Is rolling a 60 face die (D60) the same as rolling 10 6 faced ones (D6)? I understand that if you roll $10$ $6$-faced dice (D6) the different outcomes should be $60,466,176$ right?  (meaning $1/60,466,176$ probability)
but if you are playing a board-game, you don't care about that right? what matters is the sum of the numbers in the end, so it should anyway be $1/60$ instead of whatever right? because its a sum not $10$ individual rolls.
so a $60$ face die (D60) should be the same as rolling all the dice right? if not can someone please explain? I've been looking all over the net for a compelling answer but there is none. I've seen a D60 can be used as $2$ D6 but also there is no explanation for this.
if not, what would be the number of faces needed for an equivalent single die that can replace $10$ D6. I know so far the D120 is the last possible die that can be made, but it doesn't matter, I just want to know too...
EDIT: Sorry I forgot to mention in some tabletop games Dice can have $0$, so you'd treat all $6$ sided dice as going from $0$ to $5$, making minimum value of $0$ possible in all $10$ rolls. But still that changes the maximum in the same way minimum used to be, now the max you get in 6 sided dice is $50$, while on the 60 sided die it would be $59$ if we apply a $-1$ to all results... it gets messy, and this detail doesn't really change much

A simple argument is you can roll a $1$ using one $60$ sided die but you cannot using ten $6$ sided dice.
