Each 20-sided die in this case has equal chance to land on any whole-number value 1-20.
I understand that if I aimed for at least one 20 from a roll of two dice: either d1 is a 20 (1/20), or d2, or both (1/(20^2)). If if wanted to be more "strict" and aim for just one 20 between them, I combine the chances for either and subtract the chance for both.
I managed to apply this "formula" to a chance to get any value on the die in a 2d20 roll situation.
What I struggle with is the following situations: rolling 3d20 aiming for just one "success", and rolling 4d20 aiming for just one "success".
Please help me understand the logic for both the 3d20 and the 4d20 situations, let's say regarding 20 as a success.
Then please show me how you turn this into a "formula" for any d20 value as a success.
For example, when I tried a formula for 3d20: 3 times P(success per die) - 3 times P(success on 2/3 dice with no regard to die #3) - P(success on all 3).
This seemed fine until I reached values of 11 and lower, where the chances dropped instead of increased.
Thanks a lot in advance, please help me understand :)
Ever done event probabilities or the binomial distribution? Your first situation of 2 dice perfectly describes an event space.
Let $Pr(A) = d1$ getting a $20$ $Pr(B) = d2$ getting a $20$. $Pr(A \cap B) = \frac{1}{20} \times \frac{1}{20}$ as the events are independent, just like you said. This is fine when you have 2 events but when you start get more complicated events like 3 or 4 or more (ie 3+ die), you may want to consider the binomial distribution.
The binomial distribution states for every criteria in binomially distributed event space, the probability of obtaining such a criteria is
$$Pr(X = x) = {n\choose x}({p})^x({1-p})^{n-x}$$
Lets decode this. $X$ is the success criteria (what you want to achieve) (In your case rolling a 20). Small $x$ is on how many tries do you want to achieve that success (In your case just 1 "20"). $n$ is total number of tries (In your case 4 tries as each die (assuming you have 4) can roll a 20). $p$ is the probability of success (In your case, success of rolling a 20 or ($\frac{1}{20}$))
The ${n\choose x}$ part means how many ways are there to choose $x$ items from $n$ number of items.
The $({p})^x$ denotes the probability of your event occurring with respect to total number of success chosen.
The $({1-p})^{n-x}$ denotes the probability of everything else in your sample space occurring with respect to total number of success chosen. This is key as not only does your success criteria, $X$ affect your probability but also $X'$ - everything else their is to occur can also occur.
Now putting everything in terms of your event:
In your event of you having 4 die and wanting just $1$ roll of $20$ between them. Let $X$ denote the success criteria of rolling a $20$ out of 4 die and the probability of success of rolling a $20$ is ofc $\frac{1}{20}$. We just want $X = 1$ out of the 4 die (as there is 1 "20" on each).
$Pr(X = 1) = {4\choose 1}\times(\frac{1}{20})^1\times(1-\frac{1}{20})^{4-1} \approx 0.171475$