Election Fraud Probabilities
It is an election year and the media has obsessed over Trump’s inflammatory statements and HRC’s health. A recent Google search for “Hillary’s health” showed 140 million hits. Parkinson’s has been widely mentioned.
Both issues could affect the outcome of the Presidential election IF the election is honest and has not already been decided.
But are election fraud, vote “miscounts,” illegal voters, and buying the election probable? We all know the jokes about the Chicago cemetery vote, but what about election fraud in voting machines, punch cards, or manual systems?
Consider: “In 59 Philadelphia voting divisions, Mitt Romney received zero votes.” Very strange voting results were reported in the 2012 election – possible, but not likely ……
- Was the 2012 Presidential election count in Philadelphia accurate?
- Yes, those divisions vote heavily democratic. One can easily imagine a 90% democratic vote, or maybe even 99% in some divisions.
- But is it likely that 59 voting divisions voted 100.00% for Obama against Romney? Total votes 19,605 to zero.
- Were votes “miscounted” or were voting machines programmed for an Obama result?
- Is it reasonable to believe that not one person in 59 voting divisions voted for Romney out of choice, or as a vote against Obama, or even as the “lesser evil?”
Apply logic and probability math:
- Assume those voting divisions were about 99% democratic and the voters would vote about 1% Republican. That suggests Romney should have received about 190 votes, yet he received zero.
- Assume an honest election, no fraud, and statistical probability that only 1 person in 100 would vote for Romney. Math below:
- However, not one vote went for Romney. The odds against that result are staggeringly unlikely.
- For the purposes of probability math, assume exactly 1 vote went to Romney and 19,604 went to Obama in what are normally 99% democratically voting divisions. The odds AGAINST that are 1.88 times ten to the 83rd power, or a number with 83 zeros. See below and link for the math.
- For comparison, there have been about 4 times ten to the 17th power seconds in the age of the universe, about 14 billion years. The age of the universe measured in seconds is only a miniscule number compared to the odds AGAINST just 1 Romney vote in a 99% democratic division.
- Given that a 99% democratic bias creates an essentially impossible conclusion, what if we assume an extremely high 99.9% democratic bias?
- Use the same probability math and ask what are the odds against receiving only 1 vote out of 19,605 when 999 out of 1,000 will vote democratic?
- The odds are about 16.8 million to one AGAINST receiving only 1 vote, and much higher for receiving no votes.
- Receiving only 1 republican vote in 59 divisions is only somewhat more likely than winning the Powerball Lottery, even assuming a heavily biased 99.9% probability of democratic votes.
- Really? Voter fraud, “miscounts,” and programmed voting machines appear likely!
- Given the above statistics, and assuming an extreme 99.9% democratic bias in those voting divisions, it looks effectively impossible for 19,605 votes to have been honestly cast for Obama while zero votes were cast for Romney.
How Easy Is It To Hack a Voting Machine?
The “status quo” heavily supports HRC, in spite of FBI investigations, top secret security failures, possible Parkinson’s Disease, and a growing body count. But if her star is falling and massive wealth and power are at risk …
- HRC must sell the story she is okay for about two months while depending upon considerable help from her friends. Or …
- Rig the voting results! It has happened before and it could happen again. Or …
- Create a crisis, cancel or delay the election, and keep “status quo” in power. Or …
- Trump is elected, in spite of the DNC’s best efforts, and then a “lone shooter” … and you know the rest. It has happened before …
My hope and expectation is that votes are honestly cast and accurately counted and the people of the US elect the President they want in 2016. Yet the above possibilities are worth considering.
If the probability of a democratic vote is 99%, the odds against receiving only 1 republican vote in 19,605 are:
1 / (19605 x (0.99^19604) x (0.01^1)) = 1.88486 E+83 to 1. Or,
One divided by (19605 times 0.99 to the 19604 power times 0.01)
If the probability of a democratic vote is 99.9%, the odds against receiving only 1 republican vote in 19,605 are:
1 / (19605 x (0.999^19604) x (0.001^1)) = 16,819,069 to 1. Or,
One divided by (19605 times 0.999 to the 19604 power times 0.001)
The Deviant Investor