I do not know how to correctly interpret Hilbert's Irreducibility theorem
with Galois group as my aim.


Here $K$ is a number field (or simply $\mathbf{Q}$).


Scenario 1: Take a field $L$ that is a finite Galois extension of $K(t)$ ($t$ an indeterminate) with Galois group $G$. Writing $L=K(t)[X]/(f(t,X))$ for
an irreducible polynomial $f(t,X)\in K(t)[X]$, and taking a specialization
$t=a\in K$ guaranteed by Hilbert we can see the Galois group descends
and we get a $G$-Galois extension over the number field $K$ as $K[X]/(f(a,X))$.


I understand this situation well.


Scenario 2: Instead of a $G$-Galois extension we are merely provided
with an irreducible polynomial whose SPLITTING FIELD has
$G$ as Galois group, so the {\it degree of the polynomial can be less than
the order of $G$.} 


I took the following example from Malle and Matzat's book on Inverse Galois Theory. (Page 88, attributed to Beckman). (Instead of a general degree $n$ I take $n=3)$.


He claims
$f(t,X) = X^3-3tX +2t \in \mathbf{Q}(t)[X]$ is irreducible with $S\_3$ as Galois group.
(of its splitting field).


For the special value $t=4$ we get the irreducible polynomial $X^3-12X+8$, but the discriminant is a square (of 72) and we get a cubic number field as splitting field and not the expected $S\_3$ extension of $\mathbf Q$.


What mistake am I making in this scenario?


Instead of giving a degree $n$-polynomial in $K(t)[X]$ with $S\_n$ as Galois group I would be more comfortable 
with an irreducible polynomial of degree $n!$ with $S\_n$ as Galois group so that I can specialize that polynomial. Perhaps it is expecting too much.



How to apply  Hilbert's Irreducibilty theorem? 
In ``Scenario 2'', you have to take a minimal polynomial $g(t,X)$ of a primitive element of the splitting field of $f(t,X)$ over $K(t)$. Then $g(a,X)$ is irreducible for infinitely many $a\in K$ by Hilbert's irreducibility theorem, and $g(a,X)$ and $g(t,X)$ have the same Galois group over $K$ and $K(t)$, respectively.


So preservation of Galois groups follows indeed from Hilbert's irreducibility theorem, but one has to apply it to the correct polynomial.


As to the last paragraph of the question: I doubt that there is a simply shaped polynomial of degree $n!$ with Galois group $S\_n$ over the rationals. But maybe this one is good enough for the OP's purpose: $X^n-X-1$ has Galois group $S\_n$ over $\mathbb Q$, see page 42 in Serre's *Topics in Galois Theory*. The proof isn't easy, even seeing that this polynomial is irreducible (as proved by Selmer in the 50s) requires a trick. 


