Calculating the probability of a function being real when it has random variables I'm trying to calculate the probability of 
$ax^2+bx+c$ root being a real number when the variables $a,$ $b,$ and $c$ values are all randomized by throwing a standard die. 
I got to the point where I can get the probability by calculating the chance of $b^2-4ac>0$, but I'm not sure how I can conveniently carry on from here and my attempts at doing it by hand (finding every possible real occurrence and calculating them by the total outcomes) have failed me. 
In other words, the values of $a$, $b$ and $c$ are within $\{1,2,3,4,5,6\}$ and each of the three variables is randomly picked from that list with no special weighting (so $\frac 16$  chance to get any of the $6$ values). 
 A: There are $43$ triples $(a,b,c)$ with $b^2 - 4 a c \ge 0$,  so the probability is $43/6^3$.  Note that you do want $\ge$, not just $>$, unless you just want two distinct roots.
The easiest way to compute these by hand, I think, is to consider the possible values of $ac$.
A: We need $b^2 - 4ac \ge 0$.
If $b= 6$ then $ac \le 9$.  If $a =1$ then $c = 1...6$.  If $a=2$ then $c=1...4$.  If $c=3; a= 1..3$ if $a=4; c=1,2$ if $a=5,6;c =1$.  There are $6+4+3+2+1+1 = 17$ ways to do this.
If $b = 5$ then $ac \le 6$.  If $a=1$ then $c = 1...6$ if $a= 2$ then $b=1..3$ and if $a = 3; c=1,2$ and if $a \ge 4; c=1$.  So $6+3+2+3 = 14$ ways to do this.
If $b = 4$ then $ac \le 4$  So if $a= 1,2,3,4$ then $c \le 4,2,1,1$ so there are $4+2+1+1 = 8$ ways to do that
If $b =3$ then $ac \le 2$ So if $a= 1,2$ then $c \le 2,1$ so there are $2+1=3$ wways to do that.
If $b = 2$ then $ac \le 1$ so $a=c=1$.  $1$ way to do that.
$b$ can not be $1$ because $4ac \ge 4 > 1^2$.
So there are $17+14+8+3+1 = 43$ ways to do this.
So the probability is $\frac {43}{6^3} \approx 20\%$ (very slightly less.)
