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: No, the probability distribution of the sum of the faces of rolling $10$ $6$-sided dice is different than that of rolling a single $60$-sided die.
For obvious reasons, this is because you can roll any number between $1$ and $9$ with a $60$-sided die, but cannot attain such a sum with $10$ $6$-sided dice.
The probability distribution of a single $6$-sided die can be modeled with the generating function
$$\frac{1}{6}(x+x^2+x^3+x^4+x^5+x^6)$$
Where the coefficient of $x^k$ is the probability of rolling a $k$. Moreover, the probability distribution of the sum of rolling $10$ $6$-sided dice can be modeled as
$$\frac{1}{6^{10}}(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
Where the coefficient of $x^k$ is the probability of rolling a sum of $k$. It can be clearly seen that this expansion is completely different than the generating function that models the probability distribution of rolling a $60$-sided die i.e.
$$\frac{1}{60}(x+x^2+x^3+\ldots+x^{59}+x^{60})$$
However, there are some interesting ideas that you can get from using these generating functions. For example, we have that
$$\frac{1}{6}(x+x^2+x^3+x^4+x^5+x^6)$$
$$=\frac{1}{6}x(1+x+x^2+x^3+x^4+x^5)$$
$$=\frac{1}{6}x(1+x^3)(1+x+x^2)$$
$$=\frac{1}{6}x(1+x)(1-x+x^2)(1+x+x^2)$$
We can take two dice, and say that their generating function for their probability distribution is
$$=\frac{1}{6^2}x^2(1+x)^2(1-x+x^2)^2(1+x+x^2)^2$$
$$=\frac{1}{6^2}[(1+x)(1+x+x^2)x^2][(1+x)(1+x+x^2)(1-x+x^2)^2]$$
$$=\frac{1}{6^2}[x^2+2x^3+2x^4+x^5][1+x^2+x^3+x^4+x^5+x^7]$$
This means we can replace two $6$-sided dice with, for example, a die with faces of $2,3,3,4,4,5$ and a die with faces of $0,2,3,4,5,7$, and the sum from rolling these two dice will have the same probability distribution as rolling two standard $6$-sided dice. We can extend this process to the generating functions for more $6$-sided dice and create some wacky dice that still have the same probability distribution.
A: A simple argument is you can roll a $1$ using one $60$ sided die but you cannot using ten $6$ sided dice.
A: Adding onto other answers…
Rolling ten six-sided dice, the most likely sum you are to get is $\frac{1+6}{2}*10=35.$ Then, getting a sum of $34$ is slightly less likely than getting a sum of $35$. Etc… all the way down to getting a sum of $10$, which has probability just $\frac{1}{6^{10}}.$
But with a $60-$ sided die, the probability of getting a $10$ is just $\frac{1}{60}.$
A: When you roll a 60-face die, you are in effect rolling ten six-face dice but counting only one at a time, which is very different from counting the entire combination of ten independent six-face dice.
You can approximate a fair 60-face die with a red six-face die, a blue one and several white ones, call the number of white dice $n$. Interpret a roll as follows:

*

*Count the actual outcome of the red die.


*Count the blue die as 1 if its outcome is odd, 2 if even.


*Sum the white dice and divide by 5; count the remainder. You may add 1 to each remainder, or exchange a remainder of 0 for 5, if you want this number to range from 1 to 5 instead of 0 to 4.
You then have 60 ordered triples whose probabilities are nearly equal for reasonable values of $n$. The only deviation is associated with the remainder from the white-dice sum divided by $5$ being congruent with $n\bmod 5$. In that case the probability is increased by a factor of $(6^n+4)/(6^n-1)$ compared with the other remainders (which all come out equal). This corresponds to an excess of just over 2% with $n=3$ and less than 0.4% with $n=4$. Each increment of $n$ by $1$ reduces the deviation by a factor of six.
