Number of rolls with m many n-sided dice

[mathboy]

[SOLVED] Number of rolls with m many n-sided dice

How many different rolls are there with m many n-sided dice, if order is not important?

If order was important, the answer would simply be n^m. But if order is not important?

If two 6-sided dice are rolled, there are C(6,2)+C(6,1) = 21 different rolls.
If three 6-sided dice are rolled, there are C(6,3)+C(6,1)C(5,1)+C(6,1) = 56 different rolls.

Is there a general method for rolling m many n-sided dice? Or do you have to continue adding up all the cases? Perhaps a general method involving a matrix with m indices (e.g. a tensor) and simply excluding all permutations of the indices?

 

[e(ho0n3]

Wouldn't it be n^m/m!? For example, say you have a black die and a red die, both 6-sided. The number of possible rolls is 6^2. If you ignore the color of the dice, then you have to divide by 2.

 

[mathboy]

No. The number of rolls with 2 six-sided dice (if order is unimportant) is 21, not 18. Rolls with doubles are not being counted twice if order is considered important.

 

[e(ho0n3]

Oh that's right. Sorry about that. I don't know of a general method. I'm thinking that you would first have to count the number of rolls where all dice show the same side, then for m-1 dice showing the same side, then for m-2 dice, etc.

 

[mathboy]

That's exactly what I did in my opening post. But it gets awkward after a while (e.g. for 7 dice, you can have 2 doubles and 1 triple). I'm looking for a better method that gives a single formula for arbitrarily values of m and n,

e.g. m^n minus something.

 

[stewartcs]

How many different rolls are there with m many n-sided dice, if order is not important?

If order was important, the answer would simply be n^m. But if order is not important?

If two 6-sided dice are rolled, there are C(6,2)+C(6,1) = 21 different rolls.
If three 6-sided dice are rolled, there are C(6,3)+C(6,1)C(5,1)+C(6,1) = 56 different rolls.

Is there a general method for rolling m many n-sided dice? Or do you have to continue adding up all the cases? Perhaps a general method involving a matrix with m indices (e.g. a tensor) and simply excluding all permutations of the indices?

Yes.

$$\frac{(n + m - 1)!}{m!(n - 1)!}$$


CS

 

[mathboy]

Wow! That gives the right answer for all the cases I worked out.

So the answer is C(m+n-1,m), i.e. the number of ways to choose m numbers from m+n-1 numbers. Half of this makes some sense to me because we are rolling m dice. But what's the reason for the m+n-1?

 

[stewartcs]

Wow! That gives the right answer for all the cases I worked out.

So the answer is C(m+n-1,m), i.e. the number of ways to choose m numbers from m+n-1 numbers. Half of this makes some sense to me because we are rolling m dice. But what's the reason for the m+n-1?

It's just a combination that allows repetitions.

http://oregonstate.edu/~peterseb/mth232/232euler_comb.html

or

http://www.csee.umbc.edu/~stephens/203/PDF/6-5.pdf

Hope that helps.

CS

 

[mathboy]

Thank you so much. Sometimes extra knowledge is the key to solving a problem.

 

[lcscipiop]

That doesn't work for larger values because your dice cannot show more pips than n. Your formula doesn't account for that...