Home > Age of Conan, Statistics, Vanity Gear > Level 40-69 world-drop sets

Level 40-69 world-drop sets

You might remember my post from some time ago about the old world-drop armor sets. Well, I got the idea that it would be nice to have screenshots of all 12 level 40-69 sets, so I’ve lately been doing some farming in epic Fields of the Dead and eventually managed to get all items from all twelve sets. Unfortunately they take up too much inventory space, so I’ll probably delete them soonish.

Anyway, without further ado, here are the screenshots; above each screenshot, there’s the name of the set and of the class with which it has traditionally been associated.

Beatific (Priest of Mitra)

Brimstoned (Demonologist)

Corybantic (Barbarian)

Dark Ember (Herald of Xotli)

(Note: for the back picture, I had to remove the legs and shoulders as they were clipping very badly with the robe. As you can see, some clipping problems with the character’s chest remain.)

Eidolon’s (Necromancer)

Heretic’s (Dark Templar)

Pathfinder (Ranger)

Twilight (Assassin)

Vindicator (Conqueror)

Watchman’s (Guardian)

Wildsoul (Bear Shaman)

Zephyrous (Tempest of Set)

How much farming is needed?

Each of the 12 sets discussed in this post consists of 8 items, so that’s a total of 96 different items. So you’ll need to kill at least 96 bosses to get them all; but it’s likely that it will take more than that, because sometimes you’ll get the same item several times. So what can we say about the number of bosses you’ll have to kill to collect all 96 items?

This is a nice little exercise in combinatorics. Let’s say we have a loot table of n = 96 items that we’re interested in, and that every time we kill a boss, one of these items drops, and they are all equally likely. What’s the probability that we have to kill exactly k bosses to get all n items?

Killing k bosses results in a sequence of k items looted from them; since there are n possiblities for each of these items, there are a total of nk loot sequences from these k kills. They are all equally likely. Now we want to count those that contain each of the n items at least once, and where no shorter prefix of this sequence contains all n items (otherwise we could have stopped sooner than at k bosses).

So there must be some item, let’s call it x, that dropped for the first time only at the very end of the sequence; before that we have a slightly shorter sequence of k − 1 drops from among the remaining n − 1 items (i.e. all items except x; let’s call these items y1, …, yn − 1), with the additional constraint that each of these n − 1 items must occur at least once in that shorter sequence.

How many such shorter sequences are there? There are a total of (n − 1)k − 1 sequences of length k − 1 over a set of n − 1 items; from these we have to subtract those which lack y1, and also those which lack y2, and so on; and now we find that we subtracted those which lacked *both* y1 and y2 twice, instead of just once, so we have to add them back again; and so on. This is what is known as the inclusion-exclusion principle, and it leads us to the following formula:

Σ0 ≤ i < n (−1)i binom(n − 1, i) (n − 1 − i)k − 1.

This is the number of sequences of length k − 1 from a set of n − 1 items that contain each of these items at least once. Now, at the beginning we started with item x, which we could have chosen in any of n ways, so we have to multiply our sum by n; and then, to get a probability, we have to divide it by nk, which is the total number of sequences of length k from our set of n items.

Now, you might remember that half of the items we’re talking about here are actually bind-on-equip instead of bind-on-pickup, so strictly speaking you don’t have to farm them by yourself. You could just focus on farming the bind-on-pickup parts and once you have all those, you can then buy the missing bind-on-equip parts on the tradepost (assuming that some other player is selling them). How does that affect our calculations regarding the number of boss kills we need?

Let’s say that the loot table still has n items, like before, but we’re only interested in r specific items among those n items. A very similar line of reasoning like before now leads to the following formula:

Σ0 ≤ i < r (−1)i binom(r − 1, i) (n − 1 − i)k − 1.

And similarly as above, we have to multiply this sum by r and then divide it by nk to get the probability we’re looking for.

Note that our initial formula is just a special case of this second one, and you can get back to the initial formula by setting r = n.

