Crazy Pair of Dice...

Probability Level 5

Lets roll a pair of standard 6-sided dice first.

There is one way of obtaining a 2, two ways of obtaining a 3, and so on, up to one way of obtaining a 12.

Now, let there be a pair of another 6-sided dice A and B . (Both non-standard)

A and B satisfy these properties:

  • Each face has at least one dot.
  • The number of ways of obtaining each sum is the same as for the standard dice.

Let the number of dots on faces of A and B be represented by a i a_i and b i b_i respectively. ( i ∈ { 1 , 2 , 3 , 4 , 5 , 6 } ) (i\in \{1,2,3,4,5,6\})

Calculate ∑ i = 1 6 ( a i 2 + b i 2 ) + ( ∑ i = 1 6 a i ) ( ∑ i = 1 6 b i ) \displaystyle\sum_{i=1}^{6} (a_i^2+b_i^2)+\left (\displaystyle\sum_{i=1}^6 a_i\right )\left (\displaystyle\sum_{i=1}^6 b_i\right ) .


The answer is 599.

Pranjal Jain by Pranjal Jain , 23, India

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1 solution

Eilon Lavi
Feb 19, 2015

We can represent "dice" as polynomials, with x n x^n representing a face with n n dots.

The standard die is represented as

x 6 + x 5 + x 4 + x 3 + x 2 + x x^6+x^5+x^4+x^3+x^2+x

A coin with a side with one dot and a side with two dots is represented as

x 2 + x x^2+x

A die with 3 sides of 5 dots, 2 sides of 4 dots and a side of 2 dots is represented as

3 x 5 + 2 x 4 + x 2 3x^5+2x^4+x^2

When we multiply two dice, we get a representation of all possible sums of the two dice.

( x 6 + x 5 + x 4 + x 3 + x 2 + x ) 2 = x 12 + 2 x 11 + 3 x 10 + 4 x 9 + 5 x 8 + 6 x 7 + 5 x 6 + 4 x 5 + 3 x 4 + 2 x 3 + x 2 (x^6+x^5+x^4+x^3+x^2+x)^2 \\=x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2

So we need to find two other polynomials that multiply to this representation. As additional constraints, these must have all exponents ≥ 1 \ge 1 and the sum of the constants must equal 6 for each polynomial (or else it would not represent a die). Notice that this "sum of constants" property is multiplicative.

( x 6 + x 5 + x 4 + x 3 + x 2 + x ) 2 = x 2 ( x + 1 ) 2 ( x 2 + x + 1 ) 2 ( x 2 − x + 1 ) 2 (x^6+x^5+x^4+x^3+x^2+x)^2 =x^2(x+1)^2(x^2+x+1)^2(x^2-x+1)^2

Now we need to decide how to divvy up these factors to make A and B, the two new dice.

Let S ( ) S() denote the sum of constants.

S ( x ) = 1 S ( x + 1 ) = 2 S ( x 2 + x + 1 ) = 3 S ( x 2 − x + 1 ) = 1 S(x)=1 \\S(x+1)=2 \\S(x^2+x+1)=3 \\S(x^2-x+1)=1

Since each polynomial needs to make a 6, we are forced to give one copy of ( x + 1 ) (x+1) and one copy of ( x 2 + x + 1 ) (x^2+x+1) to both A and B.

Since all faces must have at least one dot, we need to give one copy of ( x ) (x) to both A and B.

Nevertheless, we can change things up by giving both copies of ( x 2 − x + 1 ) (x^2-x+1) to A.

Thus we have

A = x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x 8 + x 6 + x 5 + x 4 + x 3 + x A=x(x+1)(x^2+x+1)(x^2-x+1)^2 \\=x^8+x^6+x^5+x^4+x^3+x

B = x ( x + 1 ) ( x 2 + x + 1 ) = x 4 + x 3 + x 3 + x 2 + x 2 + x B=x(x+1)(x^2+x+1) \\=x^4+x^3+x^3+x^2+x^2+x

So die A has faces 1,3,4,5,6,8 and die B has faces 1,2,2,3,3,4.

Plugging these into the given expression, we get 599.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Nicely explained!

Pranjal Jain - 6 years, 3 months ago

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