What is the expected determinant of a symmetric $2\times2$ matrix, whose three elements are distinct and draw from$ [-n,n]$ 
The symmetric matrix has three elements:
  \begin{pmatrix} A&B\\ B&C \end{pmatrix}
  $A,B$ and $C$ are integers draw randomly from $\{-n,-n+1,\dots,n-1,n\}$, $n\ge2$ and they are distinct with each other. What is the expected determinant of this matrix?

I have no clue how to do this, this seems multiple random variables, I don't know how to attack this problem systematically. 
Edit
Computer run give the result of $n =[2,10]$:
-2.49301 -4.66934 -7.52694 -11.0413 -15.18877 -19.97767 -25.64429 -31.6045 -38.66689
Therefore we have a correct answer which is $$\displaystyle-\frac{(n+1)(2n+1)}6$$ However, an more easier to understand answer is still very welcomed.
Code:
import random
N = 100000
for n in range(2,11):
    total = 0
    for i in range(0,N):
        a,b,c=random.sample(range(-n,n+1),3)
        total += a*c-b*b
    print(total/N,end = ' ')

 A: You may calculate $E(AC)$ without thinking of $B$ (you first pick randomly $A,C$ distinct and don't worry about $B$, the first double sum below is zero):
$$ E(AC) =\frac{1}{(2n+1)2n} \left( \sum_{a=-n}^n \sum_{c=-n}^n ac - \sum_{a=-n}^n  a^2\right) =  -\frac{1}{(2n+1)2n}\frac{2n(n+1)(2n+1)}{6}= -\frac{n+1}{6}$$
and  then for $E(B^2)$ you need not worry about $A,C$, so:
$$ E(B^2) =\frac{1}{(2n+1)} \left( \sum_{a=-n}^n  b^2\right) =  \frac{1}{(2n+1)}\frac{2n(n+1)(2n+1)}{6}= \frac{2n(n+1)}{6}$$
Finally (with a non-zero probability of having done some errors along the way):
$$E(\det S)= E(AC-B^2)=-\frac{(n+1)(1+2n)}{6}$$
A: The problem is in the notation. When you write $B$ you actually mean $B$ conditional on $B\neq A=a$, which is not uniformly distributed on $[-n,n]$ but on $[-n,n]\backslash\{a\}$. Similarly, conditional on the values of $A=a,B=b$ the random variable $C$ is uniformly distributed on $[-n,n]\backslash\{a,b\}$. So, actually when you write $B$ you abuse notation for $B\mid A=a$ and when you write $C$ you abuse notation for $C\mid A=a, B=b$, plus the condition that $B\neq A\neq C$. All that said, you can solve it by conditioning, as is your intuition: 
\begin{align}\mathbb E[AC]-\mathbb E[B^2]&=\sum_{a=-n}^n\left(\mathbb E\left[AC\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)P(A=a)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(a\cdot \mathbb E\left[C\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)\end{align} with 
\begin{align}\mathbb E[B^2\mid A=a]&=\frac1{2n}\sum_{b=-n,b\neq a}^nb^2=\frac{1}{2n}\left(\sum_{b=-n}^nb^2-a^2\right)=\frac{1}{2n}\left(2\sum_{b=-n}^nb^2-a^2\right)\\[0.3cm]&=\frac{1}{2n}\left(2\frac{n(n+1)(2n+1)}{6}-a^2\right)=\frac{(n+1)(2n+1)}{6}-\frac{a^2}{2n}\end{align} and (now we must condition also on $B$):
