In this situation there are 20-sided dice (or d20) with equal chances for all values. A "success" in each scenario would be a result of a certain value or higher.
For example, if a 13 is needed to succeed, it means that 13 and all greater values are a success, total of 40% chance to succeed. If a 3 is needed to succeed, 3 and all greater values would yield a success, for a total of 90% chance to succeed - and so on.
If we use the example of 13 to understand a situation in which we roll two d20s instead of one, a single success would be calculated as follow: each individual die has the same 40% chance to yield a success, which we can sum, but we'll have to subtract the chance to succeed with both (I imagine this like a Venn diagram, where the overlap is "added twice"). The total chance would be 2 times 0.4, minus 0.4 squared = 0.64, or 64%.
What I'm struggling with is the three and four d20s situations. How do I approach this? I get confused from all of the possible "double-overlaps", I can't reach a solid conclusion.
I tried using a binomial logic, but ended up with really weird chances, even negative ones!
Please help me understand, I'd really appreciate it if you'd add an example with your explanation. Thanks a lot in advance :)