calculate combinations of n rolls with m sided dice summing up to k

Question

I have been asking this question to $\tt AI$ ( I usually ask math questions to $\tt AI$) but its so frustratingly stubborn with not giving me the answer i want. I want to know:

• If you roll a $m$-sided dice $n$ times, what is the total amount of combinations for it to sum up to $k$ ?, where the order does not matter.
• Now, the important thing is i do NOT want to count the partitions of this.
• I want to know if there is a formula to do this.
• So far the only answers i have been able to find is where the dice are not identical.
• Or where you have to list out every possibility.
• I just want to know if there is a formula for this, no matter how complicated it might be.

Answer 1

$\underline{\text{Overview}}$

The best that I can do is provide an algorithm that can routinely be converted into a computer program that takes the values of $~m, ~n, ~$ and $~k~$ as parameters. Then, you can run the program to perform any $~(m,n,k)~$ enumeration.

You want the number of non-negative integer solutions to

• $x_1 + x_2 + x_3 + \cdots + x_m = n.$

• $[1 \times x_1] + [2 \times x_2] + [3 \times x_3] + \cdots + [m \times x_m] = k.$

For $~i \in \{1,2,\cdots,m\},~$ I am construing the variable $~x_i~$ to represent the number of die throws that showed the value $~i.~$

As will be discussed in the next section, a solution will be possible if and only if:

$$n \leq k \leq (m \times n). \tag1 $$

That is, the minimum sum possible of $~n~$ die throws occurs if each number shows $~1,~$ and the maximum sum possible of $~n~$ die throws occurs if each number shows $~m.$

So, I will assume, without loss of generality that the inequalities in (1) above hold.

---

$\underline{\text{When is a Solution Possible ?}}$

Continuing the discussion from the end of the previous section, if $~n = k~$ or $~k = (m \times n),~$ then a solution is clearly possible.

So, what happens if $~n < k < (m \times n)~?$

Choose $~i \in \{1,2,\cdots, (m-1)\},~$ such that

$$[n \times i] \leq k < [n \times (i+1)].$$

Then, set

$$x_i = [n - (k - ni)], ~x_{i+1} = (k - ni) \implies $$

$$x_i + x_{i+1} = n, ~~[i ~x_i] + [(i+1) ~x_{i+1}] = k.$$

Therefore, a solution will be possible if and only if the inequalities in (1) above hold.

---

$\underline{\text{Nested Summations Strategy}}$

Assume a range of values has been determined, so that a solution will be possible if and only if $~x_m~$ is in this range.

Then, assuming that $~x_m~$ is some arbitrary specific value in this range, assume that a corresponding range of values has been determined for $~x_{m-1},~$ based on the specific value of $~x_m.$

Continuing in this manner, for each $~i \in \{2,3,\cdots,(m-2)\},~$ assume that specific in range values have been assigned for each of $~x_{i+1}, ~x_{i+2}, ~\cdots, ~x_m.~$

Now assume that based on the values assigned to each of $~x_{i+1}, ~x_{i+2}, ~\cdots, ~x_m,~$ a specific range of values is determined for $~x_i.~$

Then, once the variables $~x_m, ~x_{m-1}, \cdots, ~x_2~$ are assigned specific values, then the variable $~x_1~$ is determined.

So, the result is an enumeration that looks like:

$$\sum_{x_m ~\text{in range}} \left\{ ~\sum_{x_{m-1} ~\text{in range}} \left[ ~\sum_{x_{m-2} ~\text{in range}} \cdots \left( ~\sum_{x_2 ~\text{in range}} ~1 ~\right) \cdots ~\right] ~\right\}. \tag2 $$

So, the entire problem is reduced to determining the specific range of values allowable for $~x_m,~$ and then, for each $~i \in \{2,3,\cdots,m-1\},~$ determining the allowable range of values for $~x_i,~$ based on the assigned in range values of $~x_{i+1}, ~x_{i+2}, \cdots, x_m.$

---

$\underline{\text{Minimum and Maximum Allowable Values}}$
$\underline{\text{for} ~x_i ~: ~i \in \{2,3,\cdots,m\}}$

A specific value for $~x_m~$ will be allowable if and only if there exists at least one solution to

• $x_1 + x_2 + \cdots + x_{m-1} = n - x_m.$

• $[1 \times x_1] + [2 \times x_2] + [3 \times x_3] + \cdots + [(m-1) \times x_{m-1}] = k - (m \times x_m).$

