Is this formulation about voting correct?

[jamilk]

A Vote With Less Than 77.3% YES Portion Is Mathematically Improper

By
Jamil Kazoun

Sep 23, 2020

Abstract: The traditional voting scale begins at zero for the percentage of voters that vote YES. I prove that this is inappropriate, and that the voting scale should begin at 77.3% point of the number of voters voting YES.

Analysis:
For a vote with two mutually exclusive options, YES and NO, let:
p be sample or population mean in percent of voters voting YES
q be 1 - p
Let YES = p
NO = q
Let H = Entropy = - (p*log(p) + (1 – p)*log (1 – p)) ; log is base 2

If p is assumed as the probability of correctness, and (1 - p) as the probability of incorrectness, then, subtracting entropy from p should give the net certainty of the correctness of the vote:
Y = p – H = net certainty of the correctness of the vote
If we set this net certainty to 0 as the minimum acceptable level for a vote, then:
0 = p – H
0 = p - - (p*log(p) + (1 – p)*log (1 – p))
Solving this for p, I get
P = .773 = 77.3%
Therefore, a vote with 77.3% YES portion has zero certainty. This therefore is the starting point of the voting scale.
Conclusion: No vote bellow 77.3% should be acceptable mathematically.

Side note: Looking at a completely different derivation approach in my previous publications, I get similar results.

 

[jamilk]

Corrections:
Should be p - pH which gives p=0.5
and Variance = pq for a binomial distribution = 0.5*0.5=0.25
shifting p by Variance to the right on the p axis of the distribution graph
we get p = 0.5 + 0.25 = .75% = 75% as the minimal YES portion for a vote.

 

[jamilk]

Other functions than variance can be used. What is important is that all the area under the distribution graph is to the right of p. P becomes the end point of the left tail of the disribution graph.