Flaw in a "dice deck"? I happened to run into this product today:
https://www.thegamecrafter.com/games/ultimate-geek-dicedeck
It immediately bothered me, since (without quite claiming it) the creator seems to imply that he can simulate several consecutive rolls of a die without needing to shuffle this deck.
I was trying to think of some simple way of proving that this is, in fact, impossible with this kind of design. Assuming a die with 4 faces and a deck size of $2n$, here's the closest I could get:
# of sequences of possible die rolls of length $n$: $4^n$
# of permutations of the "dice deck": $(2n)!$
Then for the "dice deck" to perfectly simulate the appropriate odds, it must be true that $4^n = 2^{2n} \mid (2n)!$. But this is impossible via, say, de Polignac's formula.
I feel like there's some missing link in the above argument, however. Must, in fact, $2^{2n} \mid (2n)!$?
 A: As Carl Heckman speculated in comments, you can't even simulate two fair rolls of a $4$-sided die by drawing two cards at a time (without replacement).
Enumerate the $2n$ cards in the deck and consider the $2n\times2n$ matrix whose off-diagonal entries $a_{ij}$ (with $i\not=j$) are the simulated-die result of drawing the $i$th card first and the $j$th card second.  Picking two cards at random is equivalent to randomly picking one of these $2n(2n-1)$ entries.  If you are simulating a fair $4$-sided die, there must be an equal number of entries for each outcome, hence $4\mid 2n(2n-1)$, which means $n$ must be even, say $n=2m$.  But then the simulated second roll is equivalent to picking a random off-diagonal entry from a $(2n-2)\times(2n-2)$ matrix, formed by simply deleting the rows and columns corresponding to the two cards of the first draw.  No matter which rows and columns were removed, the smaller matrix has $(2n-2)(2n-3)=2(2m-1)(4m-3)$ entries, which is not a multiple of $4$, so the second draw cannot simulate a fair roll.
This works even if you start with an odd number of cards.  In an arbitrary $N\times N$ matrix, you have $N(N-1)$ off-diagonal terms.  If $4\mid N(N-1)$ then $4\not\mid(N-2)(N-3)$.
Final remark:  The video demonstation at the linked-to site shows a shuffle that results in all $n$ simulated rolls resulting in a "$1$."  It's possible what the designer has created is a deck for which, given any of the $4^n$ possible results of rolling a $4$-sided die $n$ times, there is at least one shuffle that simulates that $n$-tuple.
