P{(A U B)-->AB} = 1/2 Gives Astonishing Result

>>From Osher Doctorow mdoctorow@xxxxxxxxxxx

COPYRIGHT NOTICE
P{(A U B)-->AB} = 1/2 Gives Astonishing Result
Copyright By Owner Osher Doctorow Ph.D.
First Published 2005

Consider the equation:

1) P{(A U B) --> AB} = 1/2 for A, B statistically independent

In my previous thread, I proved that the left hand side is:

2) 2P(A)P(B) - P(A) - P(B) + 1

Hence we get:

3) 2P(A)P(B) - P(A) - P(B) + 1 = 1/2

or multiplying through by 2:

4) 4P(A)P(B) - 2P(A) - 2P(B) + 2 = 1

which is equivalent to:

5) -1 + 2P(A) + 2P(B) - 4P(A)P(B) = 0

which factors as:

6) 2P(B)(1 - 2P(A)) = 1 - 2P(A)

and therefore either P(A) = 1/2 or else:

7) P(B) = (1/2)(1 - 2P(A))/(1 - 2P(A)) = 1/2

regardless of what P(A) is as long as P(A) is not 1/2! Of course, P(A)
is the probability of A so it is between 0 and 1. You can check this
by inserting P(B) = 1/2 into (4) which is an identity!

We therefore obtain the remarkable result:

Theorem. The Universe exerts intermediate (1/2) Independent Probable
Influence on intersections of any two of its parts iff one of each of
its pairs of parts each has probability 1/2 (which is itself
intermediate between the defined range of probability, namely 0 to 1,
or in other words the mean of the uniform probability distribution on
(0, 1) or [0, 1]. In the case of intersection of two sets, this says:

8) P{(A U B) --> AB)} = 1/2 for A, B independent iff P(B) = 1/2 iff
P(B) = 1/2 and P(A) is arbitrary but not 1/2 or P(A) = 1/2 and P(B) is
arbirary and not 1/2.

Note that the theorem can be generalized to either just two parts or to
any number of parts greater than 1.

Osher Doctorow

.

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