**First of all - this post is huge. If you want the tl;dr version - check out the coloured parts.**You find an item you think it is worth using some fusings / chromatics on to get your specific mix of colours for your skills.

You try and try and after several orbs you still don't have your wanted result.

Is the chance for special combinations that low?

Is it worth using some chromatics/fusings on a special item?

I am going to calculate this through with you for 3 different cases and hope you can benefit in the future.

**First of all some mathematic basics you will need:**

0! = 1

1! = 1

2! = 2*1 = 2

3! = 3*2*1 = 6

4! = 4*3*2*1 = 24 etc

3^0 = 1

3^1 = 3

3² = 3*3 = 9

3³ = 3*3*3 = 27 etc

n% = n/100 (e.g.: 10% = 10/100 = 1/10 ; 50% = 50/100 = 1/2)

P = A/Ω - The Probability (or chance) on a certain event is are all the combinations you need divided by all the possible combinations

Let's start with the easiest case:

__1) Jeweller's Orbs__

If an item can get up to n sockets, the chance to make it 1-2-...-n sockets is 1/n

Example:

An item can have up to 2 sockets => the chance to make it 1 sox = 1/2, to make it 2 sox = 1/2

An item can have up to 3 sockets => the chance to make it 1 sox = 1/3, to make it 2 sox = 1/3, to make it 3 sox = 1/3I am not going to waste more time on this since it should be clear.

__2) Chromatic Orbs__Okay, this is really going to become a bit more complicated, but nevertheless, let me explain it, and you should understand it.

Let's start:

Every item we are interested in got a certain amount of sockets (in the future I will call these "slots"). Every slot can have 3 different colours.

Slot 1 can have 3 different colours

Slot 2 can have 3 different colours

Slot 3 can have 3 different colours etc.

So when I have n slots, the possible colour combinations are 3^n

Example:

I have 1 slot => 3^1 = 3 different combinations are possible (green, blue, red)

I have 2 slots => 3² = 3*3 = 9 different combinations are possible

I have 3 slots => 3³ = 3*3*3 = 27 different combinations are possible etcetc

Let us say we have an item with 3 different slots and we want to have the colours red blue green.

Now the question is - how many times can I expect those colours to appear? Therefore, let us see in how many ways we can allocate these colours on the 3 slots.

I can put colour 1 in any of the 3 slots => 3 possibilities (since you don't care if the green slot is the first, the second or the third, e.g. when slots are linked!)

I can put colour 2 in one of the other 2 slots (because I already filled one slot with the first colour!) => 2 possibilities

For the last colour, there is only 1 slot left => 1 possibility

So I got 3*2*1 = 6 posiibilities. 3*2*1 is called 3 factorial and is written 3!

So my chance to get this combination is 6/3³ = 6/27 ~ 22,3%

So far, so good. But what happenes when I want to have same colours?

This is going to be very important, since you always get atleast 2x the same colour when you have 4+ slots. Let's take a look at another example:

I got 4 slots and I want the colours red blue blue green.

The possible combinations are 3^4 = 81

For the wanted results: In the first step, let us treat every colour as if they were different.

Like explained above, the possible combinations for allocating 4 colours in 4 slots are 4! = 4*3*2*1 = 24

Now it is getting important - we got the same colour, 2 times blue. We cannot differ these 2 colours. There is no "blue1" and "blue2". It doesn't matter which which is.

So there are indeed not 24 different combinations. You got to look at the same colours now.

Since they are totally equal, we can exchange them with each other as we like. And how many different possibilities exist to switch these 2 colours - or better said:

How many possibilities exist to allocate those 2 colours on 2 slots?

First got 2 slots

Second got 1 slot

So I got 2! = 2 different ways to allocate them on their 2 slots.

These combinations must be excluded, so I got to divide my 24 by them.

