Is this the way to estimate the amount of lucky twins? To estimate the amount of prime twins between $3$ and $x$ we just take $x \prod_{p}(1-2/p)$ where $p$ runs over the primes between $3$ and $\sqrt x$. Lucky numbers are similar to prime numbers. Does this imply that a good way to estimate the amount of lucky twins between $3$ and $x$ is $x \prod_{l}(1-2/l)$ where $l$ runs over the lucky numbers between $3$ and $\sqrt x$?
*EDIT:*I Think I can improve the question by stating it as follows. 
$1$) If $x$ goes to infinity and $l(x)$ denotes the amount of lucky numbers between $3$ and $x$ does
 $\dfrac{x \prod_{l}(1-2/l)}{l(x)}=Constant$   ?
$2$) If $x$ goes to infinity, does
 $\dfrac{\prod_{p}(1-2/p)}{\prod_{l}(1-2/l)}=Constant$ ?
If no theoretical answer is possible are $1$) and $2$) supported numerically ?
 A: I think that to answer this question one would need to start from the paper by Bui and Keating on the random sieve (of which the lucky sequence is a particular realization) and then carefully examine the proofs that they cite of Mertens' theorems for that sieve.  The issue is whether there is a correction factor like Mertens' constant or the twin prime constant in the product that you wrote down for the number of lucky twins, and if there is, is it calculated (up to $1+o(1)$ factors) by the same formula as for the primes.  
http://arxiv.org/abs/math/0607196
The paper gives an asymptotic formula for random twin primes that does not have any twin prime constant, but the explanation there is not exactly about the same situation. It looks to me like some analysis of the proofs of the analogues of Mertens' theorems is needed to understand what product over random primes is the correct estimate of the number of random twin primes.
I cannot tell the answer to the question from Bui and Keating paper by itself, but would guess that up to multiplication by a constant, the product you wrote down is, for nearly all realizations of the random sieve (ie., with probability 1) asymptotic to the number of random-sieve twins, and this is the heuristic guess for what happens with the lucky numbers.
