DeathControl

Bone008 Author

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#1 Jan 21, 2012 at 18:42 UTC - 0 likes

Quote from kozzy68: Go
Hi, use this formula it will be much better. Lets say you keep 33% (1/3) of items and you have 2 diamonds that is 2/3=0.666 = you have 66% chance to keep that one diamond. Scenario 2 when you have 4 diamonds 4/3 = 1.33 you keep one diamond always and have 33% chance to keep also second one. Its elegant fast solution and spliting diamonds would not give you any adventage. I hope its clear how I explained it might sound bit geeky but its simple. Another idea would be not bad to have option to keep items that you are wearig.

I don't really understand the logics behind it yet, but I think it is quite what I am looking for as a replacement for the current loss system. Mind explaining it a bit more? Maybe a general forumla? And ... what would for example happen if you had a stack of 30 diamonds, and the loss-percentage would be let's say 60% (3/5) ?

Bone008 Author

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#2 Jan 22, 2012 at 14:13 UTC - 0 likes

Quote from kozzy68: Go
60% we can say its 6/10 = 0.6 * 30 = 18 diamonds to keep and 12 to drop.So there is no complication here result is integer. Tts not best example. Complcation is when you get results like keep 0.3 diamods or 1.7 diamond. We dont like to simply round it. Instead we play dice with last item in stack if we keep it or not and odds are given by reminder 0.5 is 50% 0.3 is 30% ... . so fair formula is: [%] * [items in stack] / 100 = X.y (1.33 x=1 y=0.33 ) X is integer part and y is rest. keep = X+Radom(y) result of random function is either 0 or 1 and odds are given by y

Bone008 Author

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#3 Jan 22, 2012 at 14:13 UTC - 0 likes

@kozzy68: Go

Alright, I got it. So the idea is to apply a random factor in case of an uneven outcome instead of rounding it. Not a huge thing for big stacks, I think, but it definetaly solves the reason why I switched to stack-dropping in the first place: non-stackable items would always be kept when the percentage would be below 50% and vice versa.

I'll probably change it back and to what you suggested in the next update, no promises, though ;)

kozzy68 Regular Shmoe

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#4 Jan 23, 2012 at 20:22 UTC - 0 likes

:) well non-stackable items or 1 item stacks will be keept randomly exacly same as yours original formula works ... and thats ok I think. You have one pickaxe and you can keep 90% of items so formula: 90*1*/100 = 0.9 X=0 y=0.9 keep = 0+Random(0.9) so you have 90% chance to keep pickaxe.

No one can cheat by splitting items into bunch of one stack piles :)

I see nonstackable item as stack of 1 item. not sure if MC works like that .

Last edited Jan 23, 2012 by kozzy68

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New loss percentage formula (kozzy68)