Analysis. A dice problem

"Here is an oft-posed question. Can you construct a pair of dice other than the usual ones with 1, 2, 3, 4, 5, 6 on the faces so that the resultant probabilities are the same? First, we construct the usual addition chart and its frequency chart."

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

sum	2	3	4	5	6	7	8	9	10	11	12
freq	1	2	3	4	5	6	5	5	3	2	1

"The entry 6, for example, indicates that there are 6 ways to achieve a total of 7.

Now suppose you have two die with the following face numbers: 1, 2, 2, 3, 3, 4 on one die and 1, 3, 4, 5, 6, 8 on the other die. As before we will construct the addition table and its frequency chart."

+	1	3	4	5	6	8
1	2	4	5	6	7	9
2	3	5	6	7	8	10
2	3	5	6	7	8	10
3	4	6	7	8	9	11
3	4	6	7	8	9	11
4	5	7	8	9	10	12

sum	2	3	4	5	6	7	8	9	10	11	12
freq	1	2	3	4							1

"The reader should complete this frequency chart to discover a surprising result.

Why did this happen? Will this happen with other dice configurations? Let's look at the underlying algebra."

"The generating function x¹ + x² + x³ + x⁴ + x⁵ + x⁶gives the outcomes of one die ; just look at the exponents. For two dice look at the product (or square):"

(x¹ + x² + x³ + x⁴ + x⁵ + x⁶)² =

1x² + 2x³ + 3x⁴ + 4x⁵ + 5x⁶ + 6x⁷ + 5x⁸ + 4x⁹ + 3x¹⁰ + 2x¹¹ +x¹².

"Those coefficients are exactly the numbers in the frequency chart for a pair of regular dice! On a craps table in Las Vegas 7 is a special outcome when tossing a pair of dice; there are 6 ways it can happen:"

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).

"An alternate factorization of our 12^th degree polynomial could reveal a "new dice" pair. Start with

x + x² + x³ + x⁴ + x⁵ + x⁶ =

x (1 + x + x² + x³ + x⁴ + x⁵) = x () = x(1 + x³) = x (1 + x + x²)(1 + x)(1 - x + x²).

Remember this is just for one die. A "new pair" of factors must multiply to the square of this 6^th degree polynomial or in other words, to"

x²(1 + x + x²)(1 + x)²(1 - x + x²)² .

"So we have eight factors to allocate to two die, keeping in mind that the factors have to behave. Each of the two ways of factoring must have an x to maintain positive outcomes. Each factor must have a 1 + x and 1 + x + x² so that you have 6 faces, by letting x = 1. So there are not many options. In fact, there are just two ways of allocating the remaining two factors: 1 - x + x² and 1 - x + x². Here they are:"

1. Allocate one to each die

x (1 + x + x²)(1 + x)(1 - x + x²) = x + x² + x³ + x⁴+ x⁵ + x⁶

and the same for the second die

1. Allocate both to one die

x (1 + x + x²)(1 + x)( 1 - x + x²)² = x + x³ + x⁴ + x⁵ + x⁶+ x⁸

and the second die gets

x (1 + x + x²)(1 + x) = x + 2x² + 2x³ + x⁴

"The first die has 1, 3, 4, 5, 6, 8 on its faces, and the second has 1, 2, 2, 3, 3, 4 on its six faces.

This new pair of dice is called Sicherman (1978 issue of Scientific American - Martin Gardner)dice or "weird dice".

Our tables show that rolling a particular sum is the same using either pair of dice. But this is not so when rolling doubles. With the usual pair of dice there are 6 ways to roll a double: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6), so the probability is 6/36. The "new pair" has just 4 ways to roll a double:

(1, 1), (3, 3), (3, 3) and (4, 4). The new probability is 4/36.

Also, if you allow 0 to be on a face the following pairs are produced:

011223 and 245679 ; 233445 and 023457."

Question

Label the four faces of a pair of dice in the shape of a regular tetrahedron with the integers 1, 2, 3, 4. Make both tables and reproduce the analysisabove.