TL;DR

http://i1306.photobucket.com/albums/s565/drag0re/TRIAR_zps09a5180e.jpg

http://i1306.photobucket.com/albums/s565/drag0re/trifectaalone_zps8ea94311.jpg

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After reading some of the archon crafting show off threads I have come to wonder how many trials it would take to eventually craft a pair of insanely godly gloves. I found similar attempts on a number of different threads, but there the models were oversimplified and rather dubious (e.g. some dude was trying to adapt the statistics of poker directly to stats roll, but left out many key assumptions that needed to be taken into account).

So I'm going to work out the math for the

Let's first look at the most ideal (and easiest to compute) combination of stats, regardless of their values:

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THE ASSUMPTIONS

There are affixes that add to a single main stat, and affixes that add to two main stats at once (double stat roll, max +100 for each). It’s impossible to roll a +mainstat that’s the same as a single inherent mainstat bonus EXCEPT if this +mainstat comes from a double stat roll. Mainstats (other than inherent) can't roll more than once EXCEPT if this +mainstat comes in part from a double stat roll and a single stat roll. This does not preclude double All properties have an equal probability to roll, among a pool of 25 possible stats for rare gloves; in other words, the probability that your first stat rolls +strength is the same as the probability that it rolls +pickup radius.

In addition, these events are INDEPENDENT (again getting +pickup radius on the first modifier will not affect the probability of getting +dexterity on the second roll). Of course this is not true if say +Int has already rolled twice total. The probability of getting another +Int roll would then be 0. source:

http://diablo.incgamers.com/blog/comments/diablo-3-crafting-strategy-the-new-archon-items

http://diablonut.incgamers.com/affixes/?c=adventure-and-stats&l=1&u=70&h=119253633

http://www.diablofans.com/topic/41045-spoiler-diablo-iii-item-affixes/#entry863934

http://us.battle.net/d3/en/forum/topic/7923992596

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THE MATH

So under the assumption of independence I can proceed as follows: I consider the rolls one by one, having no influence on each other except when: 1) stat already rolled in one or two of the previous rolls, in which case it cannot roll a third time.

MAIN ROLL

This one is guaranteed.

FISRT ROLL

Let's calculate the number of combinations we can get. We technically have 24 stats to choose from (25 minus the inherent stat, here Intelligence). But we also have to take into account all the possible double stat rolls: Int/Vit; Int/Str; Int/Dex; Str/Vit; Str/Dex; Dex/Vit. That gives us 6 pairs of stats. So we have a total of 24 + 6 = 30 events, each with a probability of 1/30 to occur. So let's say that

SECOND ROLL

So we're left with 23 stats to choose from (24 minus AS), in addition to the remaining 6 pairs of double rolling stats. We thus have a total of 23 + 6 = 29 events, each with a probability of 1/29 to occur. So suppose

THIRD ROLL

Now we're left with 22 stats to choose from (23 minus CC), in addition to the 6 double stat rolls.

That gives us a total of 28 events each with probability 1/28. Let's say

FOURTH ROLL

21 stats left, plus the 6 pairs. Total of 27 events each with prob 1/27. Now let's say the

FIFTH ROLL

20 stats left, plus the 6 pairs. Total of 26 events each with prob 1/26. That's when the double stat

However, AS could have rolled last, CC first and AR third, etc. There are actually X = 5! = 5*4*3*2*1 = 120 possible ways to get the perfect trifecta through these 5 rolls. Since each has a probability of Y to occur, we end up with a probability of XY = 1/142506 to get the perfect gloves.

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Fair enough. Now let's pretend that you don't care about Vitality but still want AR to maximize your EHP. Then all you want is a roll of the following type:

The calculations are similar: Y = (1/30)*(1/29)*(1/28)*(1/27)

However, instead of multiplying by 5!, here we must take into acount the number of choices for random property 1:

For example, before anything has rolled, we have 5 ways to get say AS: either on the first roll, or on the second, or the third etc.

Then 4 ways left to get CC. Then 3 ways for CD and finally 2 ways left to get AR. But for the last roll you can have anything besides AS, CC, CD, AR and +Int. That gives you 26 possibilities. So X = 5*4*3*2*26. In the end that gives you XY = 26/5481.

What does that really mean? -->

http://i1306.photobucket.com/albums/s565/drag0re/TRIAR_zps09a5180e.jpg Horizontal axis displays the number of trials. Vertical axis shows the probability that you will get at least one success after that many trials. So after about 210 trials, you have a 60% chance of success. From the graph we see that after 800 trials you're pretty much guaranteed to get your trifecta+AR.

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Now what about the probability to get just the trifecta, regardless of the last two rolls? #softcoreproblems

The math actually gets a little more complicated. A more refined calculation would actually take into account the fact that if say a double stat roll of the form +Int & +<other mainstat> occurs, then the second random property cannot be any of the remaing pairs of stat with Int, which cuts two branches from our probability tree. Now whether AS rolls before or after CD, and likewise for CC does not matter. So we really have to consider only (5 Choose 2) = 10 combinations. In other words, it's like having to pick five pebbles among a pool of 30 pebbles, three of which are black and the remaining ones are white, and asking in how many ways you can arrange them. Also note that we may now have double rolls occuring twice.

So, let's look at all the cases separately and add their respective probabilities.

EDIT: Turns out it takes an entire page of math, so HERE IS THE RESULT.

Here's the graph:

http://s1306.photobucket.com/user/drag0re/media/trifectaalone_zps8ea94311.jpg.html

Enjoy.

**Here is the graph for the TRIFECTA + AR:**http://i1306.photobucket.com/albums/s565/drag0re/TRIAR_zps09a5180e.jpg

**Here is the graph for the TRIFECTA ALONE:**http://i1306.photobucket.com/albums/s565/drag0re/trifectaalone_zps8ea94311.jpg

**Y axis: probability to craft a trifecta...**

X axis: ... after that many attempts.X axis: ... after that many attempts.

