The probability that \({N}_{x}\) is a prime is given by
\(\prod _{ i=1 }^{ { P }_{ N }<{ N }_{ x }<{ P }_{ N+1 } }{ \frac { { P }_{ i }-1 }{ { P }_{ i }}=\prod _{ i }^{ { P }_{ N } }{ (1-\frac { 1 }{ { P }_{ i } }) }} \) when \({ P }_{ N }={ N }_{ x }\) the associated probability is 1.
where \( { P }_{ N }\) is the \(Nth\) prime smaller but closest to or equal to \({ N }_{ x }\).
The expected value of \({ N }_{ x }\) statistically is then,
\(E\left[ { N }_{ x } \right] =\sum _{ 1 }^{ x\rightarrow \infty }{ ({ N }_{ x } \prod _{ i=1 }^{ { P }_{ N }<{ N }_{ x }<{ P }_{ N+1 } }{ \frac { { P }_{ i }-1 }{ { P }_{ i } } } } )\)
the summation is over all \({N}_{x}\) positive integer; and 1 not being a prime by definition.
\(E\left[ { N }_{ x } \right] =\sum _{ 2 }^{ x\rightarrow \infty }{ ({ N }_{ x } \prod _{ i=1 }^{ { P }_{ N }<{ N }_{ x }<{ P }_{ N+1 } }{ \frac { { P }_{ i }-1 }{ { P }_{ i } } } } )\)
The starting value of 1 prove to be very troublesome as far as consistency is concern. The value of \(\frac{1}{2}\) for \({P}_{2}\) can be argued as the probability of any number \({N}_{x}\) being odd or even. Strictly speaking \(2≮2\) and so \(?<2<2\) from the expression \({ P }_{ N }<{ N }_{ x }<{ P }_{ N+1 }\) does not hold.
Consider the term \( { P }_{ N }<{ N }_{ x }<{ P }_{ N+1 }\) , the numbers \({N}_{x}\) between
two consecutive primes \( { P }_{ N }, { P }_{ N+1 }\). All these numbers will share the same probability of
\({\prod _{ 1 }^{ { N }_{ x } }{ (\frac { 1 }{ 2 } )(\frac { 2 }{ 3 } )(\frac { 4 }{ 5 } )(\frac { 6 }{ 7 } )...\frac{{P}_{N}-1}{{P}_{N}}}}\)
Some may argue that since such \({N}_{x}\) are between two consecutive primes, they are not primes and so have a probability of zero being prime. The point is to calculate expected value \(E[{N}_{x}]\); \({N}_{x}\) is considered a random variable here. The expected value of \({N}_{x}\) being prime is not the average value of primes in a given set of prime numbers.
