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The Prime Number theorem asserts that an integer $m$ selected at random has roughly a $\frac{1}{\ln{m}}$ chance of being prime.Thus if $n$ is a large even integer and $m$ is a number between $3$ and $\frac{n}{2}$, then in a paper which approaching the heuristic probabilistic argument is saying one might expect the probability of $m$ and $n-m$ simultaneously being prime to be $\frac{1}{\ln{m}\cdot \ln{(n-m)}}$.
So, it assumes that the events that $m$ and $n-m$ are prime are statistically independent of each other. But how they can be independent ? I am not getting it... as we are working with a fixed even number $n$, so events $m$ and $n-m$ being prime should be dependent . . .and also this relation ignores some correlations between likelihood of $m$ and $(n-m)$ being prime. For instance, if $m$ is odd then $(n-m)$ is also odd and if $m$ is even then $(n-m)$ also even ,similarly if $3$ divides $n$ and as $m$ was a prime distinct from $3$, then $(n-m)$ would also be coprime to $3$ and thus be slightly more likely to be prime than general numbers... Can anyone explain how they are independent(if they are) or, can explain what will be the probability(closed form if possible) of being $m$ and $(n-m)$ both prime where $m\in [3,\frac{n}{2}]$ ?

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  • $\begingroup$ In fact , the numbers are not independent, the argument is only heuristic. The main dependence is that $m$ and $n-m$ are both odd, but this is no problem because it is even an advantage increasing the chance that both are prime. I assume that the paper deals with more details, such as how $m$ and $n-m$ have to be chosen to rule out a small factor. By the way, the probability $\frac{1}{\ln(m)}$ for a random integer to be prime is not at all a bad approximation, if $m$ is large. $\endgroup$ – Peter Apr 1 '18 at 7:27
  • $\begingroup$ @Peter Yeah, I am agree with the fact that $\frac{1}{\ln{m}}$ is a good approximation. But, my question is about the independence of the events of being both of them prime. Can we make a better approximation ? A better formula dealing with the flaws(e.g; I have mentioned some of them ) ... $\endgroup$ – OppoInfinity Apr 1 '18 at 11:58
  • $\begingroup$ Seems difficult. But anyway, every ruled out factor improves the chance of getting primes, so the given formula should be the worst case, or do I miss something ? $\endgroup$ – Peter Apr 1 '18 at 12:05
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    $\begingroup$ Yes this is the worst case. $\endgroup$ – OppoInfinity Apr 1 '18 at 14:10
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Selecting a random number is independent of selecting another random number.

But the probability that $m$ is prime is not independent of the probability that $n-m$ is prime.

This can be shown by finding a list of remainders of $m$ when we divide by some list of primes, and finding the remainders for $n$ divided by those too, then simply calculating wether $n-m$ is divisible by any of those primes with modulo arithmetic. Information about $m$ affects the probability that $n-m$ is prime, so they are not independent.

The heuristic you mentioned is just something to compare other theories too, to help locate important details relevant to some problem about the distribution of primes. It shouldn’t be taken as true.

A different (related) probabilistic approach to Goldbach’s conjecture is to consider the list of remainders of $n$ when divided by a list of prime numbers, then determine the possible remainder pairs of two numbers that would add up $n$, then find the probability that a pair of numbers less than $n$ satisfies those conditions.

For example; consider $n=12$. Observe that $9<12<25$ thus all composite numbers upto $12$ are divisible by a prime number less than or equal to $3$. So consider what $12$ divided by those primes are. For both of those the remainder happens to be $0$. The more interesting fact is that $m$ and $n-m$ have to have remainder $1$ and $2$ when divided by $3$, and that they have to both have a remainder of $1$ when divided by $2$. Now imagine all numbers are now represented by their remainders $(r_2, r_3)$. $m$ and $n-m$ must be equal to $(1,1)$ and $(1,2)$. Perhaps the probability of a pair of numbers fitting the representation and being less than $12$ (but more than $1$) can be calculated? In this case it can, but for larger $n$ it requires some more details to take into account to create a better probabilistic model, just as the one you have mentioned.

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  • $\begingroup$ In the example I have used, we’d need to take into account remainders when divided by 5 too. $\endgroup$ – Brad Graham Apr 1 '18 at 16:53

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