7

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.

Martin Sleziak
  • 53,687
user66733
  • 7,379
  • 3
    I am interested in the generalization to this problem where $a,b,c \in {1,2,\dots, N}$ for arbitrary $N$. – abnry Dec 25 '13 at 02:19
  • @nayrb: Yeah, post it as a separate question. I guess it could be a challenging problem. – user66733 Dec 25 '13 at 02:58
  • For the usual dice, you are expected to count, as $6^3=216$ is few enough to count, especially if you use the symmetries. For $N$, a program can count up to (tens or hundreds) of thousands in reasonable time. Above that you can approximate it well by the volume of the intersection of a paraboloid and a cube. This will take some calculus, but can be rendered as a single integral if you use the symmetry. – Ross Millikan Dec 25 '13 at 03:51
  • @nayrb: The OEIS doesn't know anything about the generalization (or any obvious variants on it), so there's probably no representation that's much nicer than Ross's approximation... – Micah Dec 25 '13 at 21:24
  • @Micah I guess this is akin to the Gauss circle problem, which is unsolved. – abnry Dec 25 '13 at 21:34
  • Not proposing to close, but just to link to the current choice of node for essentially the same math question. – Lee David Chung Lin Feb 13 '20 at 10:49

4 Answers4

6

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.

Berci
  • 90,745
  • Great idea! I'll try this right now Berci. – user66733 Dec 25 '13 at 01:02
  • OK Berci. It worked. But do you know any other way of counting them without bruteforce? – user66733 Dec 25 '13 at 01:26
  • I did the same table. I do not know how to draw a table here. So I copied each row of the table. Also, in the table I looked for equations with real roots (there are fewer). If you look at the order of entries, you can easily reconstruct the table. – user44197 Dec 25 '13 at 03:08
2

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} \ . $

enter image description here

colormegone
  • 10,842
1

Following up on Berci's answer:

You can count it out simply and arrive at the answer: $\frac{89}{108}$.

Added later: I had made an error in my counting! The answer is $\frac{43}{6^3}$

Here are the 43 possibilities: [[1,2,1,0],[1,3,1,5],[1,4,1,12],[1,5,1,21],[1,6,1,32],[ 1,3,2,1],[1,4,2,8],[1,5,2,17],[1,6,2,28],[1,4,3,4],[1,5,3,13], [1,6,3,24],[1,4,4,0],[1,5,4,9],[1,6,4,20],[1,5,5,5],[1,6,5,16] ,[1,5,6,1],[1,6,6,12],[2,3,1,1],[2,4,1,8],[2,5,1,17],[2,6,1,28 ],[2,4,2,0],[2,5,2,9],[2,6,2,20],[2,5,3,1],[2,6,3,12],[2,6,4,4 ],[3,4,1,4],[3,5,1,13],[3,6,1,24],[3,5,2,1],[3,6,2,12],[3,6,3, 0],[4,4,1,0],[4,5,1,9],[4,6,1,20],[4,6,2,4],[5,5,1,5],[5,6,1, 16],[6,5,1,1],[6,6,1,12]]

Added even more later [I need sleep!] Those 43 values give real roots. So the probability of no real roots is $1-43/216 = 173/216$.

user44197
  • 9,730
  • The final answer is apparently $\frac{173}{216}$. You've made a mistake somewhere I guess. I counted it twice, since there is $1$ at $(1,1)$ in the table the final answer must be some odd number over $216$, so there is no way that you can simplify the fraction by a factor of two. – user66733 Dec 25 '13 at 01:22
  • let me double check – user44197 Dec 25 '13 at 02:03
  • Here is the complete list. Each entry is a, b, c, b^2-4 a c: [[1,2,1,0],[1,3,1,5],[1,4,1,12],[1,5,1,21],[1,6,1,32],[ 1,3,2,1],[1,4,2,8],[1,5,2,17],[1,6,2,28],[1,4,3,4],[1,5,3,13], [1,6,3,24],[1,4,4,0],[1,5,4,9],[1,6,4,20],[1,5,5,5],[1,6,5,16] ,[1,5,6,1],[1,6,6,12],[2,3,1,1],[2,4,1,8],[2,5,1,17],[2,6,1,28 ],[2,4,2,0],[2,5,2,9],[2,6,2,20],[2,5,3,1],[2,6,3,12],[2,6,4,4 ],[3,4,1,4],[3,5,1,13],[3,6,1,24],[3,5,2,1],[3,6,2,12],[3,6,3, 0],[4,4,1,0],[4,5,1,9],[4,6,1,20],[4,6,2,4],[5,5,1,5],[5,6,1, 16],[6,5,1,1],[6,6,1,12]]. There are 43 of them – user44197 Dec 25 '13 at 02:09
  • I counted it this way: I drew a table as Berci suggested. In the ij-th entry I put the number of $b$'s that satisfy $b^2<4ac$. Then I used symmetry, so I filled the upper triangular matrix. So, I had $\frac{6 \times 5}{2}$ entries to fill in. Then I summed the diagonal entries and upper non-diagonal entries separately. I doubled the sum of upper non-diagonal entries because of symmetry and added it to the sum of diagonal entries and I arrived at $173$. – user66733 Dec 25 '13 at 02:54
1

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*}

A.G.
  • 2,781