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?
 A: 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:
$$\left[\matrix{a_{n+1}\cr b_{n+1}\cr c_{n+1}\cr d_{n+1}\cr}\right]=
\left[\matrix{9&9&9&0\cr 1&0&0&0\cr 0&1&0&0\cr 0&0&1&10\cr}\right]\ \left[\matrix{a_n\cr b_n\cr c_n\cr d_n\cr}\right]\ .$$
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
$$p={d_{10}\over 10^{10}}\doteq0.00729\ .$$
A: No, you can only add the probabilities if the events are disjoint, which they aren't here.  One way you can see that this reasoning is incorrect is to pretend you have a phone number with $n$ digits, which would seemingly make the answer $n / 10^3$.  But this clearly cannot be when $n > 10^3$.
You also seem to be counting the case where the trailing three digits are sixes multiple times.  I'm assuming we have a phone number with seven digits, which means there are only five positions for the three consecutive sixes.
The direct approach to finding this probability would be to apply the inclusion exclusion principle to $P(\cup_{i=1}^{5} \{ \text{digits $i$ to $i + 2$ are a $6$} \})$, although this looks a bit tedious.  Maybe someone else knows of a clever method.
A: We are going to count various types of five digit strings first. I write $x$ for an arbitrary decimal digit and $\dagger$ for a non-$6$ digit. The five digit strings not containing the substring $666$ can be put into three classes as follows:
$$C_0:\quad x\,x\,x\,x\,\dagger\>,\qquad C_1:\quad x\,x\,x\dagger6\>,\qquad C_2:\quad x\,x\dagger6\,6\ .$$
Denote by $a_i$ the number of strings  not containing the substring $666$ in each of these classes. Then
$$a_0=9(10^4-19)=89829,\qquad a_1=9(10^3-1)=8991,\qquad a_2=9\cdot10^2=900\ .$$
The number $a_*$ of five digit strings containing the substring $666$ therefore comes to
$$a_*=10^5-(a_0+a_1+a_2)=280\ .$$
A ten digit string is the concatenation of two five digit strings. The total number $d_{10}$ of ten digit strings containing the substring $666$ can now be computed as follows:
$$d_{10}=2\cdot a_*\cdot 10^5-a_*^2+2\cdot a_1\cdot a_2+a_2\cdot a_2=72\,915\,400\ .$$
The probability $p$ that a random ten digit phone number contains the string substring $666$ is therefore given by
$$p={d_{10}\over 10^{10}}=0.00729\ .$$
A: 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.  
