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?

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 $p_{15}\sim0.231059$.
