Let's say I have a bag with 6 different dice in it, a D4, D6, D8, D10, D12, and D20. I randomly draw 3 dice from the bag without replacement and roll each of them.
I want to know a formula for calculating the probability of at least 2 of them rolling a value of 1.
I think I'm getting hung up on the fact that the probability of each die rolling a 1 is different, along with the fact that I'm doing this without replacement so the chance is changing with each draw. It feels like a hypergeometric distribution problem, but since there's not clear "successes" for the draws (any die is OK to draw), I'm not sure how to take that into account.
I'm happy with first answering the question of "exactly 3 of them rolling a value of 1" and then expanding it to the "at least 2" version.
I understand that, at first, the combined chance for each die of drawing it and rolling a 1 is, respectively...
$\left(\frac{1}{6}\right)\left(\frac{1}{4}\right)$, $\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)$, $\left(\frac{1}{6}\right)\left(\frac{1}{8}\right)$, $\left(\frac{1}{6}\right)\left(\frac{1}{10}\right)$, $\left(\frac{1}{6}\right)\left(\frac{1}{12}\right)$, and $\left(\frac{1}{6}\right)\left(\frac{1}{20}\right)$
or
$0.041\bar6$, $0.02\bar7$, $0.0208\bar3$, $0.01\bar6$, $0.013\bar8$, and $0.008\bar3$
But then for the second draw it changes to a $\left(\frac{1}{5}\right)$ chance of drawing each die that is left. And while the chance of rolling a 1 on each individual die doesn't change, WHICH die was drawn first and is no longer available to be drawn does affect the chances that are left.
Brute forcing this with a probability tree gets large very quickly. If the problem was just drawing from a bag or just rolling dice, I feel like they'd be relatively simple, but I'm having trouble putting those two types of problems together.
For dice of type D(A),D(B),D(C) the probability of at least 2 showing 1's is
$ \begin{align} (1/A)*(1/B)*((C-1)/C) + (1/A)*((B-1)/B)*(1/C) + ((A-1)/A)*(1/B)*(1/C) + (1/A)*(1/B)*(1/C) = (A+B+C-2)/(A*B*C) \end{align}$
Moving on, We are left to choose A,B and C. Which we have to do manually and there is no other way around. All the possible 20 combinations have equal probability of being chosen. We can calculate answer for each of those, add them up and divide by 20.
I understand your concern of manually doing calculations so you could always do a little coding to help you out with the process. I have taken the liberty to code it for you.
Edit: Yes, the (-2) accounts for double counting. It would be more clear if we used the Inclusion Exclusion Principle to calculate the probability. Try it.