I am currently developing a TTRPG game system, and I am trying to determine the probability of success in various situations to see how balanced it might be.
The system is fairly simple, you roll a number of 10 sided dice, and if any of them are below your score, you succeed. However, if any of the dice you roll land on $10$ (or $0$ as notated on most d10s), you remove the lowest die rolled, which I guess you could call an imploding dice rule (as opposed to exploding dice, which would allow you to roll extra on a critical success)
For example, my score is $5$, so to succeed I need to roll a $<5$. With one dice, and without the $10$ rule, that is easy to work out as $40$% chance of success, and as I add more dice, I can work it out as by using the chance of failure:
$1-(1-0.4)^n$ where $n$ is the number of dice.
But I am struggling to work out how to introduce the imploding dice to my probability calculations. The chance of rolling a $10$ is $10$% for each die I roll, but when I do, I am essentially reducing the amount of dice I originally rolled, right?
/edit: clarification, if you roll a $10$, you only remove one die, not one for every $10$ rolled
Problem-statement: We have $n$ dice and a given $s$ score. We roll all the dice and we get results form $1$ to $9$ or a special result, called WILDCARD. If we have rolled a number of $k$ wildcards, the $k$ lowest dice are eliminated from the table. If any of the $n - k$ is less than $s$, we have WON. Otherwise, we have LOST.
Probability evaluation: The probability of having $k$ wildcards out of $n$ dice is $\frac{1}{n} \cdot \frac{1}{n - 1} \cdot \dots \cdot \frac{1}{n - k + 1} = \frac{(n - k)!}{n!} = \frac{1}{A_n^k}$. Next, from $n - k$ dice we have the probability to draw greater equal than $s$: $\frac{(10 - s)^{n - k}}{9^{n - k}}$, and the probability to have at least one dice less than $s$ will be $1 - \frac{(10 - s)^{n - k}}{9^{n - k}}$.
Result: By summating all probabilities (considering $k$ variating from $0$ to $n$): $$p = \sum_{k = 0}^n \frac{1}{A_n^k} \cdot \bigg(1 - \frac{(10 - s)^{n - k}}{9^{n - k}}\bigg)$$
That's all!