\begin{align}\mathbb E[C\mid A=a]&=\sum_{b=-n,b\neq a}^n\mathbb E[C\mid A=a,B=b]P(B=b)\\[0.3cm]&=\frac{1}{2n}\sum_{b=-n,b\neq a}^n\left(\sum_{c=-n,c\neq a,b}^n c\cdot\frac{1}{2n-1}\right)=\frac{1}{2n(2n-1)}\sum_{b=-n,b\neq a}^n\left(\sum_{c=-n}^nc-b-a\right)\\[0.3cm]&=\frac{1}{2n(2n-1)}\sum_{b=-n,b\neq a}^n(0-b-a)=-\frac{1}{2n(2n-1)}\left(\sum_{b=-n}^n(b+a)-(a+a)\right)\\[0.3cm]&=-\frac{1}{2n(2n-1)}\left(0+(2n+1)a-2a\right)=-\frac{a}{2n}\end{align} And now, we substitute in the first equation to get 
\begin{align}\mathbb E[AC]-\mathbb E[B^2]&=\sum_{a=-n}^n\left(\mathbb E\left[AC\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\right)P(A=a)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^na\cdot \mathbb E\left[C\mid A=a\right]-\mathbb E\left[B^2\mid A=a\right]\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(a\left(-\frac{a}{2n}\right)-\frac{(n+1)(2n+1)}{6}+\frac{a^2}{2n}\right)\\[0.3cm]&=\frac{1}{2n+1}\sum_{a=-n}^n\left(\frac{a^2}{2n}-\frac{a^2}{2n}-\frac{(n+1)(2n+1)}{6}\right)\\[0.3cm]&=\frac{1}{2n+1}\cdot(2n+1)\cdot\frac{(n+1)(2n+1)}{6}=-\frac{(n+1)(2n+1)}{6}\end{align} 
A: If we do not require that $A\neq B\neq C$, then the random variables $A,B,C$ are independent and distributed discrete uniformly on $\{-n,-n+1,\dots, -1, 0, 1, \dots, n-1,n \}$. Hence, $$\mathbb E[A]=\mathbb E[B]=\mathbb E[C]=\frac{n+(-n)}{2}=0$$ and $$Var(A)=Var(B)=Var(C)=\frac{(n-(-n)+1)^2-1}{12}=\frac{(2n+1)^2-1}{12}$$
Now, if we denote the symmetric matrix with $S$, we have $$\det(S)=AC-B^2$$ hence
\begin{align}\mathbb E[\det(S)]&=\mathbb E[AC-B^2]=^{A,B,C \text{ are independent} }\\[0.2cm]&=\mathbb E[A]\mathbb E[C]-\mathbb E[B^2]=0\cdot0-\mathbb E[B^2]=-\mathbb E[B^2]\end{align} Recall that 
\begin{align}Var(B)&=\mathbb E[B^2]-\left(\mathbb E[B]\right)^2\\[0.2cm] \iff-\mathbb E[B^2]&=-Var(B)-\left(\mathbb E[B]\right)^2=-\frac{(2n+1)^2-1}{12}-0=\frac{n(n+1)}{3}\end{align} to obtain 
\begin{align}\mathbb E[\det(S)]=\frac{n(n+1)}{3}\end{align}
A: I think the beginning should be like Jimmy R started : by linearity of expectation,
$$\mathbb E(AC-B^2)=\mathbb E(AC)-\mathbb E(B^2)$$
Now $A$ and $C$ are not independent ($\mathbb P(A=i)=\mathbb P(C=j)=\frac{1}{2n+1}$, but $\mathbb P(A=i \wedge C=j)=\frac{1}{2n(2n+1)}$), but as $\mathbb P(A=i \wedge C=j)=\mathbb P(A=-i \wedge C=j)$, it should be clear that $\mathbb E(AC)=0$.
Now for $\mathbb E(B^2)$, I find :
$$\mathbb E(B^2)=\sum_{k=-n}^n k^2\mathbb P(B=k)=\frac{2}{2n+1}\sum_{k=1}^n k^2=\frac{2}{2n+1}\frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)}{3}$$
So if my reasoning is correct, 
$$\mathbb E\left(\det\begin{pmatrix} A & B \\ B & C \end{pmatrix}\right)=-\frac{n(n+1)}{3}$$
for the given distribution of values $A$, $B$ and $C$.
