Take $n$ , $n$ - sided rigged dice with the probability of a die landing on the $i$th side being $p_i$ for $ i =1, ..., n$.
Throw $k$ dice ( $k \in [1,n]$), what is the probability that all of the $k$ throws will be different?
I would like to treat this problem as a sequence of events, and then use the relation:
\begin{equation} P ( \cap_{l=1}^{m} A_l) = P(A_1) P(A_2|A_1) ... P( A_m | \cap_{l=1}^{m-1} A_l )\tag{1} \end{equation}
But, I am not quite sure on how to define the events $A_l$ so that the previous relation makes sense.
Even for the simplest case: the "fair" case ( $ p_i = \dfrac{1}{n}) $ , where this becomes a "counting problem" ,the probabilty of getting $k$ different throws being:
$$ P = \dfrac{\dfrac{n!}{(n-k)!}}{n^{k}} $$
I am unsure on how to define the events in order to be able to use $(1)$...
Let $[n] = [1,n] \cap \mathbb{N} = \{1,\ldots, n\}$ be the set of integers $1$ through $n$.
For any subset $A \subset [n]$, there exists a mapping between $A$ and a set of rolls of dice where each die rolls a different number. Let's use some examples. If $n=4$ and $A = \{1\}$, this is represented by one die rolling a $1$. If $A = \{1,3\}$, this represents two dice rolling a $1$ and a $3$.
If you are rolling $k \in [n]$ dice, then you only care about $A$ if $|A| = k$. This means the set $A$ contains $k$ distinct values. So, let's look at our example with $A = \{1,3\}$. This set not only represents when we have two dice and the first rolls a $1$ and the second rolls a $3$. It also represents when the first rolls a $3$ and the second rolls a $1$. In other words, this set does not take order into account. But, the probability for either order will be the same since $p_1p_3 = p_3p_1$.
So, for a given subset $A \subset [n], |A| = k$, there are $k!$ different orders. And the probability for any individual order is given by $$\prod_{i \in A} p_i$$
So, the probability over all possible orders is $$k!\prod_{i \in A}p_i$$
And we can sum over all possible subsets:
$$k!\sum_{A \subset [n] \\ |A| = k}\left(\prod_{i \in A}p_i\right)$$
Unless you start grouping together faces with the same probability, it is unlikely there will be a way to simplify this.