Originally Posted by

**Soroban** Hello, ChaosticMoon!

I assume each die has sides numbered from 1 to 100.

There are $\displaystyle 100^3\,=\,1,\!000,\!000$ possible outcomes.

Visualize a cube graphed in the first octant of an *xyz*-system.

One vertex is at the origin; the opposite vertex is at (100, 100, 100).

Consider all the lattice points (those with integer coordinates)

. . from (1,1,1) to (100,100,100).

These represent the 1,000,000 possible outcomes.

The points whose coordinates have a sum exceeding 222

. . are "outside" the triangle with coordinates:

. . (22, 100, 100), (100, 22, 100), (100, 100, 22).

How many lattice points are contained in this tetrahedron?

The tetrahedron has 21 "levels".

Each level contains a triangular number of points.

We want the sum of the first 21 triangular numbers.

Fortunately, there is a formula for this: .$\displaystyle N \:=\:\frac{n(n+1)(n+2)}{6}$

For $\displaystyle n = 21\!:\;\;N \:=\:\frac{(21)(22)(23)}{6} \:=\:1771$

Therefore: .$\displaystyle P(\text{sum}> 222) \:=\:\frac{1,\!771}{1,\!000,\!000}$