Probability that no letter is in alphabetical order

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.

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.