I have a set of game dice that have unique faces with the following number of occurrences: 2/8 blank, 2/8 "hit", 1/8 "block", 1/8 "crit", 1/8 "wild", and 1/8 "fail".
I'm trying to determine the probability of rolling at least 2 "crits" and one "wild" (1/8 chance each) on a given roll of 9 dice.
One additional factor making this difficult for me is whenever a "crit" is rolled, it adds an additional die roll into the pool. This extra dies face is then counted towards the total pool.
So with these circumstances, how would I calculate the final chance of rolling 3 exact 1-in-8 chance faces on at least 9 dice (due to crits)?
To clear up some confusion pointed out by the comments, the additional dice rolled by a crit do not add an additional die themselves. This means the most dice you could end a roll with is double the original roll, if every original die rolled a crit face.
If you roll at least 2 crits + at least 1 wild in the first 9 dice, you are done.
If you roll exactly 1 crit + at least 1 wild in the first 9 dice, you need a crit, and then you are done.
Let's calculate these first, then discuss cases where you find extra crits and/or the wild among the bonus dice.
If we let $j$ represent the number of dice that roll "crit" and $k$ represent the number of dice that roll "wild" among the first nine dice, we have: $$\sum_{j=2}^8\sum_{k=1}^{9-j}\dfrac{9!}{j!k!(9-j-k)!}\left(\dfrac{1}{8}\right)^{j+k}\left(\dfrac{6}{8}\right)^{9-j-k} = \dfrac{26821545}{134217728}$$
$$\sum_{k=1}^8\dfrac{9!}{1!k!(8-k)!}\left(\dfrac{1}{8}\right)^{2+k}\left(\dfrac{6}{8}\right)^{8-k} = \dfrac{36766665}{1073741824}$$
Finally, we need at least 2 "crit"s and zero "wild"s among the first 9 dice, but we get at least one "wild" among the bonus dice:
$$\sum_{k=2}^9\dbinom{9}{k}\left(\dfrac{1}{8}\right)^k\left(\dfrac{6}{8}\right)^{9-k}\left(1-\left(\dfrac{7}{8}\right)^k\right) = \dfrac{557189340299017}{18014398509481984}$$
This is all, because if you only get zero or one "crit" among the first nine dice and no "wild", you will not have enough rolls to achieve 2 "crit" and 1 "wild".
Thus, the answer is:
$$\dfrac{26821545}{134217728}+\dfrac{36766665}{1073741824}+\dfrac{557189340299017}{18014398509481984} = \dfrac{4773958451953417}{18014398509481984} \approx 26.5\%$$
Note: These probabilities are only additive because I defined them to be disjoint events, and I ensured I found every event that yielded the requisite outcome by listing the possibilities to exhaustion. Let me know if you have any questions about the formulas I used.
Another way to solve this is to calculate the probability that you do not achieve at least two "crits" and one "wild". This may be an easier way to solve the problem. We can break this down by the number of "crit"s rolled among the first nine dice. Then, from the law of total probability, the probability that you do roll at least two "crits" and at least one "wild" will be the complementary probability.
Zero "crit"s: $$\left(\dfrac{7}{8}\right)^9 = \dfrac{40353607}{134217728}$$
One "crit", bonus die is not crit: $$\dbinom{9}{1}\left(\dfrac{1}{8}\right)\left(\dfrac{7}{8}\right)^9 = \dfrac{363182463}{1073741824}$$
One "crit", bonus die is crit, no wilds: $$\dbinom{9}{1}\left(\dfrac{1}{8}\right)^2\left(\dfrac{6}{8}\right)^8 = \dfrac{59049}{4194304}$$
More than one "crit", no wilds: $$\sum_{k=2}^9\dbinom{9}{k}\left(\dfrac{1}{8}\right)^k\left(\dfrac{6}{8}\right)^{9-k}\left(\dfrac{7}{8}\right)^k = \dfrac{1477466456359159}{18014398509481984}$$
Thus, the probability that you do achieve the result you want is:
$$1-\dfrac{40353607}{134217728}-\dfrac{363182463}{1073741824}-\dfrac{59049}{4194304}-\dfrac{1477466456359159}{18014398509481984} = \dfrac{4773958451953417}{18014398509481984}$$
Which is the same answer as above.