Crude verification of Goldbach conjecture [closed]

So the Goldbach conjecture says 'Every even integer greater than 2 can be written as sum of two primes'. Here is what I have roughly done to verify it, using probability. I don't say it is correct but I just want to show it.

Let a be an even integer greater than 2. The prime counting function gives number of primes n as n ~ a/ln a. Now, for every prime less than a, we generate odd numbers(except for 2)

a-p(i) where, p(i) is the ith prime below a. Thus, we have approximately n odd numbers below a.

We have n/a = 1/lna which gives the probability of prime number below a.

If we consider only the odd numbers below a, the probability becomes 2/lna, which means 2 out of lna odd numbers below a are prime numbers.

But we have generated approximately(which just excludes case for 2 and is negligible for large numbers) a/lna odd numbers.

So, from unitary method, 2/lna out of 1 odd numbers below a are prime, which yields for our case, $2 a/(lna)^2$ prime numbers out of a/lna odd numbers generated.

And we can see that, 2 a/(lna)^2 is obviously greater than 1 which even grows when a gets larger.

This shows that, among our generated odd numbers(a-p(i)) there is at least one prime number q giving,

a-p(i) = q

or, p(i) + q = a. where p(i) and q are primes, verifying Goldbach Conjecture.

Edit: I just wanted to know what is wrong with this approach(I knew there was). And I got my answer too. Thank you.

closed as unclear what you're asking by Daniel, user147263, graydad, Matt Samuel, StrantsAug 30 '15 at 6:06

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• possible duplicate of Goldbach conjecture – Morgan Rodgers Aug 29 '15 at 12:29
• i could not understand the question you mentioned. – Bibekpandey Aug 29 '15 at 13:13
• There are like 30 "check my proof of the Goldbach conjecture" questions..... – Morgan Rodgers Aug 29 '15 at 14:39
• @BeWakePandey It isn't clear what your question is. You give a crude heuristic which could be refined to a sharp heuristic, and though neither is proven. What type of answer are you expecting that you'd accept?? As it stands, there isn't any question in your question. – Erick Wong Aug 30 '15 at 1:53
• Use anything but $e$, because $e$ is already universally defined. – user253055 Sep 2 '15 at 0:08

On average, you would expect around $2e/(\log e)^2$ primes. But the $e/\log e$ numbers are a small proportion of the numbers below $e$, and it might happen, for one particular $e$, that none of them are prime.