Probability of two digit number sequence in series of numbers Given a random sequence (say $15$) of numbers I want to find the odds of finding '$90$' and '$09$' in the sequence. Looking at just two numbers in the sequence you have a $\dfrac{2}{10}$ chance of getting a '$9$' or '$0$' as the first digit, followed by $\dfrac{1}{10}$ chance of getting specifically the opposite digit that you need to complete the pair. $\dfrac{1}{10}*\dfrac{2}{10} = \dfrac{1}{50}$. Simple right?
But then I think of the repercussions of a failure case on the next adjacent pair examined in the sequence. If analyzing the first two numbers in the set has failed, it is very likely sequences adjacent to it have failed too because adjacent digits 'share' numbers.
if you have the set $4 \; 5 \; 9 \; 1 \;0 \; 5$, my steps for solving the problem fall apart when you consider that in checking '$4$' '$5$' leads into checking '$5$' '$9$', the $5$ mathematically has been accounted for as a failure and is leading towards this next pair to fail as well.
I haven't taken a probability class so maybe I'm failing to express my logic and perhaps I have some crazy misunderstanding of probability. What is this dependency that I acknowledge and how do I account for it when solving these kind of problems?
 A: Your question is not totally clear, but working from your calculation that the probability of success with a two digit string is $\frac{1}{50}$ then you seem to be looking for the probability of either $09$ or $90$ (or both) in your string.
This will be easier to consider as the complement of neither $09$ nor $90$ appearing.
So let $q_n$ be the probability that neither $09$ nor $90$ has appeared in the first $n$ digits and the $n$th digit is not $0$ or $9$, and let $r_n$ be the probability that neither $09$ nor $90$ has appeared in the first $n$ digits and the $n$th digit is $0$ or $9$.  You have the following:
$$q_1=0.8 \qquad r_1=0.2$$ $$q_{n+1}=0.8q_n + 0.8 r_n$$ $$r_{n+1}=0.2q_n + 0.1 r_n$$ 
You can solve this in closed form, but it is probably easier to find $q_{10}=0.6880531472$ and $r_{10}=0.1561083282$, making the probability you are seeking $1-q_{10}-r_{10}=0.1558385246$.
A: Denote the event that you find $09$ or $90$ in a sequence of $n\ge0$
numbers by $E_{n}$. Denote the event that the first number belongs
to $\left\{ 0,9\right\} $ by $A$. Then for $n>1$: $$P\left(E_{n}\right)=P\left(E_{n}\mid A\right)\times\frac{1}{5}+P\left(E_{n-1}\right)\times\frac{4}{5}$$
and: $$P\left(E_{n}\mid A\right)=\frac{1}{10}+P\left(E_{n-1}\mid A\right)\times\frac{1}{10}+P\left(E_{n-2}\right)\times\frac{4}{5}$$
Term $\frac{1}{10}$ corresponds with the chance of $0$ as second number after first number $9$ or vice versa. The second term with the chance on a second number that equalizes the first. The third term with the chance on a second number not belonging to $\{0,9\}$
Abbreviating $p_{n}=P\left(E_{n}\right)$ and $r_{n}=P\left(E_{n}\mid A\right)$
we have the recursive relations: $$5p_{n}=r_{n}+4p_{n-1}$$ and: $$10r_{n}=1+r_{n-1}+8p_{n-2}$$
and we can start with $p_{0}=p_{1}=0=r_{1}$. This leads to:$$50p_n=1+45p_{n-1}+4p_{n-2}$$ for $n=2,3,\dots$ where $p_{0}=p_{1}=0$.
Challenge is now to find a closed form for $p_n$. Hint on that: you can find a closed form for $f(x)=\sum_{n=0}^{\infty}p_nx^n$ and $p_{n}=\frac{f^{\left(n\right)}\left(0\right)}{n!}$. Practicizing the relations in an Excel-sheet I find $p_{10}\sim0.155839$ (agreeing with the result mentioned by Henry) and $p_{15}\sim0.231059$.
A: Let's call a string of digits "good" (as in good to go) if it ends in a digit between $1$ and $8$, and "bad" if it ends in a $0$ or $9$.  Let's let $g(n)$ and $b(n)$ count the number of good and bad strings of $n$ digits that don't have any adjacent pairs $09$ or $90$.  It's easy to see that
$$\begin{align}
g(n+1)&=8g(n)+8b(n)\\
b(n+1)&=2g(n)+b(n)\\
\end{align}$$
It follows that
$$\pmatrix{g(n)\\b(n)}=\pmatrix{8&8\\2&1}^n\pmatrix{1\\0}$$
The probability that the OP seeks is $p(n)=(g(n)+b(n))/10^n$, or
$$p(n)={1\over10^n}\pmatrix{1&1}\pmatrix{8&8\\2&1}^n\pmatrix{1\\0}$$
An exact formula analogous to the Binet formula for Fibonacci numbers can be obtained by diagonalizing the $2\times2$ matrix.  However, the eigenvalues here are
$$\lambda={9\pm\sqrt{113}\over2}$$
so the analogous result is likely to look a little ugly.
A: Odlyzko's "Enumeration of Strings" (in "Combinatorial Algorithms on Words", Apostolico and Gallil (eds), Springer, 1985) answers this kind of question (number of strings missing patterns) in great detail.
