TITLE: Probability that $ax^2+bx+c$ has no real roots after rolling 3 dice.
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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} \ . $