I'd like to come back on the Bobby's Dream Quest Chain. Initially I have mathed nothing out and randomly choose 7 required items for each of Bobby's Dream Quest. I was wondering if I could come up with weightings (drop chances) for each element from the drop tables that would be reasonable to grind for. After randomly plugging in some values I quickly came to the conclusion that having 7 different required items would result in very unreasonable chances to complete any of the quests from the proposed quest chain. To find out reasonable values I tried writing a general formula to help find appropriatie values. This is what I came up with. This formula can be used to calculate when someone has 50% chance to pull each desired element at least once out of a larger set while having a variable weight after (n) rolls from the drop table. x = Total Desired Elements from a Set a = Weight of each Element p = Chance to Pull any Desired Element from a Set n = Total Pulls from Set {x,n ∈ N| x,n>0} {a,p ∈ R| 0<p<1, a>0} When plugging in for 7 Elements of the Set A Weight of 2 for each Desired Element having(wich means having twice the drop chance over undesired Elemets from the set) 2000 pulls from the proposed Drop Table WolframAlpha gives back values for (p). Wich is the chance to pull any of the Desired Elements from the Set. Because maths we find multiple possible solutions ranging from p=0.613 to p=0.812. I can't really be bother to find out the right solution from all possible solutions that WolframAlpha gives back at the moment. But having 61.3% chance or more to pull any of these required items for the Quest Chain feel very incorrect. Consider the dopamine rush and hype for pulling a chaos scroll when you'd have 61% chance to pull one. Newly proposed values: Spoiler Intuitively I feel a=2 p=0.2 n=2000 solve for (x) When plugging these number we'll receive x=3.663 since x∈R we can only use 3 or 4 as our answers a=2 p=0.2 x=3 or 4 solving for (n) returns x=3 x=4 n=390 n=4621 Both values for (n) are either quite low or way to high. Trying to find new values for (p) a=2 x=3 or 4 n=2000 solve for (p) x=3 x=4 p=0.116 p=0.247 These values feel appropriate. tl;dr x=3 or x=4 a=2 a=2 p=0.116 p=0.247 Or at least changing the propsed amount of Quest Items to 3 or 4 instead of 7. Solving for (n)=2000 might be a large amount of pulls. However with these proposed quest items being tradeable and it being meant to incentivize trading and engaging with the Island Community. I feel it's not very unreasonable. We could endlessly plug in different values and agree on different desired values for (n). However, at the moment I do not have much experience with designing for drop tables, I might return on the subject later. Note that these items only drop from the Ore/Leather Drop Table, however both the Normal and the Ore/Leather Drop Tables return ores. Increasing values for these Ore-Drops can also mitigate the difficulty acquiring these Quest Items. My vision for the quest chain is for it to be a great accomplishment with a lot of longevity. Increasing engagement and activity. While minimaly affecting the current Maple Island. I still do not support new maps, mobs, pq or a boss and do think that this would be a good comprimise to those looking for new content and those wanting to preserve Maple Island as it is.
