Dice!
So I've had a few people ask me about the Dice question of which I posted on last. I shall now elucidate on the solution and why it's true.
First, I'm using the notation Z[x] = {[;a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 | a_i \in Z,;] n positive}
What does that mean? It's the set of all polynomials with coefficients from Z, the integers. We're going to be working mostly in Z[x] and I will point out that it's a Unique Factorization Domain. This means every polynomial over this ring of polynomials has a unique set of irreducible elements that make up said polynomial. If you think back to your days of learning these things, you used to factor polynomials in high school. For example, [;x^2 - 4;] = [;(x-2)(x+2);]. Since this is over Z[x], that is the ONLY factorization. Just think of irreducible polynomials as polynomials which don't have a solution over Z. For eg, [;x^2 + 2;] is an irreducible of Z[x], because [;\sqrt{2}i \notin Z;].
Now that's out of the way, we can start discussing the dice.
Firstly, if we take the polynomial f(x) = [;x^6 + x^5 + x^4 + x^3 + x^2 + x;], we can see that [;(f(x))^2;] will mean the coefficients will give the probabilities of the dice. That is, if you wanted to see what the probability of rolling a 7 would be, you would see that [;x^6 * x, x^5 *x^2;], etc would lead to it, and thus the coefficient of [;x^7;] would be the probability (over 36 of course) of rolling a 7. Now, hand waving time! Note that f(x) factors as:
[;f(x) = x(x + 1)(x^2 + x + 1)(x^2 - x + 1);] (If you don't believe me, multiply it out.)
Let our weird dice be denoted by:
[;P(x) = x^{a_1} + x^{a_2} + x^{a_3} + x^{a_4} + x^{a_5} + x^{a_6};]
Then P(x) has the form:
[;P(x) = x^r(x + 1)^s(x^2 + x + 1)^t(x^2 - x + 1)^u;] Where r, s, t, u all range between 0 and 2. (Note what will happen is that these are the ONLY possible factors of [;P(x);])
Now, first we will evaluate P(x) two different ways; [;P(1) = 1^{a_1} + 1^{a_2} + 1^{a_3} + 1^{a_4} + 1^{a_5} + 1^{a_6} = 6;] and [;1^r2^s3^t1^u;]. Firstly, we see s and t must be 1. Therefore the only two parameters we are dealing with that will change are r and u. As we evaluate P(0) as two different ways we know [;r \neq 0;]. Suppose r = 2. Then the smallest sum one could roll for the dice would be 3. (as every term has a multiple of [;x^2;] in it). So r = s = t = 1. So u can now fluctuate.
Suppose u = 0. Then [;P(x) = x^4 + x^3 + x^3 + x^2 + x^2 + 1;], what we call a Sicherman die.
Let u = 1. Then [;P(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x;], a standard die.
Let u = 2. Then [;P(x) = x^8 + x^6 + x^5 + x^4 + x^3 + x;], the other Sicherman die.
Since these are the only variants in the parameters we can change, we see that these are the ONLY dice that will have the same probabilities as your standard dice.
And there's another application of Algebra, ladies and gents.
First, I'm using the notation Z[x] = {[;a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 | a_i \in Z,;] n positive}
What does that mean? It's the set of all polynomials with coefficients from Z, the integers. We're going to be working mostly in Z[x] and I will point out that it's a Unique Factorization Domain. This means every polynomial over this ring of polynomials has a unique set of irreducible elements that make up said polynomial. If you think back to your days of learning these things, you used to factor polynomials in high school. For example, [;x^2 - 4;] = [;(x-2)(x+2);]. Since this is over Z[x], that is the ONLY factorization. Just think of irreducible polynomials as polynomials which don't have a solution over Z. For eg, [;x^2 + 2;] is an irreducible of Z[x], because [;\sqrt{2}i \notin Z;].
Now that's out of the way, we can start discussing the dice.
Firstly, if we take the polynomial f(x) = [;x^6 + x^5 + x^4 + x^3 + x^2 + x;], we can see that [;(f(x))^2;] will mean the coefficients will give the probabilities of the dice. That is, if you wanted to see what the probability of rolling a 7 would be, you would see that [;x^6 * x, x^5 *x^2;], etc would lead to it, and thus the coefficient of [;x^7;] would be the probability (over 36 of course) of rolling a 7. Now, hand waving time! Note that f(x) factors as:
[;f(x) = x(x + 1)(x^2 + x + 1)(x^2 - x + 1);] (If you don't believe me, multiply it out.)
Let our weird dice be denoted by:
[;P(x) = x^{a_1} + x^{a_2} + x^{a_3} + x^{a_4} + x^{a_5} + x^{a_6};]
Then P(x) has the form:
[;P(x) = x^r(x + 1)^s(x^2 + x + 1)^t(x^2 - x + 1)^u;] Where r, s, t, u all range between 0 and 2. (Note what will happen is that these are the ONLY possible factors of [;P(x);])
Now, first we will evaluate P(x) two different ways; [;P(1) = 1^{a_1} + 1^{a_2} + 1^{a_3} + 1^{a_4} + 1^{a_5} + 1^{a_6} = 6;] and [;1^r2^s3^t1^u;]. Firstly, we see s and t must be 1. Therefore the only two parameters we are dealing with that will change are r and u. As we evaluate P(0) as two different ways we know [;r \neq 0;]. Suppose r = 2. Then the smallest sum one could roll for the dice would be 3. (as every term has a multiple of [;x^2;] in it). So r = s = t = 1. So u can now fluctuate.
Suppose u = 0. Then [;P(x) = x^4 + x^3 + x^3 + x^2 + x^2 + 1;], what we call a Sicherman die.
Let u = 1. Then [;P(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x;], a standard die.
Let u = 2. Then [;P(x) = x^8 + x^6 + x^5 + x^4 + x^3 + x;], the other Sicherman die.
Since these are the only variants in the parameters we can change, we see that these are the ONLY dice that will have the same probabilities as your standard dice.
And there's another application of Algebra, ladies and gents.