The Theory of everything Part 6- Can science get it wrong?

Only a fool does not question  what they are taught by education and the existing knowledge that is programmed into a person from often an early age.  We are taught or were taught by an education discipline, a disciple that does not allow for disagreement.  A discipline that if you do not accept are forced to failure and often a -F.

We accept this discipline and agree with someone from the past, a someone who originated the theory or maths or whatever the education may be.   We follow this past with discipline and our everyday lives continue as if we know reality.

In a recent forum topic, I asked a question about maths and in-placed some of my own maths by using symbols of maths representation.

http://www.rapidtables.com/math/symbols/Statistical_Symbols.htm

https://en.wikipedia.org/wiki/List_of_mathematical_symbols

To be totally honest, I have no idea what I was doing, I was using the symbols to represent voice.

I had self learnt some maths, but was unsure if I was even correct in my understanding of the maths, at this time I wished I was back at school so someone could confirm I had learnt correctly.

A person may call me stupid for not knowing probability or basic algebra, but to learn is just to memorise somebodies else’s thoughts, to repeat and mimic what we are taught.

In short science said I was wrong,  a logical assumption and the maths I provided made no sense.  However to me, on my understanding of what I had self learnt, my maths was correct and it was science who were wrong.

Firstly the logical question, 

Does the probabilities change of an output receiving a specific variant from an individual  set of variants distributed then re-shuffled and used again to the same output, if we offered a choice of an individual set of multiple individual sets of variants, when each set has been randomly shuffled and the output always receives the left aligned of the unseen variants?

Simplified

x. = nnn = 1/3

multiple sets

X=x1 = nnn= 1/3
X=x2 = nnn = 1/3
X=x3 = nnn = 1/3
X=x4 = nnn = 1/3

I was told no and I was incorrect.

My logical argument. 

We will define a single set as x (lower case)   We will define any specific variant of 1/3 as (n).

We will define output as p1, and p1 always receives the left aligned variant after the random shuffle.

We will also add two vectors, X (upper case) and Y (upper case).  We will define P is probability and t is time.

P(n)/x=(1/3)/t

The chance of receiving any specific (n) from a single set is 1 out of 3 variants over continuous time.

P(n)/X=(1/3)/t

The chance of receiving any specific (n) from the X-axis is 1 out of 3 variants over continuous time.

So what happens if we add random choice of set of the sets baring in mind p1 receives X1 of any set?

X1/X2/X3
X1/X2/X3
X1/X2/X3
X1/X2/X3

Is it my imagination? or have we just created a Y-axis by adding choice,  and Y is not equal to X.

YX

P(n)/Y=σ2/t2

The chance of receiving (n) by random choice of set is dependent to the variance of population values by the shuffle of X, (the rows), aligning values to p1 , (the output), in a Y-axis by adding choice.

X1
X1
X1
X1

If (n)=A,B,C of each set then why cant they tell me the chance of receiving an A from the
above?

Maybe just maybe I am correct and the probability does change if we add random choice.

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