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
- n 10 sided dice are rolled.
- Any rolls $\geq$ 8 are a success.
- Any rolls of 10 are re-rolled until < 10 is obtained.
- 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.
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.