The Fibonacci numbers are hiding...

$10$ coins are flipped randomly by Bob, and then placed in a line.

For an arrangement of coins $S$ , suppose $s$ is a subset of consecutive elements of $S$ (meaning that they appear next to each other in the line). Let $U(s)$ and $D(s)$ be the number of coins facing up and down in $s$ , respectively. $S$ is considered to be $n-close$ if for any $s$ , $\left| U(s)-D(s) \right| \le n$ .

Given that an arrangement $S$ is $3-close$ , what is the probability that it is also $2-close$ ? Express your answer as $\frac { m }{ n }$ where $m$ and $n$ are relatively prime natural numbers. What is $m+n$ ?

Tip: There's a nice, simple way to solve this without casework. Don't make things too complicated!