I’m not sure if our formulas can be simplified still further (by replacing the sums with something more elegant), but in any case for the purposes of calculating these probabilities our current formula is enough. The following chart shows the resulting probability distribution at n = 96 and for r = 96 (if we want to farm all items) and r = 48 (if we want to farm just the bind-on-pickup items and will then buy whatever bind-on-equip items we’ll be missing by then):

Now that we can calculate the probability of each k, we can also calculate the expected value of k (i.e. the weighted average of all possible values of k, in which each k is weighted by its probability). In our case this turns out to be approx. 494 for r = 96, and 428 for r = 48. In other words, on average you’ll have to kill 494 bosses to get all 96 items, or 428 bosses to get just the 48 bind-on-pickup items; but you can of course be lucky and get them sooner, or you can be unlucky and require a lot more than 494 boss kills. (It turns out that if you kill 876 bosses, the probability that you’ll have obtained all n = 96 items by then exceeds 99%.) So from this point of view, the idea of focusing on farming just the bind-on-pickup parts isn’t actually a huge improvement.

[If you look at the above-mentioned formulas for P(k) more carefully, you can see that they are really a combination of r different geometric distributions. This also allows us to derive a nice direct formula for the expected value of k, which turns out to be:

E[k] = r n Σ0 ≤ i < r (−1)i binom(r − 1, i) / (i + 1)2.

Thus at r = 1, where we’re interested in just one item among n possible drops, the expected value of k turns out to be, unsurprisingly, simply n. At r = 2, it turns out that the expected value of k is 3n/2.]

In practice, of course, we’ll have to kill a bit more bosses than our calculations so far have shown, because their loot tables also contain some items that we aren’t interested in (e.g. weapons, or armor pieces that aren’t part of sets). We could in fact take this into account with our second formula by setting n to some value greater than 96 (to include the “uninteresting” items from the loot tables) and then keeping r at either 96 or 48, as before; but I haven’t tried to do so as I don’t have any clear idea of how many such uninteresting items there are.

I haven’t tried to keep track of how many bosses I actually had to kill to get my 96 items, but my impression is that it was probably less than 494, so I think I was relatively lucky there. And in fact the last item I got was a chestpiece, so the plan to focus on farming the bind-on-pickup parts (and buying any missing bind-on-equip parts at the end) wouldn’t have done me any good in this concrete case, because by the time I got my last bind-on-pickup part, I already had all the bind-on-equip parts as well.

[P.S. Another reason why I found this calculation interesting is because I was wondering if I should try collecting the level 70-80 sets as well. There, killing each boss takes quite a bit more time (unless you’re doing it with a group), so I wanted to get a sense of how many kills would be required to complete all the sets. I don’t think I want to try solo killing level 70-80 group bosses 494 times. :P]

So far all our calculations have been done for n = 96; but it is also interesting to ask what happens if you want to collect all items from a loot table of some other size. For example, here are distributions for n = 5 and 10. At n = 5, we’re dealing with sufficiently small values of k that it’s still easy to see the discrete nature of the distribution, so we’ve shown it as a bar chart.

Another very interesting question is: how does k (the required number of boss kills) depend on n (the size of the loot table)? The following chart shows some answers to this question. It shows the expected value of k (as a function of n), as well as the number of kills required if you want to have a certain chance (25%, 50%, 75%, 90%, 99%) of discovering all n items.

Or, if we zoom in a little on the smaller n‘s:

  1. Shaidar
    July 6, 2012 at 22:10

    That is one impressive post 🙂
    Looking forward to the next email post!

  2. AoC Nerd
    July 4, 2013 at 23:11

    Hey, awesome blog you have there! wondering if you might know something about this item:

    you got any idea where it drops and if it is a single piece or theres whole set for it?

    • July 5, 2013 at 09:45

      It drops from Toirdealbach in Toirdealbach’s Tomb. AFAIK it isn’t part of any set – certainly the other drops in Toirdealbach’s Tomb seem to be a random assemblage of odd bits and pieces of armor and accessories for various classes.

  1. September 12, 2012 at 20:35
  2. September 29, 2012 at 07:58
  3. September 29, 2012 at 08:06
  4. October 10, 2012 at 18:21
  5. June 19, 2013 at 22:03
  6. January 25, 2014 at 16:43
  7. September 18, 2014 at 21:09

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