Lua Function percent

[maikons]

Someone could help me.. I need a percent function to make this

function onUse(cid, item, fromPosition, itemEx, toPosition)
  doRemoveItem(item.uid)
   
  light magic missile runes,2287,50%
  intense healing rune,2265,30%
  heavy magic missile runes,2311,10%
  great fireball runes,2304,5%
  explosion rune,2313,3%
  ultimate healing rune,2273,2%
  sudden death rune,2268,1%
  return true
end

On use this item check percent and add item rare or normal...

 

[Omni Cloud]

If 100.0 is 100% guess what 50, 30, 10, 5, 3, 2 or 1 is?

 

[Elime]

If I understand you correctly you want something like this:

function chanse(percent)
    return percent >= math.random(1, 100)
end

if chanse(50) then player:addItem(2287, 1) end
if chanse(30) then player:addItem(2265, 1) end
if chanse(10) then player:addItem(2311, 1) end
if chanse(5) then player:addItem(2304, 1) end
if chanse(3) then player:addItem(2313, 1) end
if chanse(2) then player:addItem(2273, 1) end
if chanse(1) then player:addItem(2268, 1) end

 

[maikons]

If I understand you correctly you want something like this:

function chanse(percent)
    return percent >= math.random(1, 100)
end

if chanse(50) then player:addItem(2287, 1) end
if chanse(30) then player:addItem(2265, 1) end
if chanse(10) then player:addItem(2311, 1) end
if chanse(5) then player:addItem(2304, 1) end
if chanse(3) then player:addItem(2313, 1) end
if chanse(2) then player:addItem(2273, 1) end
if chanse(1) then player:addItem(2268, 1) end

Was that, i needed a random function!
TY!