Heuristic Proof of Hardy-Littlewood Conjecture for 3-term Arithmetic Progressions The Hardy-Littlewood Conjecture for 3-term arithmetic progressions is that
$$
\#  \{ x,d \in \{1,\ldots,N\} \, | \, x,x+d,x+2d \text{ are all prime} \} \sim
\frac{3}{2} \prod_{p > 2} \left(1+\frac{1}{(p-1)^2}\right) \frac{N^2}{(\log N)^3}.
$$
In this (piece) of a paper (http://www.claymath.org/publications/Gauss_Dirichlet/green.pdf), Ben Green outlines a heuristic argument for the (k-term version) of the conjecture that I am trying to understand.  I will repeat the most important parts here.
For large $N$, the probability that an arbitrary integer $\leq N$ is prime is
$$
\mathbb{P}(x \text{ is prime} | 1 \leq x \leq N) \approx \frac{1}{\log N}
$$
by the Prime Number Theorem.
Choose $x,d \in \{1,\ldots,N\}$ at random among the $N^2$ choices and write $E_j$ for the event that $x+jd$ is prime.  If the events $E_0, E_1, E_2$ were independent, we would expect that 
$$
\mathbb{P}(x,x+d,x+2d \text{ are all prime}) =  \mathbb{P}(E_0 \cap E_1 \cap E_2) \approx \frac{1}{(\log N)^3},
$$
and so
$$
\# \{x,d \in \{1,\ldots,N\} \, | \, x,x+d,x+2d \text{ are all prime} \} \approx \frac{N^2}{(\log N)^3},
$$
which is the correct result up to a constant factor.
Green says that the correct constant can be obtained by discarding the incorrect assumption of independence and taking account of the fact that the primes $> q$ fall only in those residue classes $a(\text{mod }q)$ with $a$ coprime to $q$.  He gives no more details.
I've been trying to figure out how to do this, but haven't been successful.
Could someone please help me out or point me to a reference where it is done?
Thanks.

Very roughly speaking, the probability for a number $x$ to be divisible by a prime $p$ is $1/p$. We can't take this literally because it stops making sense when $p$ gets close to $\sqrt x$, but it will make more and more sense for more and more primes as $x$ increases. Thus, even though we can't calculate the probability of a number being prime as $\prod_p(1-\frac1p)$ (which would be zero), we can calculate corrections to the independence assumption as if we were correcting the factor $1-\frac1p$. If the product of the corrections from primes above some limit $q$ converges to $1$ as $q\to\infty$, we can expect the correction to be asymptotically correct, since the part of it that doesn't make sense will not matter asymptotically.
Now if $E_0$, $E_1$ and $E_2$ were independent, the contribution from a prime $p$ to the probability of the triple being prime would be $(1-\frac1p)^3$. To find the correction for $p\gt2$, we can consider the two cases where $d$ is or isn't divisible by $p$. If $d$ is divisible by $p$, which happens with probability $\frac1p$, then $x$, $x+d$ and $x+2d$ all have the same residue $\bmod p$, and this is non-zero with probability $1-\frac1p$; whereas if $d$ is not divisible by $p$, which happens with probability $1-\frac1p$, then $x$, $x+d$ and $x+2d$ all have different residues $\bmod p$, and the probability of none of them being zero is $1-\frac3p$.
Thus, in total, the probability of none of the three numbers being divisible by $p$ is
$$\frac1p(1-\frac1p)+(1-\frac1p)(1-\frac3p)=1-\frac3p+\frac2{p^2}=(1-\frac1p)(1-\frac2p)\;.$$
Thus the correction with respect to $(1-\frac1p)^3$ is
$$\frac{(1-\frac1p)(1-\frac2p)}{(1-\frac1p)^3}=\frac{1-\frac2p}{(1-\frac1p)^2}=\frac{p^2-2p}{(p-1)^2}=1-\frac1{(p-1)^2}\;,$$
so we should expect a minus sign where you have a plus sign, and indeed the paper you cite has a minus sign.
The case $p=2$ has to be treated separately because there aren't three different residues $\bmod2$. In this case, $(1-\frac1p)^3$ is $\frac18$, whereas the correct probability is $\frac12\cdot\frac12=\frac14$, so the correction is $2$, and indeed the paper you cite has $2$ where you have $\frac32$.
