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
 A: 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.

Note that the formulas can be simplified a bit, e.g. the one with a double sum is $$\left(\frac{29}{30}\right)^{12}\left(\frac{1}{12}\frac{19}{20}\frac{1}{20}\right)
\sum_{k=0}^7
\left(\frac{11}{12}\frac{19}{20}\right)^k\cdot
\sum_{l=0}^{9-k}
\left(\frac{19}{20}\right)^l $$
A: For $i\in \{1,2,3\}$, let $D_{i,0}$ be the event that die $i$ is not discarded and $D_{i,1}$ be the event that die $i$ is discarded.
We wish to calculate
$$p = \mathbb P\big(D_{1,1}\cap D_{2,1} \cap D_{3,1}\big).$$
Complimentary probability and De Morgan's laws yield that
\begin{align}
1-p
&=
\mathbb P\left(\left(\bigcap_{i=1}^3\, D_{i,1}\right)^\complement\right)
\\&= 
\mathbb P\left(\bigcup_{i=1}^3\,{D_{i,1}}^\complement\right)
\\&=
\mathbb P\left(\bigcup_{i=1}^3\,D_{i,0}\right)
\end{align}
We write $\cup_{i=1}^3\,D_{i,0}$ as a disjoint union of events:
\begin{align}
\bigcup_{i=1}^3\,D_{i,0}
\quad=\quad&
\left(D_{1,0} \cap D_{2,0} \cap D_{3,0}\right)
\\\cup\,\,&
\left(D_{1,0} \cap D_{2,0} \cap D_{3,1}\right)
\cup \left(D_{1,0} \cap D_{2,1} \cap D_{3,0}\right)
\cup \left(D_{1,1} \cap D_{2,0} \cap D_{3,0}\right)
\\\cup\,\,&
\left(D_{1,1} \cap D_{2,1} \cap D_{3,0}\right)
\cup \left(D_{1,1} \cap D_{2,0} \cap D_{3,1}\right)
\cup \left(D_{1,0} \cap D_{2,1} \cap D_{3,1}\right)
\end{align}
Do you think you can take it from here?

The probabilities for
$$\left(D_{1,0} \cap D_{2,0} \cap D_{3,0}\right)\\
\left(D_{1,0} \cap D_{2,0} \cap D_{3,1}\right)\\
\left(D_{1,0} \cap D_{2,1} \cap D_{3,0}\right)\\
\left(D_{1,1} \cap D_{2,0} \cap D_{3,0}\right)$$
should be easy enough to calculate.
The probability for events of type $\left(D_{i,1} \cap D_{j,1} \cap D_{k,0}\right)$ is a bit trickier.
Regardless of when each die, $i$ and $j$, roll $1$, die $k$ will definitely be rolled $12$ times.
It must not be discarded, so that gives us a factor of ${(1-1/d_k)}^{12}$ already, where $d_k$ is the number of sides of die $k$.
At this point, we need only consider the probability of $\left(D_{i,1} \cap D_{j,1}\right)$.
We can make this precise with conditional probability.
We have
$$
\Bbb P\left(D_{i,1} \cap D_{j,1} \cap D_{k,0}\right)
= \underbrace{\Bbb P\left(D_{k,0}\mid D_{i,1} \cap D_{j,1}\right)}_
{{(1-1/d_k)}^{12}}
\cdot \Bbb P\left(D_{i,1} \cap D_{j,1}\right)$$

Calculating $P\left(D_{i,1} \cap D_{j,1}\right)$ can be done similarly.
We use complimentary probability and De Morgan's laws again:
\begin{align}
\Bbb P\left(D_{i,1} \cap D_{j,1}\right)
&=
1 - \Bbb P\left(\left(D_{i,1} \cap D_{j,1}\right)^\complement\right)
\\&=
1 - \Bbb P\left({D_{i,1}}^\complement \cup {D_{j,1}}^\complement\right)
\\&=
1 - \Bbb P\left(D_{i,0}\cup D_{j,0}\right)
\end{align}
We write $D_{i,0}\cup D_{j,0}$ as a disjoint union of events:
\begin{align}
D_{i,0}\cup D_{j,0}
=\quad &
(D_{i,0}\cap D_{j,0}\cap D_{k,0})\cup (D_{i,0}\cap D_{j,0}\cap D_{k,1})
\\\cup\,\,&
(D_{i,0}\cap D_{j,1}\cap D_{k,0})\cup (D_{i,0}\cap D_{j,1}\cap D_{k,1})
\\\cup\,\,&
(D_{i,1}\cap D_{j,0}\cap D_{k,0})\cup (D_{i,1}\cap D_{j,0}\cap D_{k,1})
\end{align}
Can you see how this will yield a linear system relating the $\left(D_{i,1} \cap D_{j,1} \cap D_{k,0}\right)$?
