skyblock VS icepool

pip install git+https://github.com/peter-hunt/skyblock.git

My Icepool Python package can do this:

Icepool is open-source. You can run Icepool anywhere you like: on a server, on your own computer, on your client's computer, on your phone, even offline. And if you're interested in understanding how it works, you can read through the source code, though for this purpose my paper on the dice pool algorithm may be a better choice.

Still, I chose Python for my Icepool dice probability package; compared to JS, Python has BigInts by default (you can append n to literals in JS, but it is a little annoying); and operator overloading, which allows for more compact and readable syntax.

While the main focus is dice, my Icepool Python package can do some types of card problems. Here is a simple example that computes the chance of dealing at least one H and one K from a deck:

I have some articles, but they do require some background in probability, and I haven't pursued these lines of inquiry in a while. More recently, I've been working on Icepool, a Python probability package. You may be interested in the collection of runnable notebooks. I also have a paper on the dice pool algorithm, which can handily outpace multiset enumeration systems such as AnyDice and Troll on a variety of problems, but I wouldn't exactly call it accessible to beginners.

Here's mine.. Depending on how far you want to go, you may also be interested in the accompanying paper.

And my own Icepool, which includes a high-efficiency pool algorithm.

My Icepool Python library has some deck support.

If you are interested in learning more, you can read the preprint of my paper on the subject, or try my Icepool Python package, which is what powers the apps in the first two links.

You can compute the analytic solution efficiently with the right algorithms. I've implemented one such algorithm in my Icepool Python package, which reduces the computation to just a few lines and seconds. Here's an example of computing the distribution of the difference of the best and worst highest score for a party of 4.