After 200,000 die rolls for each, I got this result.
That's a "Monte Carlo" simulation, but a) I've figured out the exact probability based on all possible combinations of dice rolls, and b) you're including Spectacularly Interesting results along with the Spectacular Successes and Failures. I extracted those. The computer was only told to be on the lookout for rolls with at least 3 1's, but wasn't further told to check for at least 3 6's as well. The correct thing to do is figure out things exactly and subtract the Spectacularly Interestings from the Spectacular Successes (or, equally, from the Spectacular Failures) to show the "pure" Spectacular Successes or Spectacular Failures with no Spectacularly Interestings included.
But it's a nice brute-force approach to help confirm that I didn't make some arithmetic error. Are you able to expand your total number of rolls into tens of millions?
DANGER! HERE BE MATH!
The formulas I'm using (trying to write them in-line) are
Spectacular Success: Odds of rolling at least 3 1's on xD: Sum(from i=3 minimum to x)[C(x,i)*(1/6)^i*(5/6)^(x-i)]
In other words, for 3 1's there is a 1/6 chance of getting a 1 for each of 3 of the dice, and a 5/6 chance for all the other dice of ending up as something else. But then you must figure that dice can be rolled in any order to get 3 1's and the number of arrangements is C(x,i). You must also add in additional terms for 4 1's, 5 1's, etc. up to the maximum number of dice you have, since they also satisfy the "Spectacular Success" condition.
Spectacular Failure: Same odds as Spectacular Success only for at least 3 6's.
Spectacularly Interesting: Odds of rolling at least 3 1's and 3 6's on xD: Sum(from i=3,j=3 minimum up to x-3 maximum)[C(x,i)*(1/6)^i*C(x-i,j)*(1/6)^j*4^(x-i-j)]
C(a,b) means the Combination function ("a choose b" or the number of possible ways to choose "b" non-repeating objects out of "a" objects total). For example, C(6,0)=1, C(6,1)=6, C(6,2)=15, C(6,3)=20, etc. and going back down again since C(6,a) = C(6,6-a). It is also shown as "a" over "b" in long parentheses. A binomial distribution ("Pascal's Triangle") can give you these values. Or you can calculate it as C(a,b) = a!/[b!(a-b)!] where ! is "factorial" (multiply the integer with every integer below it, down to 1. So C(10,6) is 10!/[6!4!] which (taking out 6! on both upper and lower) cancels to 10*9*8*7/4*3*2*1 or 210.
For example: the first Spectacularly Interesting rolls require at least 6D. For 6D, you have only one possible combo: 3 1's and 3 6's. You can't have any more because then you would exceed 6D. However, the 1's can occur in different ways such as the first, second and third die, or first, second and fourth, etc. There are only C(6,3) or 20 arrangements of the 3 1's. Wherever they occur, they constrain the 6's to occupy the remaining spots so the possible different arrangements of the 6's are C(6-3,3) or 1 only (no freedom to rearrange, in this case.)
So Spectacularly Interesting probability for 6D: 20*(1/6)^3*C(3,3)*(1/6)^3*4^0 = 20*1/216*1*1/216*1 = 20/46,656.
The "4^0" term, equal to 1, results from the fact there are no extra dice on 6D that can assume any values other than 1 or 6 so that term disappears. For larger amounts of dice, there are these other dice rolled and they can assume values from 2 to 5, hence 4^(x-i-j) showing all the possible rolls of these, increasing the odds of occurrence. However, the exact values of those other dice don't matter to whether it is a Spectacularly Interesting roll.
The probability is EXACTLY that, but will only be approached with a very large number of fair dice-rolls. It should approach 0.042866941...%.
With more dice, you have to figure different combinations that have at least 3 1's and 3 6's, and sum the probabilities together. For 8 dice it can be (I'm showing possible result sets here: ) (3,3), (3,4), (3,5), (4,3), (4,4) and (5,3).
I'm no textbook author, but I hope I was sufficiently clear on my method.
