case.

If a candidate's Recorded count is higher than the Intended vote due to fraud (as we claim is the case for Bush in 2004), then F would be a negative number for that candidate. It would contain the net gain in votes due to fraud but expressed as a negative number.

So F should be defined as the net loss in votes by a candidate due to fraud. When F is negative then you can interpret its absolute value as the net gain by a candidate due to fraud.

If the 2004 election was stolen by flipping votes from Kerry to Bush then F for Kerry is positive and F for Bush is negative.

A vote that was flipped from Kerry to Bush shows up as an addition to F for Kerry and a subtraction from F for Bush.

An intended Kerry vote that was thrown away into the ether by way of fraud shows up as an addition to F for Kerry. In this case, F for Bush is not affected.

A vote that appeared out of the ether and was fraudently credited to Bush shows up as a subtraction from F for Bush. In this case, F for Kerry is not affected.

While we're at it, S should include any innocent mistake that records a vote that wasn't intended, such as a chad that spontaneously pops out when the voter intended to not vote for any candidate. An innocent recording of a vote that wasn't intended would be reflected as a subtraction from S.

So S should be defined as the net loss in votes by a candidate due to innocent mistakes. When S is negative (doesn't usually happen) then you can interpret its absolute value as the net gain by a candidate due to innocent mistakes.

Sorry that this post is on the pedantic side but the last few posts seem to be touching on this point but not quite coming out and saying it.

One final comment for those who don't agree that this is a proof of anything. What it does for sure is reframe the issue. Mitofsky is attempting to talk real fast and gloss over the issue. He pretends that the fraud component doesn't exist. Once we get the correct equation out in the open, then we can have a logical approach to the question instead of a fast talking, nothing to see here kind of approach.

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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Fri Mar-11-05 08:58 AM
Response to Reply #24
27. You are right. Using nets is important....

and both "F" and "S" can be negative.

(Since you are including "M"is-cast votes in "S": At a precinct level, "S" could easily be negative, especially for a Republican candidate at a heavily Democratic polling place with multiple precincts using different punched-card machine configurations.)
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eomer Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 09:46 AM
Response to Reply #27
29. Agreed
(Since you are including "M"is-cast votes in "S": At a precinct level, "S" could easily be negative, especially for a Republican candidate at a heavily Democratic polling place with multiple precincts using different punched-card machine configurations.)


Agreed, so long as the mixup was innocent. If the card was intentionally shuffled over to a pile for a different precinct (or some other intentional fraud) then we subtract one from F instead of from S.

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kiwi_expat Donating Member (526 posts) Send PM | Profile | Ignore Fri Mar-11-05 09:09 PM
Response to Reply #24
35. Does Mitofsky pretend that the spoiled- ballot component doesn't exist?
Edited on Fri Mar-11-05 09:18 PM by kiwi_expat
Or that the uncounted-provisional-ballot component doesn't exist?

If so, wouldn't it be useful to show those as separate variables in the equation - in addition to an "other innocent mistakes" variable? And, if you wish, the fraud variable could be labeled "other deliberate mistakes", to highlight that some of the spoiled-ballot and uncounted-provisional-ballot component is due to fraud.

However if you prefer aggregate variables, why not just:
I = R + (M)istakes


(I am embarrassed to admit that I am unable to download NEP's pdf report.)
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FreepFryer Donating Member (1000+ posts) Send PM | Profile | Ignore Fri Mar-11-05 09:35 PM
Response to Original message
36. Thanks for another clear and understandable post.
Ignore personal attacks and concentrate on facts.
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