Probability of sum of combo dice roll, with a twist. I'm trying to figure out the probability of each sum possibility of a dice roll.
Given n s-sided dice, the probability of summing to p is shown on this stack overflow question. Fairly simple to compute.
However, my problem has a twist. If any of the dice roll a 1, the "sum" is 1.
As an example, for rolling 2 3-sided dice (1, 2, 3), the probability outcomes are as follows:
"Sum" | P (x/9)
------|--------
    1 | 5
    2 | 0
    3 | 0
    4 | 1
    5 | 2
    6 | 1

I also tried this with 2 4-sided dice, but hand writing out 3 4-sided dice or 2 5-sided dice is going to get a little bit dicey. I seem to be noticing a little bit of a pattern, number 1 is pretty trivial to calculate for two dice (P that either dice is 1 - P that both are 1). I don't know how to do this for 3 though. The rest seem to start at 4 and go up from 1 then back down.
Can anybody weigh in and provide a closed-form mathematical expression for each P(n), and explain how you got this?
Thanks!
 A: The probability of getting at least one $1$ result when rolling $k$ different fair $n$-sided dice would be
$$\sum\limits_{i=1}^k\frac{(-1)^{i+1}\binom{k}{i}}{n^i} = \frac{k}{n}-\frac{\binom{k}{2}}{n^2}+\frac{\binom{k}{3}}{n^3}-\dots\pm \frac{\binom{k}{i}}{n^i}\pm \dots$$ seen easily by inclusion-exclusion.  In the case of $k=2$ and $n=3$ for instance, that would be $\frac{2}{3}-\frac{1}{9}=\frac{5}{9}$ as you found manually.
Alternatively, even better as pointed out in the comments might be to approach by complementary events and seeing that the probability of getting at least one $1$ is one minus the probability of getting no $1$'s and is
$$1-\left(\frac{n-1}{n}\right)^k$$
As for getting a particular sum while avoiding $1$'s on any of the dice, we can count how many possibilities lead to that by imagining that each die had one fewer side, and each number on the die was one less and adding $k$ to the final result.
In other words, we are recognizing that $(x^2+x^3+\dots+x^n)^k = x^k\cdot (x^1+x^2+\dots+x^{n-1})^k$.  From here, we can use exactly the same techniques as before and then shift the results and divide the number of outcomes leading to that sum by $n^k$ to find the probability.
A: Suppose we have $k$ dice, each with $n$-sides, and let $S$ be the random variable that counts your modified sum.
The total number of dice roll configurations is $n^k$, so to calculate probabilities $\Bbb P(S=s)$ it's enough to calculate the count $C(s)$ of dice roll configurations for which $S=s$.
Of course, we'll then have $\Bbb P(S=s) = C(s)/n^k$.
The minimum value greater than $1$ that $S$ can take is $2k$, which happens when all $k$ dice roll $2$.
The maximum value of $S$ is of course $nk$.
For each $s$ with $2k\leqslant s \leqslant nk$, a combination of dice rolls that add up to $s$ corresponds uniquely to a solution of
$$x_1+x_2+\dots + x_k = s,\tag{1}$$
where $2\leqslant x_i\leqslant n$ is the value of the $i$-th die.
We can rewrite this as 
$$x_1'+x_2'+\dots + x_k' = s-2k,\tag{1$'$}$$
where $x_i' = x_i - 2$, and hence $0\leqslant x_i' \leqslant n-2$.
Using this representation, one can derive a closed form expression with nested sums, but we can do better.
We can count solutions of $(1')$, and hence of $(1)$, via inclusion-exclusion in a manner similar to this answer.
Consider equation $(1')$, except we require only that $x_i\geqslant 0$.
The number of solutions is given by stars and bars and equals $\binom{s-k-1}{k-1}$.
Now, let $A_i$ be the set of solutions to $(1')$ in which $x_i'>n-2$.
Then, by inclusion-exclusion:
$$\bigcup_{i=1}^k |A_i| = \sum_{\emptyset \neq J \subset\{1,\dots,k\}} (-1)^{|J|+1}\left|\bigcap_{j\in J}A_j\right|.$$
By symmetry, $\left|\bigcap_{j\in J}A_j\right|$ depends only on $|J|$, and hence we may simplify the calculation above to
$$\bigcup_{i=1}^k |A_i| = \sum_{j=1}^k (-1)^{j+1}\binom kj \alpha_j,$$
where $\alpha_j$ is the size of any intersection $\left|\bigcap_{j\in J}A_j\right|$ with $|J| = j$.
In the spirit of the linked answer, we substitute $x_i' \mapsto y_i + n - 1$ for each $i=1,\dots,j$ and consider hence
$$y_1+y_2+\dots+y_j + x_{j+1}'+\dots+x_k'  = s - 2k - (n-1)j.$$
By stars and bars, the number of solutions to this is
$$\alpha_j = \binom{s-k-(n-1)j-1}{k-1}.$$
Of course, for many values of $n$ and $j$, this will simply be $0$ as the numerator will be $<k-1$.
It follows that
$$\begin{align}
C(s)
&= \binom{s-k-1}{k-1} - \sum_{j=1}^k (-1)^{j+1}\binom kj \binom{s-k-(n-1)j-1}{k-1}
\\&= \sum_{j=0}^k (-1)^{j}\binom kj \binom{s-k-(n-1)j-1}{k-1}
\end{align}.$$
The only thing that's left is $C(1)$, but that's the easy part.
The total number of dice roll configurations is $n^k$, and and the number of dice roll configurations that have no $1$s is $(n-1)^k$.
It follows that
$$C(1) = n^k - {(n-1)}^k$$

