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The alternate dice are numbered [1 2 2 3 3 4] and [1 3 4 5 6 8]
The probability distrbution for standard dice can be generated by powers of the polynomial d(x)=(x+x^2+x^3+x^4+x^5+x^6). The coefficient of x^k gives the frequency out of 6^n.
For two dice: (x+x^2+x^3+x^4+x^5+x^6)^2 = x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^10+2x^11+x^12
An alternate factorization of the polynomial is (x+x^3+x^4+x^5+x^6+x^8)*(x+2x^2+2x^3+x^4)
The first factor represents a die with faces [1 3 4 5 6 8]. The other factor represents a die with faces [1 2 2 3 3 4].
Gamer found a site with a similar solution here. |
