What percentage of numbers is divisible by the set of twin primes? What percentage of numbers is divisible by the set of twin primes $\{3,5,7,11,13,17,19,29,31\dots\}$ as $N\rightarrow \infty?$
Clarification
Taking the first twin prime and creating a set out of its multiples : $\{3,6,9,12,15\dots\}$ and multiplying by $\dfrac{1}{3}$ gives $\mathbb{N}: \{1,2,3,4,5\dots\}.$ This set then represents $\dfrac{1}{3}$ of $\mathbb{N}.$
Taking the first two: $\{3,5\}$ and creating a set out of its multiples gives: $\{3, 5, 6, 9, 10\dots\}.$ This set represents $\sim \dfrac{7}{15}$ of $\mathbb{N}.$
Taking the first three: $\{3,5,7\}$ and creating a set out of its multiples gives: $\{3, 5, 6, 7, 9, 10, 12, 14\dots\}.$ This set represents $\sim \dfrac{19}{35}$ of $\mathbb{N}.$
What percentage of $\mathbb{N}$ then, does the set consisting of all divisors of all twin primes $\{3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18\dots\}$ constitute? (ie This set $\times \ ? \sim \mathbb{N}$)
 A: As an addendum to mjqxxxx's excellent answer, I present a different approach which offers a minor improvement in accuracy (although a difference of $\approx 2\%$ is sufficiently large to be notable, considering how slowly the product converges at large $N$).
Let $\mathcal {P} (\mathbb{P}_2) $ represent the power set of all 
twin primes, $\mathcal {P} (\mathbb{P}{_2}(N)) $ the power set of the first $N$ 
twin primes, and $\mathcal {P}_\kappa (\mathbb{P}{_2}(N)) $ the set of subsets 
of cardinality $\kappa.$ Also let $\mathcal {P}_\kappa \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)}$ represent the subset of the products of reciptocals of twin primes with the specified cardinality.
For example, $A=\{3,5,7,11\},\ \mathcal {P}_2 {\left(\prod\frac{1}{A}\right)}$ would represent the set $\left\{\frac{1}{15},\frac{1}{21},\frac{1}{33},\frac{1}{35},\frac{1}{55},\frac{1}{77}\right\}.$ 
Since
\begin{align}
&\quad \prod_{p\in \mathbb{P}_2}^{N}\left(1-\frac{1}{p}\right)&=&\quad 1-\sum^{N} \left (\large\mathcal {P} _ {\text {odd}} \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)} - \large\mathcal {P} _ {\text {even}} \small{\left(\prod\frac{1}{p\in \mathbb{P}_2}\right)} \right) \\
\end{align}
as can be seen easily in the case $N=3:$
\begin{align}
&\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}-\frac{1}{15}-\frac{1}{21}-\frac{1}{35}+\frac{1}{105}\right)= 1-\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\\
\end{align}
it follows that
\begin{align}
&\quad \prod_{p\in \mathbb{P}_2}\left(1-\frac{1}{p}\right)&=
&\quad \quad \sum _{p\in \mathbb{P}_2} \frac{1}{p}\\
&&&-\quad \frac{1}{2} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p}\right)^2-\frac{1}{2} \sum _{p\in \mathbb{P}_2} \frac{1}{p^2}\\
&&&+\quad \frac{1}{6} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p}\right)^3-\frac{1}{2} \left(\sum _{p\in \mathbb{P}_2} \frac{1}{p^2}\right) \sum _{p\in \mathbb{P}_2} \frac{1}{p}+\frac{1}{3} \sum _{p\in \mathbb{P}_2} \frac{1}{p^3}\\
&&&-\quad \dots
\end{align}
where the coefficients are given in Table 24.2 in 1 (multinomials M2) multiplied by $(-1)^{q},$ where $q$ is the number of elements in the corresponding integer partition.
This representation turns out to be beneficial computationally, since the power sums converge so rapidly. $\infty$ in the sum can be replaced by a reasonably small $N,$ and $\sum _{p\in \mathbb{P}_2} \frac{1}{p}$ replaced by Brun's constant (see note below), to give the slightly improved approximation of $\approx 83.83 \%.$

Note: As Erick Wong notes in the comments below, the current "known" value of Brun's constant is based on a Heuristic argument (Hardy & Littlewood) that $\pi_2(x) \approx 2C_2 \int_2^x \frac{dt}{\left(\log t \right)^2},$ where $C$ is the twin prime constant. Nicely gives here an estimate for $B_2$ of $1.9021605823\pm 8\times10^{-10}.$


*

*Milton Abramowitz and Irene A. Stegun (eds.), Handbook of mathematical functions, 9th ed.,
Dover Publications, New York, 1972.

A: According to Brun's theorem, the twin primes constitute a small set.  That is, the sum of their reciprocals,
$$
\left(\frac{1}{3} + \frac{1}{5}\right) + \left(\frac{1}{5}+\frac{1}{7}\right) + \left(\frac{1}{11}+\frac{1}{13}\right) + \cdots,
$$
is convergent.  Numerically it is estimated to be about $1.902$, so the sum with the single duplicate $(1/5)$ removed is $B\approx 1.702$.  The probability that a large random number is not divisible by any twin prime approaches
$$
\prod_{p\in P_2}\left(1-\frac{1}{p}\right)=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\cdots < \exp\left(-\sum_{p\in P_2}\frac{1}{p}\right)=e^{-B},
$$
and so the natural density of twin-prime-divisible numbers is at least $1-e^{-B}\approx 0.8177.$  (Estimating the product using the twin primes through two million gives density $> 0.806$.)
As a sanity check on this, the number of twin-prime-divisible numbers from $10^5$ through $2\times 10^5 - 1$ is exactly $81714$, for a density of $0.8171$; and this should increase slightly for larger numbers.
