Calculating probility of dice rolls with conditional rerolls for specific target numbers

Question

I have searched for an answer but quite frankly, I am not sure what to even search for. As such, I apologize if this has already been asked and answered.

Given a single die with 10 sides labeled 0-9, where 0 = 10, I can calculate most of what I need in terms of probability.

However, there is a specific rule that is complicating the calculations.

The Rules

1. n 10 sided dice are rolled.
2. Any rolls $\geq$ 8 are a success.
3. Any rolls of 10 are re-rolled until < 10 is obtained.
4. Any such re-rolls from rule #3 that are $\geq$ 8 count as a success for each additional roll that occurred.

Definition of terms

$p_s$ = probability of success

$p_f$ = probability of failure

$n$ = dice rolled

$k$ = desired successes

The math I have without re-rolls

Probability of a single success:

$p_s = \frac{3}{10} = 0.3$

Probability of x success:

$p_s^x$

Probability of at least 1 success:

$1-p_f^x$

Probability of at least k successes with n dice:

$1 - \displaystyle\sum_{i=1}^{k} {n \choose i}(p_s^i)(p_f^{n-i})$

The problem

Assuming my above math is correct (and please feel free to point out if and where I went wrong), how do I account for the rules of re-rolling?

A single 10 sided die would have a 30% chance of being a success and a 10% chance of being re-rolled; the re-roll would have a 30% chance of being an additional success with a 10% chance to re-roll, and so on and so forth.

I just do not know how to express that mathematically.

Answer 1

Here is one way to think about it. For each individual die, the trial is to roll it repeatedly until the first time that the result is $<10$. The result of this trial is therefore a number from $1$ to $9$, and each of those possibilities is equally likely. So, this trial is exactly the same as rolling a 9-sided die, and your problem can be simplified: Rule (1) $n$ 9 sided dice are rolled; Rule (2) any rolls $\ge 8$ are a success. Rules (3) and (4) are no longer relevant.