Dice probabilities

[Thomas Klemm]

(05-01-2024 06:56 PM)John Keith Wrote:  Simpler than I thought using ADD instead of DOLIST.

It appears to be similar to my Post #5.

(05-01-2024 06:56 PM)John Keith Wrote:  And now a dumb question: Why can trinomial coefficients be used for any number of dice and faces per die? Are there cases where one may need tetranomial or higher order coefficients?

They can't. Member dm319 initially asked for dice like these:



We list their faces as [0 0 1 1 2 2] and encode them as \(2x^2 + 2x + 2\).
But since the coefficients are all the same we can use \(x^2 + x + 1\) instead.

If however you use one of these:



There are multiple possibilities for quadrinomial coefficients:

• Tetrahedron: [1 2 3 4]
• Octahedron: [1 1 2 2 3 3 4 4]
• Dodecahedron: [1 1 1 2 2 2 3 3 3 4 4 4]
• Icosahedron: [1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4]
  

All of them can be encoded with \(x^4 + x^3 + x^2 + x\).
Or then we rather use \(x^3 + x^2 + x + 1\) to match A008287.
Again this doesn't matter much as the coefficients are only shifted.

Check in Post #8 how "for extra fun (…) such a formula could be extended to rolling a pool of non-standard, mixed dice".