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After reading some of the archon crafting show off threads I have come to wonder how many trials it would take to eventually craft a pair of insanely godly gloves. I found similar attempts on a number of different threads, but there the models were oversimplified and rather dubious (e.g. some dude was trying to adapt the statistics of poker directly to stats roll, but left out many key assumptions that needed to be taken into account).

So I'm going to work out the math for the

**Archon Gauntlets of Intelligence**as an illustration. I might do the same thing for the amulet / shoulder at some point.Let's first look at the most ideal (and easiest to compute) combination of stats, regardless of their values:

**+Inherent Main Stat (Int)**

+Int & Vit

+AR

+AS

+CC

+CD

+Int & Vit

+AR

+AS

+CC

+CD

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THE ASSUMPTIONS

*and*single stat rolls from occuring twice, so long as the above condition is still satisfied.

- THIS ASSUMPTION MIGHT BE FLAWED, as I've read several threads were the OP reported the occurences of main stats vs. other type of modifiers on a sample of 100 crafted items, and some mainstats seemed to have a slightly better chance to roll than others, in addition to the fact that mainstats also seemed to have a higher chance to roll than other modifiers. So for now the following resuls are be taken as an upper bound for the number of trials it would take on average until you can get the desired roll.

http://diablo.incgamers.com/blog/comments/diablo-3-crafting-strategy-the-new-archon-items

http://diablonut.incgamers.com/affixes/?c=adventure-and-stats&l=1&u=70&h=119253633

http://www.diablofans.com/topic/41045-spoiler-diablo-iii-item-affixes/#entry863934

http://us.battle.net/d3/en/forum/topic/7923992596

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THE MATH

So under the assumption of independence I can proceed as follows: I consider the rolls one by one, having no influence on each other except when: 1) stat already rolled in one or two of the previous rolls, in which case it cannot roll a third time.

MAIN ROLL

**+[201-230] Intelligence.**This one is guaranteed.

FISRT ROLL

Let's calculate the number of combinations we can get. We technically have 24 stats to choose from (25 minus the inherent stat, here Intelligence). But we also have to take into account all the possible double stat rolls: Int/Vit; Int/Str; Int/Dex; Str/Vit; Str/Dex; Dex/Vit. That gives us 6 pairs of stats. So we have a total of 24 + 6 = 30 events, each with a probability of 1/30 to occur. So let's say that

**+Attack Speed**rolls first.SECOND ROLL

So we're left with 23 stats to choose from (24 minus AS), in addition to the remaining 6 pairs of double rolling stats. We thus have a total of 23 + 6 = 29 events, each with a probability of 1/29 to occur. So suppose

**+Critical Chance**rolls second.THIRD ROLL

Now we're left with 22 stats to choose from (23 minus CC), in addition to the 6 double stat rolls.

That gives us a total of 28 events each with probability 1/28. Let's say

**+Critical Hit Damage**rolls as the third property.FOURTH ROLL

21 stats left, plus the 6 pairs. Total of 27 events each with prob 1/27. Now let's say the

**+AR**roll occurs.FIFTH ROLL

20 stats left, plus the 6 pairs. Total of 26 events each with prob 1/26. That's when the double stat

**Int & Vit**roll happens. Now this has a total probability of Y = (1/30)(1/29)(1/28)(1/27)(1/26) = 1/570,024 of occuring.However, AS could have rolled last, CC first and AR third, etc. There are actually X = 5! = 5*4*3*2*1 = 120 possible ways to get the perfect trifecta through these 5 rolls. Since each has a probability of Y to occur, we end up with a probability of XY = 1/142506 to get the perfect gloves.

**That's 1 in every 142,506 trials.**●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

Fair enough. Now let's pretend that you don't care about Vitality but still want AR to maximize your EHP. Then all you want is a roll of the following type:

**+inherent main stat (Int)**

+random property 1

+AR

+Trifecta

+random property 1

+AR

+Trifecta

The calculations are similar: Y = (1/30)*(1/29)*(1/28)*(1/27)

However, instead of multiplying by 5!, here we must take into acount the number of choices for random property 1:

For example, before anything has rolled, we have 5 ways to get say AS: either on the first roll, or on the second, or the third etc.

Then 4 ways left to get CC. Then 3 ways for CD and finally 2 ways left to get AR. But for the last roll you can have anything besides AS, CC, CD, AR and +Int. That gives you 26 possibilities. So X = 5*4*3*2*26. In the end that gives you XY = 26/5481.

**That's about 1 in every 210 trials. Not too bad.**What does that really mean? -->

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Now what about the probability to get just the trifecta, regardless of the last two rolls? #softcoreproblems

The math actually gets a little more complicated. A more refined calculation would actually take into account the fact that if say a double stat roll of the form +Int & +<other mainstat> occurs, then the second random property cannot be any of the remaing pairs of stat with Int, which cuts two branches from our probability tree. Now whether AS rolls before or after CD, and likewise for CC does not matter. So we really have to consider only (5 Choose 2) = 10 combinations. In other words, it's like having to pick five pebbles among a pool of 30 pebbles, three of which are black and the remaining ones are white, and asking in how many ways you can arrange them. Also note that we may now have double rolls occuring twice.

So, let's look at all the cases separately and add their respective probabilities.

EDIT: Turns out it takes an entire page of math, so HERE IS THE RESULT.

**On average you will get a trifecta once in every 110 trials.**Here's the graph:

http://s1306.photobucket.com/user/drag0re/media/trifectaalone_zps8ea94311.jpg.html

Enjoy.

Edited by dragore#1339 on 3/25/2013 10:19 PM PDT