Probability problem about dice ( I need help ASAP!)

[megadeth95]

Hello guys, I really need some help with this problem, I'm completely lost and don't know what to do.


Your problem is to design a pair of dice. The rules are:

1. The dice must be cubic in shape ( regular-shaped dice with 6 faces each)
2. Use only 2 dice
3. Every whole number 1,2,3,4,5,6,7,8,9,10,11,12 must be able to occur as a sum when the two dice are tossed.
4. Each of the whole numbers 1,2,3,4,5,6,7,8,9,10,11,12 must have the same probabilty of occurring.

The only thing that I did so far is the first 2 bullets, and I think the probability of occurring must be 1/36.

Thanks for your help guys, I really appreciate it

Megadeth95

 

[soroban]

Hello, megadeth95!

Please give us the original wording of the problem.
As given, it makes no sense.

Your problem is to design a pair of dice. The rules are:

. . The dice must be cubic in shape (regular-shaped dice with 6 faces each).
. . Use only two dice.
. . Every whole number 1,2,3,4,5,6,7,8,9,10,11,12 must be able to occur
. . . . as a sum when the two dice are tossed. [1]
. . Each of the whole number 1,2,3,4,5,6,7,8,9,10,11,12
. . . . must have the same probability of occurring. [2]

The only thing that I did so far is the first 2 bullets, "bullets?"
and I think the probability of occuring must be 1/36. . The probability of WHAT occuring?

[1] .With a pair of standard dice the possible sums are: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

To get a sum of 1, one die must have a zero (0) and the other must have a 1.

The two dice could be: .\(\displaystyle \begin{bmatrix}\{0,2,3,4,5,6\} \\ \{1,2,3,4,5,6\} \end{bmatrix}\)


[2] .Are these whole numbers the sum of the two dice?

If so, then we want: .[\(\displaystyle \begin{bmatrix}P(\text{sum of 1}) &=& \frac{1}{12}\\ P(\text{sum of 2}) &=& \frac{1}{12} \\ P(\text{sum of 3}) &=& \frac{1}{12} \\ \vdots &\quad &\vdots \\ P(\text{sum of 12}) &=& \frac{1}{12}\end{bmatrix}\)

And that is a very tall order . . .

 

[tkhunny]

Multiply these polynomials and see what you get:

\(\displaystyle \left(1+x+x^{2}+x^{6}+x^{7}+x^{8}\right)\cdot\left(3x+3x^{4}\right)\)

Ponder These Dice

[0,1,2,6,7,8]
[1,1,1,4,4,4]

Factoring the Polynomials, you may be able to create other solutions:

\(\displaystyle \left[\left(1+x^{2}\right)\cdot\left(1+x+x^{2}\right)\cdot\left(1-x^{2}+x^{4}\right)\right]\cdot \left[3x\cdot\left(1+x\right)\cdot\left(1-x+x^{2}\right)\right]\)

Of course, you still get only six faces, so don't go too crazy.

Keep this in mind.

\(\displaystyle \left[(2)(3)(1)\right]\cdot \left[3(1)(2)(1)\right]\)

This means you cannot move the first two on the left or the middle one or the constant on the right. Try moving the other terms around and see what you get.

Wasn't that fun?!

On second thought, maybe you can move a (2), but not unilaterally. You have to swap it with the other (2). Same for the (3)s.