Probability that no letter is in alphabetical order Given a random string of distinct letters, find the probability that none of the letters are in order. A letter is in order when every letter preceding it is of lower alphabetical value and every letter after it is higher.
having trouble with the combinatorial approach because there seems to be no easy way to avoid over-counting possibilities.
Example of none in order:

dbac
d: bac should precede it - not in order
b: a should precede it and cd should come after - not in order
a: bac should come after - not in order
c: ab should precede it, d should come after it - not in order


Let's replace letters with numbers.
Without loosing in generality we can assume that they are the number $1,2,\cdots,n$.
We call $P(n)$ the sought number of permutations in which none of the numbers are ordered, according to your definition.    
Then we shall have
$$ \bbox[lightyellow] {   
\eqalign{ 
  & P(n) = {\rm N}{\rm .}\,{\rm of}\,{\rm permutations}\;{\rm of}\left[ {1, \cdots ,n} \right]\;:  \cr  
  & 1 = \prod\limits_{1\, \le \,k\, \le \,n} {\neg \left( {\prod\limits_{1\, \le \,j\, \le \,k - 1} {\left[ {x_{\,k - j}  < x_{\,k} } \right]} \prod\limits_{1\, \le \,j\, \le \,n - k} {\left[ {x_{\,k}  < x_{\,k + j} } \right]} } \right)}  =   \cr  
  &  = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {1 - \left( {\prod\limits_{1\, \le \,j\, \le \,k - 1} {\left[ {x_{\,j}  < x_{\,k} } \right]} \prod\limits_{k + 1\, \le \,j\, \le \,n} {\left[ {x_{\,k}  < x_{\,j} } \right]} } \right)} \right)}  \cr}  
} \tag{1}$$
where $[X]$ denotes the Iverson bracket.    
Taking the complement of the above
$$ \bbox[lightyellow] {   
\eqalign{ 
  & Q(n) = n! - P(n) = {\rm N}{\rm .}\,{\rm of}\,{\rm permutations}\;{\rm of}\left[ {1, \cdots ,n} \right]\;:  \cr  
  & 1 = \neg \prod\limits_{1\, \le \,k\, \le \,n} {\neg \left( {\prod\limits_{1\, \le \,j\, \le \,k - 1} {\left[ {x_{\,k - j}  < x_{\,k} } \right]} \prod\limits_{1\, \le \,j\, \le \,n - k} {\left[ {x_{\,k}  < x_{\,k + j} } \right]} } \right)} \quad  \Rightarrow   \cr  
  &  \Rightarrow \quad 0 < \sum\limits_{1\, \le \,k\, \le \,n} {\left( {\prod\limits_{1\, \le \,j\, \le \,k - 1} {\left[ {x_{\,j}  < x_{\,k} } \right]} \prod\limits_{k + 1\, \le \,j\, \le \,n} {\left[ {x_{\,k}  < x_{\,j} } \right]} } \right)}  \cr}  
} \tag{2}$$  
For a single product in the sum above to be greater than $0$, we need that all the terms lower than $x_k$ be before it, and all the higher ones come after it.
That means a permutation into two "separated" (non-overlapping) cycles.
In the matrix representation of the permutation it means that $x_k$ divides the matrix into two blocks which are permutations in their own. And since the permutation matrix has only a $1$ for each row and each column
then $x_k$ must be a fixed point.
We get the situation represented in the sketch below.   

To count $Q(n)$ avoiding over-counting, we put that $x_k$ be the first fixed point. The block above is therefore accounted by $P(k-1)$,
while the block below is a general permutation. So
$$ \bbox[lightyellow] {  
Q(n) = n! - P(n) = \sum\limits_{1\, \le \,k\, \le \,n} {\left( {n - k} \right)!\;P(k - 1)}  = \sum\limits_{0\, \le \,j\, \le \,n - 1} {\left( {n - 1 - j} \right)!\;P(j)} 
} \tag{3}$$
which means
$$ \bbox[lightyellow] {  
P(n) = n! - \sum\limits_{0\, \le \,j\, \le \,n - 1} {\left( {n - 1 - j} \right)!P(j)} 
} \tag{4}$$
This reproduce the sequence already indicated by Marko.
