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

 A: This is more of a comment. To help get started with the study of these
numbers  we compute  them  by  inclusion-exclusion, which  immediately
yields
$$S_n = \sum_{q=0}^n (-1)^q [z^{n-q}]
\left(0!\times z^0 + 1!\times z^1 + 2! \times z^2
+ \cdots + n!\times z^n\right)^{q+1}.$$
This is the sequence
$$1, 0, 1, 3, 14, 77, 497, 3676, 30677, 285335, 2928846, \ldots$$
which  is OEIS  A052186, where  additional
references await.
There is  this Maple  code to help  clarify the  problem definition
that was used.

with(combinat);

ENUM :=
proc(n)
option remember;
local res, perm, pos, stfx, idx;

    res := 0;
    perm := firstperm(n);

    while type(perm, `list`) do

        for pos to n do
            for idx to pos-1 do
                if perm[idx] > perm[pos] then
                    break;
                fi;
            od;

            if idx = pos then
                for idx from pos+1 to n do
                    if perm[idx] < perm[pos] then
                        break;
                    fi;
                od;

                if idx = n+1 then
                    break;
                fi;
            fi;
        od;

        if pos = n+1 then
            res := res + 1;
        fi;

        perm := nextperm(perm);
    od;

    res;
end;

S := n ->
add((-1)^q*coeff(expand(add(p!*z^p, p=0..n)^(q+1)),
                 z, n-q), q=0..n);

Addendum. The  inclusion-exclusion here works on  a poset whose
nodes represent  permutations having a given  set $Q$ or more  of what
the OEIS calls  strong fixed points.  The weights on  the nodes of the
poset are $(-1)^{|Q|}$.   Now the permutations having  no strong fixed
points are only  included in the bottom node of  the Hasse diagram for
zero  or more  strong fixed  points and  hence have  weight one.   All
others having exactly the set $P$ where $|P|\ge 1$ of fixed points are
included in the nodes $Q$ for all $Q\subseteq P$ and hence have weight
$$\sum_{Q\subseteq P} (-1)^{|Q|}  
= \sum_{q=0}^{|P|} {|P|\choose q} (-1)^q = 0,$$
which is precisely the required weight. Observe that the set of strong
fixed  points  in  $Q$  places   the  following  restrictions  on  the
permutations. First, the strong fixed points are ordinary fixed points
and must be  in place. And second,  the values that go  into the $q+1$
gaps between them  including leading and trailing  gaps are determined
by the these strong points and may  be permuted any way we like.  This
is what the generating function  from the introduction does, we choose
the lengths  of the $q+1$ gaps  which determines the $q$  strong fixed
points  where  the  total  length  of the  gaps  must  be  $n-q.$  The
multiplier $p!$ on the coefficient $[z^p]$  in the sum being raised to
the power  $q+1$ then  accounts for the  possible permutations  of the
values in the corresponding gap. In  the end the factor $(-1)^q$ gives
the weight of the node in the poset.
A: 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.