Based on the previous analysis, which verified the inequalities in (1) above, a solution will exist if and only if

$$(n - x_m) \leq [k - (m \times x_m)] \leq [(m-1) \times (n - x_m) \iff $$

$$k - [(m-1)n] \leq x_m \leq \left\lfloor \frac{k - n}{m-1}\right\rfloor. \tag3 $$

Note
If $~k < [(m-1)n],~$ then the lower bound for $~x_m~$ is $~0.$

Similarly, for $~i \in \{2,3,\cdots,m-1\},~$ assume that specific in range values have been assigned to the variables $~x_{i+1}, ~x_{i+2}, \cdots, x_{m-1}, ~x_m.$

Let

$$S = [ ~n - (x_m + x_{m-1} + \cdots + x_{i+1}) ~],$$

and let

$$T = k - \{ ~[m \times x_m] \\+ [(m-1) \times x_{m-1}] + \cdots + [(i+1) \times x_{i+1}] ~\}.$$

Then, a specific value for $~x_i~$ will be allowable if and only if there exists at least one solution to

• $x_1 + x_2 + \cdots + x_{i-1} = S - x_i.$

• $[1 \times x_1] + [2 \times x_2] + [3 \times x_3] + \cdots + [(i-1) \times x_{i-1}] = T - (i \times x_i).$

From the previous analysis in this section, that led to the result in (3) above, the allowable range for $~x_i~$ is

$$T - [(i-1)S] \leq x_i \leq \left\lfloor \frac{T - S}{i-1}\right\rfloor. \tag4 $$

Note
If $~T < [(i-1)S],~$ then the lower bound for $~x_i~$ is $~0.$

Answer 2

We can use a generating function approach to derive such a formula for the number of combinations, based on this blog post, where the number of dice combinations for $k$ is equal to the coefficient of $z^k$ in the function's power series.

We generally consider the unnormalized generating function of a single die to be:

$$F(z)=\sum_{i=1}^mz^i$$

Which we can rewrite:

$$F(z)=\sum_{i=1}^mz^i=\frac{z(z^m-1)}{z-1}=M(z)N(z)O(z)$$

for $n$ dice, we take this to the $n$th power:

$$[F(z)]^n=\frac{z^n(z^m-1)^n}{(z-1)^n}=[M(z)]^n[N(z)]^n[O(z)]^n$$

At which point we can start breaking this up into it's constituent functions. Let's start with the simplest, $[M(z)]^n$:

$$[M(z)]^n=z^n$$

Then let's look at $[N(z)]^n$, a clear binomial if ever I've seen one:

$$\begin{align*}[N(z)]^n&=(z^m-1)^n\\ &=\sum_{k=0}^{n}\binom{n}{k}z^{mk}(-1)^{n-k} \end{align*}$$

We can multiply $[M(z)]^n$ and $[N(z)]^n$ together:

$$[M(z)]^n[N(z)]^n=\sum_{k=0}^{n}\binom{n}{k}z^{mk+n}(-1)^{n-k}$$

Now, we can look at the denominator, $[O(z)]^n$, which if we look closely is a negative binomial:

$$\begin{align*}[O(z)]^n&=\frac{1}{(z-1)^n}=(z-1)^{-n}\\ &=(-1)^{-n}(1-z)^{-n}\\ &=(-1)^{-n}\sum_{j=0}^{\infty}\binom{n+j-1}{n-1}z^j \end{align*}$$

Thus (with a little simplification around the common$(-1)$):

$$[F(z)]^n=\left(\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}z^{mk+n}\right)\left(\sum_{j=0}^{\infty}\binom{n+j-1}{n-1}z^j\right)$$

Because the two sums have differing parameters, we normally wouldn't be able to combine these, but since the binomial expansion can be considered an infinite sum (just one where it will equal zero where $k > n$), we can take the Cauchy product of the two series to get the coefficients (which is our goal here). As such, we can view the final result as:

$$[F(z)]^n=\sum_{k=0}^{\infty}\left ( \sum_{j=0}^{\left \lfloor \frac{k-n}{m}\right \rfloor}(-1)^j\binom{n}{j}\binom{k-nj-1}{n-1} \right )z^k$$

Since we're only looking for the coefficient, we can break that out of our power series as:

$$f(n,m,k)=\sum_{j=0}^{\left \lfloor \frac{k-n}{m}\right \rfloor}(-1)^j\binom{n}{j}\binom{k-nj-1}{n-1}$$

And to turn this into a probability mass function, we just need to normalise by dividing the result by $\frac{1}{m^n}$:

$$P(K=k)=\frac{1}{m^n}\sum_{j=0}^{\left \lfloor \frac{k-n}{m}\right \rfloor}(-1)^j\binom{n}{j}\binom{k-nj-1}{n-1}$$