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Suppose we choose two random distinct numbers from $\{1,2,\cdots,n\}$.

I learned that the approximate number of primes less than or equal to $n$ is given by

$\frac{n}{ln(n)}$.

Assuming that there are approximately $2n$ possible sums, the number of primes is roughly

$\frac{2n}{ln(2n)}$

and so the probability that the sum of two random numbers is prime is given by

$\frac{2n}{ln(2n)}\cdot\frac{1}{2n} = \frac{1}{ln(2n)}$.

However, it seems like the probability is actually closer to $\frac{1}{ln(n)}$.

I think I might be misunderstanding something.

Any advice would be appreciated.

Thank you for reading.

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    the list of possible sums is hardly uniformly distributed. Just think about the possible sums from rolling two fair dice. – lulu Sep 16 '22 at 15:40
  • See this near duplicate – lulu Sep 16 '22 at 15:42
  • The sum of two Uniform distributions on ${1, ..., n}$ is Normally distributed in the limit. You can read more about this here https://www.sciencedirect.com/topics/mathematics/discrete-uniform-distribution – ilaK Sep 16 '22 at 15:47
  • $$\dfrac 1 {\ln(2n)} = \dfrac 1 {\ln(2)+\ln(n)} \mapsto \dfrac 1 {\ln(n)} \text{ for } n >>0.$$ – Rócherz Sep 16 '22 at 17:28
  • Are the numbers we sum up random ? Then, the sum is random as well , but if we , for example , sum up two odd numbers greater than $1$ , we never get a prime. We do not get a prime either if the gcd of the summands is greater than $1$. – Peter Sep 18 '22 at 10:38

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