Dice probabilities

[Thomas Klemm]

We might rather write it: \(x^2+x^1+x^0\)
Thus for an ordinary die we would use: \(x^6+x^5+x^4+x^3+x^2+x^1\)
If these terms get multiplied, their exponents are added.
We're using the distributive law.
Every possible case gets combined with all the others.
And the exponents are added up for us.

I assume that the reason we add up the coefficients of \(x^{10} \cdots x^7\) is clear.