For the sake of condensing the notation, we are going to assign the four letters, A,B,C and D to
each precinct’s Early and Mail-in Vote totals for Berkbigler and Hill.
Let A = Berkbigler’s Early Vote at a precinct.
Let B = Hill’s Early Vote at the same precinct.
Let C = Berkbigler’s Mail Vote at the same precinct.
Let D = Hill’s Mail Vote at the same precinct.
Let K=A+B+C+D, which is the sum of all four above votes.
Let G =A/(A+D); g = , which is the percentage of votes that belong to Berkbigler A
A+D
amongst the sum of Berkbgiler’s Early Vote and Hill’s Mail-in Vote at the same precinct.
Let H=C/(C+B); h = , which is the percentage of votes that belong to Berkbigler amongst C
C+B
the sum of Berkbgiler’s Mail Vote and Hill’s Early Vote at the same precinct.
Let Alpha=(A+C)/(A+B+C+D); α = , which is the percentage of all voters that voted A+C
A+B+C+D
for Berkbigler Early or by Mail.
Let Lambda=(A+D)/(A+B+C+D), λ = , which is the percentage of all voters that A+D
A+B+C+D
either voted for Berkbigler Early or for Hill by Mail. Observe that (1 − λ) =
C+B
A+B+C+D
There is a universal tautology concerning those four numbers, A,B,C and D, and those four rations, G,H, Alpha and Lambda. This tautology says:
Alpha=G(Lambda)+(1-Lambda)H; α = gλ + (1 − λ)h
Proof: Q.E.D A+C
A+B+C+D =
A
( A+D )
A+D
( A+B+C+D ) +
C+B
( A+B+C+D )
C
( C+B ) =
A+C
A+B+C+D
So why is this tautology α = gλ + (1 − λ)h important? Because it tells us that we cannot
solve for α , which is Berkbigler’s total share of the vote, knowing only g and h . In a fair
election, we need to know all three variables g, h AND λ in order to solve for α .
However, in Washoe County, we can solve for α with no knowledge of λ at any precinct, using only g and h and the same formula with an R
2 = 0. 994
α = 0. 059785 + 0. 422213h + 1. 535061g
2 − 1. 211691g
3
Suppose you are blindfolded. You don’t know A,B,C or D in a precinct, but I do. However, I provide to you the g and h percentage and the total sum of ballots cast, K.
So, if I tell you that g = 30%, this means that A and (A + D) are in a 3 to 10 ratio. There is no way for you to resolve the individual value of A or D from this information. It could be , or
3
10
or , etc. 21
70
300
1000
Knowledge of g does not impart knowledge of A, D or (A + D), therefore you remain
blindfolded even after I tell you g.
Now I tell you h = 54%, this means that C and (C + B) are in a 54:100 ratio. Again, there is now way to determine the individual values of C, B or (C + B) from this information.
And without Lambda, you cannot know α, which is Berkbigler’s total percentage of the ballots.
Allow me to give you an example.
In Precinct One:
g = , from which we know , since .
A
A+D = 30% =
30
100 D = 70 100 − 30 = 70
h = , from which we know , since .
C
C+B = 54% =
54
100 B = 46 100 − 54 = 46
λ = .
A+D
A+B+C+D = 50% =
30+70
200
α =
A+C
A+B+C+D = 42% =
30+54
200 = gλ + (1 − λ)h = (30%)(50%) + (50%)(54%)
In Precinct Two:
g = , from which we know , since .
A
A+D = 30% =
300
1000 D = 700 1000 − 309 = 70
h = , from which we know 184, since .
C
C+B = 54% =
216
400
400 − 216 = 184
λ = .
A+D
A+B+C+D = 71. 4285% =
1000
1400
α =
A+C
A+B+C+D = 36. 857142% =
516
1400 = gλ + (1 − λ)h = (30%)(71. 42%) + (28. 57%)(54%)
Notice that in both precincts, g = 30% and h = 54%; however both precincts have a different value for α. In Precinct One α = 42% and Precinct Two α = 36. 85%. Hence, you cannot solve for α knowing only g and h.
Thus, the fact that we can solve for α, without λ, knowing only g and h, in every precinct, with the equation…
α = 0. 059785 + 0. 422213h + 1. 535061g ,
2 − 1. 211691g
3
…means that the election is rigged by definition, since it violates the universal tautology of α = gλ + (1 − λ)h = , which says that A+C
A+B+C+D =
A
( A+D )
A+D
( A+B+C+D ) +
C+B
( A+B+C+D )
C
( C+B )
the fractions and alone ( and alone), cannot solve for .
A
( A+D )
C
( C+B ) g h α
Q.E.D.
Here is an example using Precinct Reno-Verdi 1033. Remember that this is blindfold. I have all the information in the table below; however, I will only provide g, h and K, where K is the
total ballots cast.
g = and K=180+164+150+382=876 180
180+382 = 0. 32028; h =
150
150+164 = 0. 47770
λ =
180+382
876 = 0. 64155
R_0 Pname R A B C D
P# Precinct Registered
Berkbigler
Early Hill Early
Berkbigler
MiV Hill MiV
30
RENO-VERDI
1033 1085 180 164 150 382
Now I provide those the values g = 0. 32028; h = 0. 47770 and K=876
We first calculate α = 0. 059785 + 0. 422213h + 1. 535061g
2 − 1. 211691g
3
g
2 = (0. 32028)(0. 32028) = 0. 1025792784
g
3 = (0. 32028)(0. 32028)(0. 32028) = 0. 032854091285952
α = 0. 059785 + 0. 422213(0. 4777) + 1. 535061(0. 10258) − 1. 211691(0. 032854)
α = 0. 059785 + 0. 20169 + 0. 15746 − 0. 039809
α = 0. 379126
We now multiply α and K to get Berkbigler’s Total Vote.
- 379126(876) = 332. 11, rounded to the nearest integer is 332.
Observer that A + C = 180 + 150 = 330, which was the actual total vote for Berkbigler,
a residual difference of only two votes.
You just predicted Berkbigler’s total vote with a blindfold, knowing only g, h and K. Notice that you did this without knowing λ = 0. 64155 !!!
Remember Alpha is the Sum of Berkbigler’s Early Vote and Berkbigler’s Mail-In Vote divided by the sum of all 4 categories.
You just solved the impossible and were only off by 2 votes!
If the math is still difficult for you, grab your high school student or his teacher and show them this. They should quickly be able to see that every precinct has a predictable outcome, and you can solve it impossibly only knowing a few of the needed variables to being able to solve it.
Thus proving Berkbigler won, not Hill.
For anyone thinking this is solvable because the person has all the data, that’s simple to remedy, Just give the user g, h and K and they will be able to do the impossible and predict each precinct without knowing alpha!
Remember Alpha is the Sum of Berkbiglers Early Vote and Berkbiglers Mail-In Vote
divided by the sum of all 4 categories. This proves that the election was predetermined and no matter how many votes Berkbigler received, Hill would always win.