>>11582322Assume that I possess X amount of dice that are hidden from me at any time. Assume also that there are Yn sides to each of the X dice that I have. The number of possible permutations Z of any possible configuration or number of dice that I could possess is Z=((Y1)^(W1))*((Y2)^(W2))...*((YX)^(WX)) where W1+W2+...WX=X
Beginning with a two-sided dice (Y1=2) (coin?) and increasing the number of sides of the dice +1 for each iteration of n, I would have... (assume I only have one of each dice)
Z=2*3*4*5...*n=n!
Assuming also that I roll the dice randomly and position them in random order, it becomes apparent that the actual number of different "rolled" dice is...
Zr=n!!
However, if I didn't know how many of these increasingly complex n-dice I had, but there was some type of 1/0 binary bit associated with each possible outcome, then the number of binary coupled outcomes becomes Zrb=2^(n!!)
If I continued to apply some type of binary outcome to this system each time (q times), it would become Zrbq=2^2^...^2^(n!!)
Assuming that q=n for bigness, and n is 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
then you would have a very big number
