[geofft]

I assume you mean P(c) = P(a) P(b) + P(b) P(a) (i.e., you can get to p+6 as p+2+4 or as p+4+2) = 2 P(a) P(b)? But even so, that's not true, because you're messing with conditional probabilities in a weird way.

Let k be any integer. Consider the probabilities a) k+2 is a multiple of 10, b) k+4 is a multiple of 10, and c) k+6 is a multiple of 10. P(a), P(b), and P(c) are all 0.1, but your approach would claim that P(c) is 0.02. And P(k + 30 is a multiple of 10) is also still 0.1.

[lohankin]

No, I meant something different. Let's consider primes, say, between 1 and 1000000. a) Compute the number of cases when p is prime AND p+2 is prime; b) number of cases when p is prime AND p+4 is prime; c) number of cases when p is prime AND p+6 is prime. My claim is that a is approx. equal to b, and c is approx. equal to 2a (which is the same as 2b). You can easily test it (which I did), and it's easy to see why.