"Money is always evenly distributed, regardless of loot setting."
Unless you happen to be in a group, and loot 1c, then you will get that 1c, and you party members get nothing. OwlBoy 19:17, 22 Mar 2005 (EST)
two methods of rolling
"Some like to use a custom way of handling loot (in a group, not in a raid) :
you want to sell the item : type s on the group chat you want to use the item : type u on the group chat When everybody has typed s, everybody rolls. When some have typed u, they roll and the others skip.
An alternate method of handling loot (in a group or raid) to the above method:
Type 'g' in party/raid chat when you want the item (greed) but do not need it. Type 'n' in party/raid chat when you need the item. When everyone has typed 'g' in party/raid chat, everybody rolls. When anyone types 'n' in party/raid chat, only those who type 'n' roll, everyone else skips. "
Why does the first one work only for groups while the second works for both? they're the same except the letters used.
This is completly stupid: just use the "Group loot" function and you will have all this done automatically.
This method is monumentally stupid as it's exactly emulating the built in group loot function. If you're in a group or guild that actually does this, you should probably consider getting the hell out as they're obviously complete morons. -Luziphir
This is currently listed under the "Notes" section:
- 1st kill: There is a 10% (0.1) chance that your item will drop and a 90% (0.9) chance that it won't.
- 2nd kill: Assuming your first kill didn't drop the item the probability that the second kill does not drop is (0.9 x 0.9 = 0.81) 81%. From this we can calculate that the probability that your item will drop is (1 - 0.81 = 0.19) 19%.
Which is garbage and should be removed. The chance for the item to drop is 10% on the first kill, and the chance for it to drop is 10% on the second kill. It will be 10% on the 100th kill as well. You could say that the author means the chance of your "unlucky streak" to continue would decrease, but the chance for that unlucky streak to start was 90%, and it will be 90% that it continues on the 100th kill, too. You can calculate what the chance of an item dropping within X attempts is that way; but the drop rate does not change and will remain 10%, no matter what. Lyom (talk) 14:08, 4 August 2009 (UTC)
- The game may look at each kill as a separate probability of dropping, but the likelihood (or probability) you will get it (and the odds against it not dropping) rises as you keep getting drops from the mob. -- ( • ) 14:25, 4 August 2009 (UTC)
- Having just looked over my maths, I've realised that I have in fact posted incorrect numbers. The one's listed are the probability that the item will drop within X number of kills, not on X kill (it's been a long time since I did probability). I will update the article. Atraignis (talk) 14:29, 4 August 2009 (UTC)
- After a third look at my maths and reading up on probability in an effort to provide a citation, I'm not sure my maths is correct and my language is possibly misleading. Will do some more research and update (unless someone better at probability than me beats me too it :)). Atraignis (talk) 15:09, 4 August 2009 (UTC)
It is true that probability of a specific item dropping from a specific mob will always be a constant for every single kill. But that is considering that you look at the kills as separate individual events. It is not wrong but it is not meaningful either. You have to look at it as an experiment with many trials, with the aim of getting an event to occur in at least one of the trials; not as individual experiments with single trials. Using the heads and tails example, the probability of getting 10 tails in an experiment with 10 trials is different from the probability of getting 1 tail in an experiment with 1 trial, even though the probability for every single trial is the same. Using this idea, and given a fixed probability of the random event occurring, an expected number of trials to get the event to occur can be calculated. The probability of the event occurring given the number of trials will then be a normal probability density function centered on the expected value. Refer to http://en.wikipedia.org/wiki/Expected_value#Discrete_distribution_taking_only_non-negative_integer_values --Octanion (talk) 18:04, January 23, 2010 (UTC)