"Same problem"
Posted by Nederland
http://www.democraticunderground.com/discuss/duboard.php?az=view_all&address=203x341540#342280I = 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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Nederland, thanks for your post regarding the equation:
I = R + S + F.
where:
I = intended vote
R = recorded vote
S = spoiled vote factor
F = fraud factor
You have just proven that:
1) the Recorded vote is wrong
and
2) the Final exit poll which matches to the Recorded vote is wrong.
Here is the proof:
Since, you state that R and S are known,
then R > 0 and S > 0,
and we must have
I > R and I - R > 0
Therefore,
the Intended vote (I) is greater than the Recorded vote (R).
It logically follows that the Recorded vote must be bogus, as it does not match the Intent of ALL of the voters.
And you didn't need to solve for the two unknowns to see that.
Talinkg it a step further, Mitofsky's FINAL exit poll must also be bogus, since he adjusts the Preliminary Exit poll weights in order to match the (incorrect) Recorded vote.
Nederland, you are half way there.
Thanks for proving my point.
The final unknown is the Fraud. If the preliminary exit poll was essentially correct, then we can approximate F, although we would still need to add another unknown the Disenfranchisement (D) factor, which is not picked up in the preliminary or final exit poll.
Now, if you will accept that programmed fraud did occur when the touchscreens switched Kerry votes to Bush (in 99% of these documented "glitches"), you will finally reach the promised land of the TRUTH.