Dice roll probability game

I have a chance game I am creating and the best way to explain the problem is using a dice game analogy.
The game consists of 8 rounds. Each round involves throwing 3 dice. One die has 12 sides. One die has 20 sides and the last die has 30 sides.
During a round if a die is rolled lands on a 1 then that die is discarded from the rest of the rounds and an extra 2 bonus rounds are added to the game. The objective of the game is to discard all 3 dice before you complete all rounds. A game can have up to 12 rounds when you include the bonus rounds.
How can I calculate the probability of a game discarding all 3 dice?
The trick to this game is knowing that after I have one die discarded the other two have an extra 2 rounds to roll their 1. If the second die rolls a 1 then the final die has a further 2 rounds again. It is of course possible but unlikely that all die get discarded in the first round.
Thanks

Answer

The probability of never discarding any die is $$\left(\frac{11}{12}\frac{19}{20}\frac{29}{30}\right)^8$$

The probability of discarding the $12$-die only is $$\sum_{k=0}^7\left(\frac{11}{12}\frac{19}{20}\frac{29}{30}\right)^k\left(\frac{1}{12}\frac{19}{20}\frac{29}{30}\right)\left(\frac{19}{20}\frac{29}{30}\right)^{9-k}$$

The probability of discarding the $12$-die first and the $20$-die second is $$\sum_{k=0}^7\sum_{l=0}^{9-k}\left(\frac{11}{12}\frac{19}{20}\frac{29}{30}\right)^k\left(\frac{1}{12}\frac{19}{20}\frac{29}{30}\right)\left(\frac{19}{20}\frac{29}{30}\right)^l \left(\frac{1}{20}\frac{29}{30}\right)\left(\frac{29}{30}\right)^{10-k-l}$$

The probability of discarding the $12$-die and the $20$-die simultaneously, and never discarding the $30$-die is $$\sum_{k=0}^7\left(\frac{11}{12}\frac{19}{20}\frac{29}{30}\right)^k\left(\frac{1}{12}\frac{1}{20}\frac{29}{30}\right)\left(\frac{29}{30}\right)^{11-k}$$

From these formulas and their symmetric versions (discarding the $20$-die only, etc.) you can work out the probability of the complementary event.

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Note that the formulas can be simplified a bit, e.g. the one with a double sum is $$\left(\frac{29}{30}\right)^{12}\left(\frac{1}{12}\frac{19}{20}\frac{1}{20}\right)
\sum_{k=0}^7
\left(\frac{11}{12}\frac{19}{20}\right)^k\cdot
\sum_{l=0}^{9-k}
\left(\frac{19}{20}\right)^l $$