6-socket Orb of Fusing Community Log (done)

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ScrotieMcB wrote:

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linkstatic wrote:

About 300 fusings.

6 : 4L
0 : 5L
1 : 6L

Everyone, please stop making posts with "about" or "approximately" X number of Fusings. I can't use them.

coldmu's data is being added now.

Just counted my stock, updated my post :) 320 fusings.

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linkstatic wrote:

Just counted my stock, updated my post :) 320 fusings.

Added, along with OP edit to include quote from Chris that heavily implies 6L chance is 1/1500 or greater.

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arcueid715 wrote:

1400 fusings

5L:7 times
6:0 times

Another 250

5L:1 time
6L:0

This is a great project.

Can't wait to see where those probabilities land when the margins get small.

Probably need about 5x more data.

I'll chip in


6 Fusings was used to make it a 6L


6L: 180 fusions were used to make this a 6L

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True_Apathy wrote:

I'll chip in

6 Fusings was used to make it a 6L

6L: 180 fusions were used to make this a 6L

Are we playing the same game? cus the luck seem to be 1 sided here

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Xmu wrote:

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True_Apathy wrote:

I'll chip in

6 Fusings was used to make it a 6L

6L: 180 fusions were used to make this a 6L

Are we playing the same game? cus the luck seem to be 1 sided here

I'd like 5L data before adding these to the OP.

We actually should expect about 2/3 of the 6L reports to take less attempts than the mean. If this seems impossible to you, read up on geometric distributions like this one, specifically regarding medians; keep in mind that only reports that eventually succeed in becoming 6L are part of this geometric distrubution. Additionally, the following formulas, which are all equivalent but solved for different variables, might be helpful to you; n represents a concept I'd describe as "unluckiness on a linear scale" (.99 is 1-in-100 unlucky, .5 is average luck and yields the median), x is the number of Orbs used, p is the chance of success, and e is Euler's number.

n=1-(1-p)^x
x=ln(1-n)/ln(1-p)
p=1-e^(ln(1-n)/x)

Regarding this specific point, if you replace p with its expression in terms of the mean, 1/x, you get n=1-(1-1/x)^x, which graphs like this... pretty much no matter what the chance of success is on your geometric distribution, 63.2% of the results will be less than the mean. (edit: as x approaches infinity, that number approaches 1-1/e, or 63.212%)

1050 Fusing used.

5L = 9 times
6L = 0

my QQ thread can be found here http://www.pathofexile.com/forum/view-thread/180908

Yeah I got the PM I'll update if I remember correctly here ~ Once again


6 fusings were used on this

They were no 5L's the 6th was the 6L obviously


This was 180 fusions used with a total of 2 5Links with the 180th fusion being the 6L

Thanks, True_Apathy. Yours and xmu's have been added.