I have three dice: 8 sided, 3 sided, and 4 sided.
I roll them in order 8, 3, 4.
I only roll the next one if the previous result is a '1'.
What is the probability of getting 1, then 1, then 1?
Is p = (1/8)x(1/3)x(1/4) correct?
Would there be a 1.04% chance of that result?
Is there a correct name or way to reference this type of probability where you only stack on an additional probability based on the outcome of the previous outcome?
I used "stack probabilities" out of naivety.
Let $A_1$ be event that you have $1$ on the $8$-sided die, $A_2$ be event that you have $1$ on $3$-sided die and let $A_3$ be event that you have $1$ on $4$-sided die. We will use chain rule here (which is often called product rule):
$\mathbb{P}(A_1 \cap A_2 \cap A_3) = \mathbb{P}(A_3 |A_2\cap A_1)\cdot\mathbb{P}(A_2|A_1) \cdot \mathbb{P}(A_1)$.
We assume that every die is fair so you have equal probability to get any one of its sides.
So $\mathbb{P}(A_1) = \frac{1}{8}$.
And if you had got $1$ on the first die then you have probability $\mathbb{P}(A_2|A_1) = \frac{1}{3}$ that you would have $1$ on the second die.
If you got $1$ on the first die and $1$ on the second die then $\mathbb{P}(A_3 |A_2\cap A_1) = \frac{1}{4}$ is the probability that you will get $1$ on the last die.
Stacking all together we get your answer:
$\mathbb{P}(A_1 \cap A_2 \cap A_3) = \mathbb{P}(A_3 |A_2\cap A_1)\cdot\mathbb{P}(A_2|A_1)\mathbb{P}(A_1) = \frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{8} = \frac{1}{96}$.