The "covering rate" of cosets $7 \Bbb{Z} + a$ by some coset $b \Bbb{Z} +c$. The covering rate of the cosets $7\Bbb{Z} + a$ by the coset $13\Bbb{Z} + 2$:
$$
7(0) -4 = 13(-1) + 2 \\
7(1) -5 = 13(0) + 2 \\
7(2) + 1 = 13(1) + 2 \\
7(4) + 0 = 13(2) + 2 \\
7(6)  -1 = 13(3) + 2 \\
7(8)  -2 = 13(4) + 2 \\
7(10)  -3 = 13(5) + 2 \\
7(12) - 4 = 13(6) + 2 \\
7(14) -5 = 13(7) + 2 \\
$$
It takes minimally an offset of $14$, i.e. $7(x + 14) + a = 13y + 2$ before the cycle starts over again (look at the residue cycling: $-4,-5,1,0,-1,-2,-3,-4,-5, \dots$, the offset separating the two $-5$'s is $14$.
Is $\dfrac{1}{14}$ or for all fixed $x, a \in \Bbb{Z}$, the limit
$$\lim\limits_{y \to \infty} \dfrac{\#(\{7x + a, 7(x+1) + a, \dots, 7(x + y) + a\} \cap (13 \Bbb{Z} + 2))}{y - x} = \dfrac{1}{14}$$.
The covering rate of $7\Bbb{Z} + a$ by $19 \Bbb{Z} + 3$:
$$
7(3) + 1 = 19(1) + 3\\
7(5) + 6 = 19(2) + 3 \\
7(8) + 4 = 19(3) + 3 \\
7(11) + 2 = 19(4) + 3 \\
7(14) + 0 = 19(5) + 3 \\
7(16) + 5 = 19(6) + 3 \\
7(19) + 3 = 19(7) + 3 \\
7(22) + 1 = 19(8) + 3 \\
7(24) + 6 = 19(9) + 3 \\
$$
is therefore $\dfrac{1}{19}$.  How can we compute this rate without going through the trouble of enumerating?
 A: The natural density of a set $S\subseteq\mathbb{N}$ of naturals is defined as
$$ \mu(S)=\lim_{N\to\infty} \frac{|S\cap[1,B]|}{N}. $$
Not all subsets of $S$ have a density - indeed we may construct $S$ so that $|S\cap[1,N]|$ oscillates indefinitely if we want. This density is also not countably additive the way probability measure would be, so for instance every singleton $S=\{n\}$ has density $0$ but their union has density $\mu(\mathbb{N})=1$. It's also worth pointing out it is difficult to prove things about the natural density of certain sets, however it is upgraded by something called the Dirichlet density, defined by a limit with Dirichlet series at $s=1$: whenever the natural density exists, so does the Dirichlet density and they are equal, but the Dirichlet density may exist even when the natural density doesn't and in practice it is easier to work with the Dirichlet density for sets encountered and under scrutiny within analytic number theory.
It should be clear how to extend the definition of natural density to all of $\mathbb{Z}$ if we want a more "symmetric" treatment of cosets. The natural density of a coset is simply $\mu(a+m\mathbb{Z})=\tfrac{1}{m}$. You are interested in the relative density $\mu(A\cap B)/\mu(B)$ for two cosets $A=a+7\mathbb{Z}$ and $B=2+13\mathbb{Z}$.
Fortunately, the intersection of two cosets is itself a coset (or empty). The elements of $A\cap B$ where $A=a+m\mathbb{Z}$ and $B=b+n\mathbb{Z}$ are solutions to the system of congruences
$$ \begin{cases} x \equiv a \mod m \\ x \equiv b \,\mod n \end{cases} $$
The Chinese Remainder Theorem says this is empty if $a\not\equiv b\mod \gcd(m,n)$, or else it is a residue mod $\mathrm{lcm}(m,n)$ (which can be constructed explicitly using a version of Lagrange interpolation and modular inverses, which I can explain if you want). Therefore we have
$$ \mu\big((a+m\mathbb{Z})\cap(b+n\mathbb{Z})\big) =
\begin{cases}
\frac{1}{\mathrm{lcm}(m,n)} & \textrm{ if } a\equiv b\mod\gcd(m,n) \\[5pt] 
~~~~~0 & \textrm{ if } a\not\equiv b \mod \gcd(m,n)
 \end{cases} $$
