There may well be a more elegant approach than this answer, which uses Inclusion-Exclusion. See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
This answer will also employ Stars and Bars analysis. See
this article and
this one.
Re-label the socks 1,1,2,2,3,3,4,4,5,5.
Using the syntax from the second Inclusion-Exclusion link, for $k \in \{1,2,3,4,5\}$, let $S_k$ denote the subset of arrangements that feature socks $(k,k)$ together.
Then, the desired probability is
$$\frac{|S_1 \cup S_2 \cup \cdots \cup S_5| \times 2^5}{(10)!}. \tag1 $$
Therefore, the problem reduces to computing
$$|S_1 \cup S_2 \cup \cdots \cup S_5|.$$
Following the syntax in the 2nd link, for $r \in \{1,2,3,4,5\}$, let $T_r$ denote the $~\displaystyle \binom{5}{r}~$ terms given by
$\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 5} \left[S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}\right].$
Then, in accordance with Inclusion-Exclusion theory,
$$|S_1 \cup S_2 \cup \cdots \cup S_5| = \sum_{r=1}^5 (-1)^{r+1}T_r. \tag2 $$
$\underline{\text{Computation of} ~T_1}$
First compute $|S_1|$. There are $(9)$ different positions for the pair $(1,1)$. So, you start with a factor of $(9)$, and then assume that pair $(1,1)$ start in position $(1).$
Now, the other $4$ pairs can occur anywhere in positions $(3)$ through $(10)$. This generates a factor of $~\displaystyle \frac{8!}{2^4}.$
Therefore, $~\displaystyle |S_1| = 9 \times \frac{8!}{2^4}.$
By symmetry, $|S_1| = |S_2| = \cdots = |S_5|$.
Therefore,
$$T_1 = 5 \times 9 \times \frac{8!}{2^4} = 113400. \tag3 $$
$\underline{\text{Computation of} ~T_2}$
First compute $|S_1 \cap S_2|$. This computation will have two parts. The first part will use Stars and Bars Theory to determine the number of ways of positioning pairs (1,1) and (2,2). Then, a second factor will enumerate the number of ways of positioning the remaining 3 pairs of socks.
The two pairs, (1,1) and (2,2) can be permuted in $(2!)$ ways. So, start with a factor of $(2!)$, and now assume that (1,1) precedes (2,2).
Consider the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 = 6 ~: ~\binom{8}{2} = 28 ~~\text{solutions}.$$
In the above equation, the three variables refer to the regions before (1,1), between (1,1) and (2,2), and after (2,2). So, you now have an additional factor of $(28)$.
Therefore, in the computation of $|S_1 \cap S_2|$, the first part of the computation yields:
$$(2!) \times (28).$$
For each of the $(56)$ ways of positioning (1,1) and (2,2), there will be $(6)$ remaining positions for the $(6)$ remaining socks. This generates a factor of $~\displaystyle \frac{6!}{2^3}.$
Therefore,
$$|S_1 \cap S_2| = (2!) \times \binom{8}{2} \times \frac{6!}{2^3}.$$
By symmetry, each of the other $~\left[\binom{5}{2} - 1\right]~$ terms in the enumeration of $~T_2~$ will have the same computation.
Therefore,
$$T_2 = (2!) \times \binom{8}{2} \times \frac{6!}{2^3} \times \binom{5}{2} = 50400. \tag4 $$
$\underline{\text{Computation of} ~T_3}$
The analysis in this section will parallel the analysis in the previous section. To compute $|S_1 \cap S_2 \cap s_3|$, you start with a factor of $(3!)$.
Then, you consider the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 + x_4 = 4 ~: ~\binom{7}{3} ~~\text{solutions}.$$
Then, you apply a factor of $~\displaystyle \frac{4!}{2^2}~$ to account for the $(2)$ remaining pairs of socks.
Therefore,
$$|S_1 \cap S_2 \cap S_3| = (3!) \times \binom{7}{3} \times \frac{4!}{2^2}.$$
By symmetry, each of the other $~\left[\binom{5}{3} - 1\right]~$ terms in the enumeration of $~T_3~$ will have the same computation.
Therefore,
$$T_3 = (3!) \times \binom{7}{3} \times \frac{4!}{2^2} \times \binom{5}{3} = 12600. \tag5 $$
$\underline{\text{Computation of} ~T_4}$
The analysis in this section will parallel the analysis in the previous sections. You start with a factor of $(4!)$.
Then, you consider the number of non-negative integer solutions to
$$x_1 + x_2 + x_3 + x_4 + x_5 = 2 ~: ~\binom{6}{4} ~~\text{solutions}.$$
Then, you apply a factor of $~\displaystyle \frac{2!}{2^1}~$ to account for the $(1)$ remaining pairs of socks.
Then, you apply the factor of $~\displaystyle \binom{5}{4}.$
Therefore,
$$T_4 = (4!) \times \binom{6}{4} \times \frac{2!}{2^1} \times \binom{5}{4} = 1800. \tag6 $$
$\underline{\text{Computation of} ~T_5}$
There is no wiggle room here, since the 5 pairs (in some order) must specifically go into positions [1,2], [3,4], ..., [9,10].
Therefore,
$$T_5 = 5! = 120.\tag 7$$
$\underline{\text{Final Computations}}$
$$|S_1 \cup \cdots \cup S_5| = \left[T_1 + T_3 + T_5\right] - \left[T_2 + T_4\right]$$
$$= \left[113400 + 12600 + 120\right] - \left[50400 + 1800\right] = 73920.$$
Therefore, the desired probability is
$$\frac{73920 \times 2^5}{(10)!} = \frac{88}{135}.$$
