Prime Probability

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.