Probability of Sum of Values in throwing dice

 

Problem

 

          Suppose we throw 3 dice. What is the probability of getting a sum of k, where  3 £ k £ 18 ?

 

 

Number of arrangements

 

          It is not queer coincidence, but the number of arrangements for getting a sum of k corresponds to coefficient of xk in the expansion of :

 

                  (x6 + x5 + x4 + x3 + x2 + x)3

          =      x18 + 3x17 + 6x16 + 10x15 + 15x14 + 21x13 + 25x12 + 27x11 + 27x10 +

                  25x9 + 21x8 + 15x7 + 10x6 + 6x5 + 3x4 +x3

 

The sum of all the coefficients

= 1 + 3 + 6 + 10 + 15 + 21 + 25 + 27 + 27 +25 + 21 + 15 + 10 + 6 + 3 +1

= 216

= 23   which is the total number of arrangements in throwing 3 dice.

 

\             The probabilities of getting a sum of 3, 4, 5, …., 17, 18 are

         

 

 

Why it works?

 

          The key reason is the index law :  xm xn = xm+n. Getting a sum of k corresponds nicely with the sum of indices of x in expanding the product, and we therefore look for the coefficients. Investigate yourselves.

 

 

Getting the coefficients

 

          It is tedious to get the coefficients, here is a possible easy way:

         

Row A

1

1

1

1

1

1

 

 

 

 

 

 

 

 

 

 

Row B

1

2

3

4

5

6

5

4

3

2

1

 

 

 

 

 

Row C

1

3

6

10

15

21

25

27

27

25

21

15

10

6

3

1

 

          Row A are the coefficients of  x6 + x5 + x4 + x3 + x2 + x.

          Row B are the coefficients in the expansion of (x6 + x5 + x4 + x3 + x2 + x)2.

         Each number can be got by adding six numbers just above that number and then to the left. You may assume the value is 0 if there is an empty cell or no such number exists.

          Similarly Row C are the coefficients in the expansion of (x6 + x5 + x4 + x3 + x2 + x)3.

          Each number can be got from Row B in the same way.

          As an example, 5 + 6 + 5 + 4 + 3 + 2 ( in yellow ) = 15 ( in pink )

 

 

The Prime dice

 

          The face values for normal dice are  1, 2, 3, 4, 5, 6.  However, we can extend this method to other face values. Suppose we have dice with prime values : 2, 3, 5, 7, 11, 13. Then we can get the number of arrangements in getting all possible sums. If we throwing two of such prime dice, the coefficients in the expansion of :

                  (x13 + x11 + x7 + x5 + x3 + x2)2

          =     x26 + 2x24 + x22 + 2x20 + 4x18 + 4x16 + 2x15 + 3x14 + 2x13 +

                  2x12 + 3x10 + 2x9 + 2x8 + 2x7 + x6 + 2x5 + x4

 

          Note that the number of arrangement to get a sum of 23 is 0.

As an example, the number of arrangements to get a sum of 18 is 4 (coefficient of x18-term)

The probability of getting a sum of 18 in throwing 2 prime dice = 4/36.

 

 

Polyhedron dice

 

          From geometry, we know that there are exactly 5 regular polyhedra which are collectively know as the Platonic Solids:

          Tetrahedron (4 faces with equilateral triangles),

          Octahedron (8 faces with equilateral triangles),

          Cube (6 faces with squares),

          Icosahedron (20 faces with equilateral triangles)

          Dodecahedron (12 faces with regular pentagons)

 

          If you are throwing 3 fair tetrahedron dice with face values 1, 2, 3, 4, the number of arrangements of the sum can be got from the coefficients in the expansion:

 

          (x4 +x3 + x2 + x)3.