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Finding the largest average from a set of unknown distributions

I have been made aware that this is a multi-armed bandit problem I'm describing.

Let's say I have a function that rolls weighted 20-sided dice, and each set of arguments rolls a die with different weighing. We don't know the precise distributions for these dice but they have a peak or two somewhere and other possible numbers are likely nearby but not always adjacent. Only some of the arguments are numerical, but it's mostly the non-numerical ones that are problematic.

The problem I want to solve is how to efficiently find the set of arguments that gives the best average result (so, find "the best die") and an estimate for its average return value.

I'll try to illustrate. Our example function takes two arguments, a die type string and an effect size. Inspiration for my example of die types: https://kingdom-come-deliverance.fandom.com/wiki/Dice#Dice_Effects

Let's take the "lu" die as our type and have an effect size of 1 for it. That makes the number 20 19%-points more probable to be the outcome and the other 19 numbers will be 1%-point less probable. Let's call this set of arguments Die-A. We get an average die roll of 12.4. Let's have another die but of the "ul" type that has an "opposite" effect from that of the lu die. With an effect size of 1, the result of 1 is 19%-points more probable to be the outcome and the other numbers are 1%-point less probable. This Die-B has an average die roll of 8.6. In this example, we would return an identifier for the Die-A and an estimate for the average it gives, close to the actual 12.4.

Of course in a real scenario there are many more "dice" and "rolling" them is done by a very complex algorithm that basically necessitates solving for the average results stochastically. How do I solve for the best die while giving less attention to less promising candidates to improve efficiency? I feel there must be some algorithm or another for this but what do I look for? My code base is in JavaScript so code examples preferably in that.

about 4 years ago · Juan Pablo Isaza
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