Probability that $ax^2+bx+c$ has no real roots after rolling 3 dice. Suppose that I roll $3$ dice and write down the outcome as the coefficients $a,b,c$ in the polynomial $ax^2+bx+c$ respectively. What is the probability that this polynomial has no real roots?
So, I have to count the number of triples $(a,b,c)$ such that $b^2 < 4ac$, where $a,b,c \in \{1,2,3,4,5,6\}$. I'm not sure how I can do that. Please give me a hint first. This problem is from a high school probability course, so I think it must have a very basic solution.
 A: If you don't have any further idea, probably the easiest is to draw a $6\times 6$ table of values $4ac$ for all possible $a,c$ pairs. (It will be symmetric.) Then you can conclude the number of $b$'s for a given pair $(a,c)$, and sum up all these.
A: A three-dimensional lattice would be preferable for investigating this, but we can get by with a two-dimensional grid.  Since we wish to count the ordered triples for which $ \ b^2 \ < \ 4ac \ $ , we can set up a lattice for $ \ a \ $ and $ \ c \ $  running from 1 to 6  (contained in the light green square), and plot level curves for  $ \ \frac{1}{4}b^2 \ $ with $ \ b \ $ also running from 1 to 6 .  At each integer value of  $ \ b \ $ , we can now easily count the number of lattice points "below" or on each level curve to find the number of combinations which do produce a discriminant value $ \ b^2 - 4ac \ \ge \ 0 \ $ .  (The red diagonal line is added to indicate the symmetry in the choices for  for $ \ a \ $ and $ \ c \ $ , which can simplify the counting.) These are then to be discarded from the count of desired outcomes.
We find that the number of combinations for each value of $ \ b \ $  is
b = 1 :  36
b = 2 :  36 - 1  =  35
b = 3 :  36 - 3  =  33
b = 4 :  36 - 8  =  28
b = 5 :  36 - 14  =  22
b = 6 :  36 - 17  =  19
As found by the other solvers here, the total number of combinations leading to complex-valued zeroes is 173 , making the probability $ \ \frac{173}{216} \ . $

A: Also following up on @Berci's answer, here is a variation that you can do on some spreadsheet. First you set up a 6x6 table with $4ac$ in each cell. For each cell you want the number of b's that are integer, $\leq6$ and such that $b^2<4ac$. As you are working with integers you can replace that with $b^2\leq 4ac-1$, i.e. $b\leq\sqrt{4ac-1}$. That number (of b's) turns out to be the minimum of 6 and $\lfloor \sqrt{4ac-1}\rfloor$ (largest integer $\leq \sqrt{4ac-1}$. Computing that for each cell yields:
\begin{array}{cccccc}
 1 & 2 & 3 & 3 & 4 & 4 \\
 2 & 3 & 4 & 5 & 6 & 6 \\
 3 & 4 & 5 & 6 & 6 & 6 \\
 3 & 5 & 6 & 6 & 6 & 6 \\
 4 & 6 & 6 & 6 & 6 & 6 \\
 4 & 6 & 6 & 6 & 6 & 6 \\
\end{array}
which then sums up to 173, thus the probability of 173/216.
To follow up on @nayrb's question, this generalizes to $n$-face dice in a straightforward way :
\begin{align*}
p(n)=&\ \sum_{a=1}^n\sum_{b=1}^n \min(n,\lfloor \sqrt{4ac-1}\rfloor)
\end{align*}
The limit for $n\rightarrow\infty$ is the continuous case :
\begin{align*}
p(\infty)=&\ \int_{a=0}^1\int_{b=0}^1 \min(1,\sqrt{4ac})= \frac{31-6\log 2}{36}\approx0.745587
\end{align*}