So our wanted combinations are 4!/2! = 24/2 = 12, and our chance to get the combination red blue blue green in 4 slots is 12/81 ~ 14,8%

If you got n slots you are looking at, and you need a certain colour combination with a*red, b*blue and c*green, the chance to get it by using Chromatic Orbs isAnother example:

I got 6 slots and I need 3 red, 2 blue and 1 green.

n=6; a=3; b=2; c=1

Ω = 3^6 = 729

A = 6! / (3!*2!*1!) = 720/12 = 60

The chance is 60/729 ~ 8,2%

You can see from this one that it can take very long to get for example this combination.

Furthermore, the more often you want to have the same colours, the lower is the chance to get them!

For example you want all 6 slots to be blue - there is only 1 possible combination, and the chance is 1/729 ~ 0,14%

__3) Fusing orbs__Fusings - or links - are connections between sockets and follow a certain path:

If you got n slots, you can have max n-1 links (this should be clear, 4 slots can have max 3 links)

Once again let's get back to our formula P = A/Ω and calculate:

**Ω: When using an orb of fusing, a link can appear or dissapear. So we got 2 different possibilities what can happen with a link - it can be there or not.**

When I got 1 link, it can be there or not = 2 combinations

When I got 2 links, both can be there, both can not be there, the first can be there or the second = 2² = 2*2 = 4 combinations

When I got 3 links, I got 2³ = 2*2*2 = 8 combinations

When I got n links, I got 2^n combinations

A: I advise you to not calculate this with a formula, but to count.

When I got 2 slots and I want them to be connected, I got 1 wanted combination and 2 possible combinations, so my chance is 1/2 = 50% (you probably knew that before).

When I got 3 slots and I want them to be connected, 1 got 1 wanted combination and 4 possible combinations, so my chance is 1/4 = 25%

When I got 4 slots and I do NOT care about he colours yet, and I want 2 of them to be linked, you got to count a bit more:

My wanted combinations are:

Slot 1,2,3 are connected

Slot 2,3,4 are connected

Slot 1,2,3,4 are connectedSo you got 3 wanted combinations and 2³ possible, and your chance to get it is 3/8 = 37,5%

4) Eveything else like "how high is the chance to get 5 wanted mods out of the pool by using Chaos Orbs when i got 6 mods" would be way too complicated. It is luck. End of the story.

__5) Expectancy value__Well, the most important question you might be interested in is: after how many orbs can I expect my wanted combination?

Therefore we got the Expectancy value. It says more or less "after xx tries I expect my wanted combination".

More or less, your Expectancy value = 1/P (one divided by the chance).

If you got a chance of 2% for a certain combination (Remember: 2% = 2/100), your expectation value = 1/2/100 = 100/2 = 50.

You expect to get your combination after 50 orbs.There is one thing you cannot express mathematically - the luck/bad luck.

Of course you can get a hard combination by using 1 Orb. Of course you can miss a simple combination after using 50 orbs.

You can express that with deviation, but it is not possible to say you WILL need xx orbs and will for sure get the wanted result.

(Well okay, you can say after infinite tries it is 100% sure to get atleast your wanted combination once, but well, who got infinite orbs).

I am very happy that luck exists in this game, this makes it more unexpected and way more fun to play it.

But you can say with the so called Gaussian distribution:

In most cases, I will get my wanted result with using more or less the number of orbs the Expectancy value tells me. Using way less is luck, using way more is bad luck.

__Final note:__I want you all to know that there may be mistakes in. I an pretty familiar with maths, but this all only works if every event got the same chance.

For example, the chance to get a blue colour is the same as getting a red colour.

The chance to get a link between slot 1 and 2 is the same as getting a link between slot 2 and 3.

If this should not be the case, you can forget about all what I wrote!

But I think it should be fine.

I hope you could follow me and benefit in the future for your wanted combinations.

Please forgive me some typos/bad english, but this is not what I was focusing on.

If you got any questions, improvements or calculations you don't understand, feel free to PM me

Regards, Redentor