What is the probability of two telephone numbers having 666 in them. My girlfriend and I both have phone numbers that are 10 digits long. They both happen to have three consecutive sixes in them (at different spots). We take it as a sign that we're meant for each other. 

What is the probability that this actually occurs, if we can use digits 0-9. Is this reasoning correct?: 

There are seven spots for the sixes to be placed:



The probability of placing the six in the first spot is $$\frac{1}{10}$$. Then the other two sixes need to be placed in the following spots, so the probability that this occuring, where the six is placed first is $$\frac{1}{10^3}$$

This can occur in seven more ways, so the total probability is $$\frac{7}{10^3}$$

Is that correct?
Denote by $D_n$ the set of decimal strings of length $n$. We are told to compute the probability $p$ that a random string from $D_{10}$ contains the substring $s_*:=666$. In order to avoid all sorts of inclusion-exclusion-traumata we argue as follows: Let

$a_n:\quad$the number of $s\in D_n$ not containing $s_*$, and not ending with $6$,

$b_n:\quad$the number of $s\in D_n$ not containing $s_*$,  and ending with a single $6$,

$c_n:\quad$the number of $s\in D_n$ not containing $s_*$, and  ending with $66$,

$d_n:\quad$the number of $s\in D_n$  containing $s_*$.

One easily sees that
$$a_3=900,\quad b_3=90,\quad c_3=9,\quad d_3=1\ .$$
The $a_n$, $b_n$, $c_n$, $d_n$ satisfy the following recursion:
$$\left[\matrix{a_{n+1}\cr b_{n+1}\cr c_{n+1}\cr d_{n+1}\cr}\right]=
\left[\matrix{9&amp;9&amp;9&amp;0\cr 1&amp;0&amp;0&amp;0\cr 0&amp;1&amp;0&amp;0\cr 0&amp;0&amp;1&amp;10\cr}\right]\ \left[\matrix{a_n\cr b_n\cr c_n\cr d_n\cr}\right]\ .$$
We therefore compute $$A^7\left[\matrix{900\cr 90\cr 9\cr 1\cr}\right]=(8942446170, 895052349, 89586081, 72915400)'$$and read off $d_{10}=72\,915\,400$. It follows that
$$p={d_{10}\over 10^{10}}\doteq0.00729\ .$$
Are you willing to accept, say, ABC-DEF-6666, that is are you counting FOUR sixes in a row as included in three sixes?  If so then, yes, this is correct.