Recent content by unmantle

Yesssssss

Don't have the notebook in front of me, but I did a 5^3 in the passenger seat and I think it went: 1, 3, 6, 10, 15, 18, 19, 18, 15, 10, 6, 3, 1 Somewhat interesting to note, if all 3 sites(n) rolled unique states(q), there are 6 combos; if one duplicate, 3; if triplets, 1. Didn't do a n=4...

What makes you think that would be true? I don't see an easy formula off the top of my head, but I'd be awfully surprised if there wasn't a somewhat elegant way to phrase it. But maybe you're seeing something that makes you think it's not possible?

microstates = q^n, where q = number of states the dice can be in, and n = number of dice Totally not sure about this, but if we're assuming that dice are always numbered using integers starting at 1 and moving upward sequentially, and label each macrostate with its dice-sum, it seems like it...