I'm doing it the way I thought you were trying to do it.
If you do it your proposed way..
Code:
local config = {
[1] = {chance = 20, item_id = 2160, item_amount = 5}, -- 20%
[2] = {chance = 20, item_id = 2160, item_amount = 3}, -- 20%
[3] = {chance = 20, item_id = 2160, item_amount = 1} -- 20%
}
function onKill(cid, target, damage, flags)
if not isPlayer(cid) or not isPlayer(target) then
return true
end
if getTileInfo(getThingPos(target)).pvp then
return true
end
if getPlayerLevel(target) >= 200 and getPlayerLevel(target) < 1000 then
for i = 1, #config do
if math.random(100) <= config[i].chance then
doPlayerAddItem(cid, config[i].item_id, config[i].item_amount)
break
end
end
end
return true
end
If you don't add in a 100% chance somewhere, then there is a possibility of the player getting nothing.
Also, for every 20% chance you add in, the next item in the list will actually only have a percentage of a percentage of a chance of actually being picked.
First roll, chance = 20 -- 20% chance to get
Second roll, chance = 20 -- IF the first item doesn't get chosen, this one have 20% chance.
Third roll, chance = 20 -- IF the first item AND IF the second item doesn't get chosen, this one have 20% chance.
The chance to get lower items on the list, with the same percent chance get's lower and lower to be chosen, hence the chance you've defined is inaccurate. (except for the first item in the list.)
-- Edit for more explanation.
Let's pretend we are rolling a 6 sided die.
What's the probability of rolling a six on roll one, two, three, four (etc.) rolls?
Let's examine this roll by roll.
In one roll, the probability of rolling a 6 is 1/6.
For two rolls, there is a 1/6 probability of rolling a six on the first roll.
If this occurs, we've satisfied our condition.
There is a 5/6 probability that the first roll is not a 6.
In that case, we need to see if the second roll is a 6.
The probability of the second roll being a 6 is 1/6, so our overall probability is 1/6 + (5/6)*(1/6) = 11/36.
Why did I multiply the second 1/6 by 5/6?
Because I only need to consider the 5/6 of the time that the first roll wasn't a 6.
As you can see the probability is slightly less than 2/6.
For three rolls, there is a 1/6 probability of rolling a six on the first roll.
There is a 5/6 probability that the first roll is not a 6.
In that case, we need to see if the second roll is a 6.
The probability of the second roll being a 6 is 1/6, giving us a probability of 11/36.
There is a 25/36 probability that neither of the first two rolls was a 6.
In that case, we need to see if the third roll is a 6.
The probability of the third roll being a 6 is 1/6, giving us a probability of 1/6 + (5/6)*(1/6) + (25/36)*(1/6) = 91/216.
Again, this is less than 3/6.
thanks
