\(10\) 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!

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