Vtyjkyuk

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

The probability of never discarding any die is $$left(frac1112frac1920frac2930right)^8$$

The probability of discarding the $12$-die only is $$sum_k=0^7left(frac1112frac1920frac2930right)^kleft(frac112frac1920frac2930right)left(frac1920frac2930right)^9-k$$

The probability of discarding the $12$-die first and the $20$-die second is $$sum_k=0^7sum_l=0^9-kleft(frac1112frac1920frac2930right)^kleft(frac112frac1920frac2930right)left(frac1920frac2930right)^l left(frac120frac2930right)left(frac2930right)^10-k-l$$

The probability of discarding the $12$-die and the $20$-die simultaneously, and never discarding the $30$-die is $$sum_k=0^7left(frac1112frac1920frac2930right)^kleft(frac112frac120frac2930right)left(frac2930right)^11-k$$

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 $$left(frac2930right)^12left(frac112frac1920frac120right)
sum_k=0^7
left(frac1112frac1920right)^kcdot
sum_l=0^9-k
left(frac1920right)^l $$

For $iin 1,2,3$, 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

$$p = mathbb Pbig(D_1,1cap D_2,1 cap D_3,1big).$$

Complimentary probability and De Morgan's laws yield that

beginalign
1-p
&=
mathbb Pleft(left(bigcap_i=1^3, D_i,1right)^complementright)
\&=
mathbb Pleft(bigcup_i=1^3,D_i,1^complementright)
\&=
mathbb Pleft(bigcup_i=1^3,D_i,0right)
endalign

We write $cup_i=1^3,D_i,0$ as a disjoint union of events:

beginalign
bigcup_i=1^3,D_i,0
quad=quad&
left(D_1,0 cap D_2,0 cap D_3,0right)
\cup,,&
left(D_1,0 cap D_2,0 cap D_3,1right)
cup left(D_1,0 cap D_2,1 cap D_3,0right)
cup left(D_1,1 cap D_2,0 cap D_3,0right)
\cup,,&
left(D_1,1 cap D_2,1 cap D_3,0right)
cup left(D_1,1 cap D_2,0 cap D_3,1right)
cup left(D_1,0 cap D_2,1 cap D_3,1right)
endalign

Do you think you can take it from here?

The probabilities for

$$left(D_1,0 cap D_2,0 cap D_3,0right)\
left(D_1,0 cap D_2,0 cap D_3,1right)\
left(D_1,0 cap D_2,1 cap D_3,0right)\
left(D_1,1 cap D_2,0 cap D_3,0right)$$

should be easy enough to calculate.

The probability for events of type $left(D_i,1 cap D_j,1 cap D_k,0right)$ 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 $left(D_i,1 cap D_j,1right)$.
We can make this precise with conditional probability.
We have

$$
Bbb Pleft(D_i,1 cap D_j,1 cap D_k,0right)
= underbraceBbb Pleft(D_k,0mid D_i,1 cap D_j,1right)_
(1-1/d_k)^12
cdot Bbb Pleft(D_i,1 cap D_j,1right)$$

Calculating $Pleft(D_i,1 cap D_j,1right)$ can be done similarly.
We use complimentary probability and De Morgan's laws again:

beginalign
Bbb Pleft(D_i,1 cap D_j,1right)
&=
1 - Bbb Pleft(left(D_i,1 cap D_j,1right)^complementright)
\&=
1 - Bbb Pleft(D_i,1^complement cup D_j,1^complementright)
\&=
1 - Bbb Pleft(D_i,0cup D_j,0right)
endalign

We write $D_i,0cup D_j,0$ as a disjoint union of events:

beginalign
D_i,0cup D_j,0
=quad &
(D_i,0cap D_j,0cap D_k,0)cup (D_i,0cap D_j,0cap D_k,1)
\cup,,&
(D_i,0cap D_j,1cap D_k,0)cup (D_i,0cap D_j,1cap D_k,1)
\cup,,&
(D_i,1cap D_j,0cap D_k,0)cup (D_i,1cap D_j,0cap D_k,1)
endalign

Can you see how this will yield a linear system relating the $left(D_i,1 cap D_j,1 cap D_k,0right)$?