A drop rate is a probability, not a guarantee
WoW Insider has a short but great post on the math behind drop rate percentages. This may not be obvious to some people and the math behind the percentages will make it much clearer.
While Wowhead may tell you the drop rate of a Badge of Justice from Shattered Sun Supplies is 10%, you are not guaranteed a badge in 10 supplies. You actually have a 34.87% chance of not finding a single badge from opening 10 supplies.
Here’s the math:
- Take the percentage chance of not receiving the item you are after (in this case, 100% - 10% = 90% or 0.9)
- Raise that to the nth power, where n is the amount of times looted (in this case, 0.9^10 = 0.3487 or 34.87%)
- The result is your chance to not get the item in the number of times you’re looting (or 100% - 34.87% to get the percent chance you will get the item, 65.13%)
This math applies to items opened, corpses looted, herbs picked, ores mined, gems prospected, and everything else with a drop percentage attached to it.
And speaking of prospecting, we have a great article coming up that will break it down and give you an idea of how profitable it might be to buy up your server’s Adamantite Ore and spend the day prospecting. Stay tuned!
April 29th, 2008 at 11:48 pm
On one aspect this is a good post with the math and all, but it seems like this was pretty much just ripped from WoW insider. The one plus to this article though is it is written in a much more organized way! All-in-all good post. Cant wait for the prospecting!
April 30th, 2008 at 5:00 am
What you have calculated is that the chance of not getting the item exactly 10 times in a row, 34.87%.
This is a very specific statistic, afterall the the chance of not getting the item 100 times in a row is 0.00002656% and the chance of not getting the item 2 times in a row is 81%.
So what? I’m not opening the bag exactly 10 times (or 100 times or 2 times) and then quitting wow, I open them all the time, it could be any random number.
Ultimately the only figure that really matters to me is that my chance of getting the item each individual time is still 1 in 10.
Your maths is correct, but I don’t see the point.
April 30th, 2008 at 6:31 am
Note that this percentage is pretty much the same for any item with a 1/n drop chance - you’ll always have approximately a 34% chance of not getting one in n attempts (for large n, anyway). So, as an example, the chance of not getting one of the drops with 0.1% drop chance (the whelplings, some of the rare mounts) in 1000 attempts is also ~35%.
The reason for this is simply that the series (1-1/n)^n converges against 1/e (with e ~= 2.71828183).
As a consequence, you can calculate the approximate chances for not getting one in a fraction of n attempts quite simple as well - for example, not getting one of the 0.1% drop chance items in 500 attempts (or generally, 1/n in n/2 attempts) is approximately Sqrt(1/e) = 0.6.
April 30th, 2008 at 8:08 am
Garf, I don’t see how the percentage is pretty much the same for any item with 1/n? I’ll admit that I don’t really understand your point about convergence but surely,
1 in 10 drop rate, 10 tries = 0.9^10 = 0.3486 = 34.86% chance to miss
1 in 1000 drop rate, 10 tries = 0.999^10 = 0.9900 = 99.00% chance to miss
That seems to make sense to me, what am I missing?
April 30th, 2008 at 4:16 pm
You missed the “n attempts” vs “10 attempts”. So you’ll have to compare 0.9^10 to 0.999^1000 (which happens to be ~36.7%).
May 1st, 2008 at 1:07 am
Whilst interesting from a math point of view, i was waiting for some insight to apply this to lining my coffers with gold.
I suppose its an education to people farming rare pets BoE items on what they should expect as a likely return.
Still nice to read something on the train to work that makes you think. Makes me wonder if there is room for an add-on that tracks published drop rates versus player loot and present you with a probability calc ingame. Hmmmm
May 1st, 2008 at 10:00 am
Traex, as the author of the WoW Insider post, I appreciate the link; I don’t feel like it was ripped off at all
This is one of my favorite up-and-coming WoW blogs anyway.
Garf, thanks! I knew that 35% figure looked familiar; I must have seen it in other calculations before. 1/e, huh? Who would’ve guessed…
May 1st, 2008 at 1:55 pm
WoW Economist recommends some sketchy gold-making guides as advertisers -.-
One of the one ones on the side even had some BS popup ‘live agent’ when I tried to leave the page. Sketch & rude.
May 1st, 2008 at 2:02 pm
We’ve gotta pay the server bill somehow, Andrew. I don’t think they are sketchy at all. I own both of the guides and they are great for new WoW economists. They go over many different gold making strategies, most of which are outside of playing the AH.
Which one had the pop-up? Neither of them had one when I left the page.
Our other option would be to add Google ads, which are usually cluttered with gold selling websites. I figured you guys would rather see us promoting legitimate gold making guides than illegal gold selling sites.
May 2nd, 2008 at 10:35 am
Yeah it doesn’t seem to be doing the irritating pop-up thing anymore, but it was this one;
http://woweconomist.com/recommends/WarcraftRiches
I was using Firefox 2.0 on a Mac, Adblock enabled.
I agree, I’d rather see the gold making guides than gold-selling sites, I just have a reaction to the way their sites are presented. They’re structured identically to the bordering-on-illegal get-rich-quick scheme sites for real moneys.
May 22nd, 2008 at 11:27 am
Just as an aside, I think its generally easier to work with expectations. IMHO, the question you are trying to answer here is “OH GOD, how many times do I have to do this?”. Unless you’ve got some crazy “a beautiful mind shit” going on, you probably can’t do the logs in your head. But, since the trials are iid, and since the expectation of a sum is the sum of the expectations:
E [sum_{i=1}^n x_i] = sum_{i=1}^n E [x_i] = n * E [x]
So, if I want to know how many times I should have to do something (n) on expectation to get some number of an item (k) with drop rate r, its just:
n = k / r.
September 12th, 2008 at 3:36 am
I know, I’m really late to the party, still thought I’d share my 2c.
The math is mostly correct, except for the 65.13% you’ll get the item - that’s actually the chance you’ll get _at least_ 1 item. You could just as well get 20, if the RNG gods really like you.
Anyway, great blog. How come I never heard of it till now?