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From: Osher Doctorow on 7 Jul 2010 22:02
From Osher Doctorow

Lets look at:

1) P(B) = nP(A) - 1, n > 2 integer.

As an example, suppose n = 3. Then we have:

2) P(B) = 3P(A) - 1.

But probabilities are always between 0 and 1 inclusive, and therefore:

3) 0 < = 3P(A) - 1 < = 1

The left half of this inequality says:

4) P(A) > = 1/3

The right half says:

5) P(A) < = 2/3.

Compare this with the equation:

6) P(B) = 2P(A) - 1, which yields 2P(A) - 1 > = 0 or P(A) > = 1/2,
and 2P(A) - 1 < = 1 or P(A) < = 1.

You can see already that (6) gives a domain of [1/2, 1] for P(A),
while (2) gives a domain of [1/3, 2/3] for P(A).

In the case of:

7) P(B) = 4P(A) - 1,

we get similarly 4P(A) - 1 > = 0 or P(A) > = 1/4, and 4P(A) - 1 < = 1
or P(A) < = 2/4 = 1/2. Therefore, here again the domain has shrunken
in length from 1/2 in [1/2, 1] to 1/4 in [1/4, 2/4]. In (2), the
domain's length was 1/3 in [1/3, 2/3].

The general pattern can now be obtained for arbitrary positive n, but
the point is that except for n = 2, the domain of A or P(A) shrinks
below 1/2 as n increases, presumably converging to length 1/n --> 0 as
n --> infinity.

The Probabilistic Mechanical Advantage of P(B) or P(AB) = 2P(A) - 1 is
optimal for the longest interval of A among nonnegative integer n of
form P(B) or P(AB) = nP(A) - 1, and the interval shrinks as n -->
infinity or even as n increases. It is obviously almost useless to
have an optimum on [1/10^10, 2/10^10] for example and nowhere else,
except for certain "exotic" purposes.

Osher Doctorow
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