Here's some Python code to validate the counts.
from scipy.special import comb

k = 2
n = 3

s_min = 2*k
s_max = n*k
counts = dict()
counts[1] = (n**k) - (n-1)**k
for s in range(s_min, s_max+1):
    counts[s] = 0
    for j in range(0, k+1):
        counts[s] += ((-1)**j) * comb(k, j, exact = True) * comb(s-k-(n-1)*j-1, k-1, exact = True)
for key in counts:
    print(key, counts[key])

The sanity check against $k=2$ and $n=3$ works out and outputs:
1 5
4 1
5 2
6 1

Other tests of mine have always had the counts add up to $n^k$, so I'm fairly confident at this point.
They also point in the direction that the counts for $s>1$ have an increasing and then decreasing pattern of repetitions (like, say, the binomial coefficients).
Indeed, to each solution of $(1)$ we can associate a unique dual solution of
$$X_1 + X_2 +\dots + X_k = (n+2)k - s$$
via the transformation $X_i = (n+2) - x_i$, so the counts must be the same.
This means we can restrict the calculations to the range
$$2k \leqslant s \leqslant \left\lfloor\frac{(n+2)k}2\right\rfloor,$$
and counts for higher values of $s$ can be obtained from these via the dual association.
A: Let $P(n,s,t)$ be the probability that $n$ dice with $s$ sides have a sum to $t$, which is formula you already understand. We can express your modified sum probabilities in terms of the previous one:
$$
\mathbb P(\text{$n$ dice, with $s$ sides, have modified sum of $t$})
=
\begin{cases}
1-(1-1/s)^n & \text{if }t=1
\\
0 & \text{if }2\le t < 2n
\\
\left(1-1/s\right)^n\cdot P(n,s-1,t-n) & \text{if }t\ge 2n
\end{cases}
$$
For the first line: the only way that the sum is not one is for all of the dice to be greater than $1$. The probability the sum is not one is therefore $(1-1/s)^n$, and the probability the sum is one is $1$ minus this.
For the second line, we first consider the probability that none of the dice are $1$, which is $(1-1/s)^n$. Given that none of the dice are one, we are now dealing with $s-1$ sided dice, since the value of $1$ is no longer possible. However, these dice are numbered $(2,3,4,\dots,s)$ instead of $(1,2,3,\dots,s-1)$. Therefore, to find the probability these dice numbered $2$ to $s$ sum to $t$, we compute the probability that $n$ normal dice numbered $1$ to $s-1$ sum to $t-n$. 
