People arranged in a queue, probability of $A$ always being in front of $B$. There are $n\geq 2$ people including $A$ and $B$. They form a line uniformly at random and we want to find the probability that $A$ is in front of $B$.
So far I've determined my sample space to be $|\Omega|=n!$.
I also let $A$'s position in the queue to be $k$, and determined the number of ways that $A$ is in front of $B$ is $(n-2)!(n-k)$. (Or (n-1)!(n-k) ?) 
Intuitively I know that the number of ways that $A$ is in front of $B$ is the same as that if $B$ is in front of $A$ and the probability I should be getting is $\frac{1}{2}$ but I'm not sure how to get there from the information I have so far.


Intuitively I know that the number of ways that A is in front of B is the same as that if B is in front of A and the probability I should be getting is 1/2 but I'm not sure how to get there from the information I have so far.

Your intuition is the way to go.   The other people are a distraction.   All you care about is the where to place $A$ and $B$.
But if you must consider them, note that $\binom n2(n-2)!$ counts ways to select two from the $n$ places to put $A$ and $B$ such that $A$ precedes $B$, then to arrange the remaining $n-2$ people into the remaining places.
Divide and calculate.$$\mathsf P(A\prec B)=\dfrac{\binom n2(n-2)!}{n!}\\~~=\dfrac{1}{2}$$

PS: and of course $\sum_{k=1}^{n-1}(n-k) =\sum_{j=1}^{n-1} j= \dfrac{n(n-1)}2$
