[Release] Formulla to calculate equal droprates

Name of Release: Uhm, Reb3lzrr's Equal Droprates?
Description: Well, I wanted to (re)make the droprates, and I came by the events. Here I wanted that ALL flags ( like 15 items ), have equal droprates ( all 100/15 = 6.666 percent chance to drop ), but was a f*cking long formula.

So I created a formula which tells which number you have to put in in the ItemGroup.txt. So I putted this formula in exel for people who aren't as good in math as I am lol
Features: for droplists
Download-Mirror 1: [ Here ]

Last Word:
* Using these numbers would result in a itemgroup where an item HAS to come out
* I'm btw not responsible for any damage to your server or droplists

can u post the formula please?

http://www.reb3lzrr.com/downloads.ph...download_id=14 this is the whole Formula^^

The formulla is in the exel file

but ill post in anyway:

Code:

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=(B21*10)/((1-(D21/1000))*(1-(E21/1000))*(1-(F21/1000))*(1-(G21/1000))*(1-(H21/1000))*(1-(I21/1000))*(1-(J21/1000))*(1-(K21/1000))*(1-(L21/1000))*(1-(M21/1000))*(1-(N21/1000))*(1-(O21/1000))*(1-(P21/1000))*(1-(Q21/1000))*(1-(R21/1000))*(1-(S21/1000))*(1-(T21/1000))*(1-(U21/1000))*(1-(V21/1000)))

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* The D/E/Fetc 21 are variables, and B21= the requested equal droprate ( for example 5% )

To be honest, I don't belive they use a formula, and far from one as complex as you posted.
They use simple math, in a very wrong way though. Pacd allready proven this long time ago.

So I doubth it will be possible to calculate equal droprates, only something close too.

When you have 4 items, wouldnt it be "250 500 750 1000" instead of "250 333 500 1000" ????

No, they don't use logic.

Quote:

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Originally Posted by DeathArt

Pacd allready proven this long time ago. too.

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I made my formula based on this guide, written by pacd,

Code:

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=(B21*10)/((1-(D21/1000))*(1-(E21/1000))*(1-(F21/1000))*(1-(G21/1000))*(1-(H21/1000))*(1-(I21/1000))*(1-(J21/1000))*(1-(K21/1000))*(1-(L21/1000))*(1-(M21/1000))*(1-(N21/1000))*(1-(O21/1000))*(1-(P21/1000))*(1-(Q21/1000))*(1-(R21/1000))*(1-(S21/1000))*(1-(T21/1000))*(1-(U21/1000))*(1-(V21/1000)))

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Well This is the formula which will apply on the 19th item in an itemgroup,
And it has to be this long, because everytime an item gets loaded, you will take a factor over the remaining chance, just take a look at this:

Code:

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(1-(NUM/1000))

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this calculates the factor which will be applied to get the remaining chance

in example:
250: (1-(250/1000)) = 0.75
500: (1-(500/1000)) = 0.5
so everytime a new item gets added the formulla gets longer. And adds this:"(1-(NUM/1000))" statement



I will use pacd's example to make clear my formula is working
This examples uses the numbers: "500 443 470"

the first item has 500% chance to drop
the second item has (1000*(1-(500/1000))*(1-(433/1000) = 283.5 ( 28.35% chance to drop )
the third item has (1000*(1-(250/1000))*(1-(500/1000)*(1-(470/1000)) = 150.255 ( 15.025% chance to drop )

*My numbers are more precies because pacd rounded off.

Quote:

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Originally Posted by BjornVH

When you have 4 items, wouldnt it be "250 500 750 1000" instead of "250 333 500 1000" ????

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Hmm not realy,
the first item has 25% chance to drop
the second item has (1000*(1-(250/1000))*(1-(500/1000) = 375 ( 37% chance to drop )
the third item has (1000*(1-(250/1000))*(1-(500/1000)*(1-(750/1000)) = 93.75 ( 9,375% chance to drop )
the forth item has (1000*(1-(250/1000))*(1-(500/1000)*(1-(750/1000))*1 ( because (1-(1000/1000) equals 0 , but it takes the 100% remaining chance. So we need to load the last enty, and do this times 1. ) = 93.75 ( 9,375% chance to drop )

In my way it would be like this:
the first item has 25% chance to drop
the second item has (1000*(1-(250/1000))*(1-(333/1000) = 250 ( 25% chance to drop )
the third item has (1000*(1-(250/1000))*(1-(333/1000)*(1-(500/1000)) = 250 ( 25% chance to drop )
the forth item has (1000*(1-(250/1000))*(1-(333/1000)*(1-(500/1000))*1 = 250 ( 25 chance to drop )



just trust me lol I wont release sh*t :D

Well Izu says you're wrong.

LOL I say im right,

This formulla is based on pacd's way of calculating this. You all say he is right, but I'm wrong w/e; beleave what you want but dont say im wrong

Actually i disagreed with PACD.
He is wrong.
I agree with KING IZU

A better way of explaining how they work as follows:

Code:

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Item Groups work like a 1000 sided dice.

when a itemgroup is called by a monsters death the die is rolled twice.
(For each bag that drops)

The first dice roll determines what group of items will be picked.

The second dice roll determines what group of items will be picked.

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Example 1

Code:

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(itemgroup (index 2000) (group (500 2001)(600 2002)(800 2003)(950 2004)))

 Broken Down:

Group 2001 is assigned the 1st 500 sides of the dice (50%)
Group 2002 is then assigned the remaining sides up-to and including 600 (10%)
Group 2003 is then assigned the remaining sides up-to and including 800 (20%)
Group 2004 is then assigned the remaining sides up-to and including 950 (15%)

as the last group was only 950 there is 50 sides that have
nothing attibuted to them and therefore if chosen they drop nothing (5%)

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Example 2

Code:

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(group (index 2001) (money (400 200) (500 8000))(item (550 730)(700 731)(1000 732)))

Broken Down:

The Groups of items work the same way as itemgroups do but reference items
rather than groups

200 geons is assigned the 1st 400 sides of the dice (40%)
8,000 geons is then assigned the remaining sides up-to and including 500 sides (10%)
Item 730 is then assigned the remaining sides up-to and including 550 sides (5%)
Item 731 is then assigned the remaining sides up-to and including 700 sides (15%)
Item 732 is then assigned the remaining sides up-to and including 1000 sides (30%)

As the last item was assigned up to and including 1000 all sides are used and
therefore there is no chance of getting no drop.

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But hey it's your server files, so you can stuff it up all you want. I just explained how they actually work. take it or leave it

LoL nice guide

It's still a fact that none of you know the precise dropsystem in details :p
We can only estimate how it works, and build up the login from that.

Totally true