How difficult is it to find/roll off-color sockets?

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ualac wrote:

yep they ought to. so you have a 10% chance for socket one (P1), and 10% chance for socket two (P2). since they dont effect one another the probabilities are multiplied.

in the four socket item case the result is different as many 4 socket colour rolls could satisfy the condition of 2 being green. this is all discrete probability, ie. a finite set of event outcomes each with varying chance, many of which satisfy the conditions. the total would be the sum of all these event chances.

(which probably ends up at 1/20 knowing my luck)

Ah, my bad - I read your first post wrong :)

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ualac wrote:

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Elynole wrote:

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ualac wrote:

but thats not correct.

the chance of rolling 2 green on a 2 socket STR/INT item would be 1/100 assuming theres a 10% chance of each. the events are unrelated, therefore the chances are multiplicative.

however on a 4 socket item there are many combinations that would suffice to be considered 2 green sockets. so the total chance is the summation of all events that generate 2 of the 4 sockets as green.

Hmm, I was always under the impression that each socket rolled separately?

yep they ought to. so you have a 10% chance for socket one (P1), and 10% chance for socket two (P2). since they dont effect one another the probabilities are multiplied.

in the four socket item case the result is different as many 4 socket colour rolls could satisfy the condition of 2 being green. this is all discrete probability, ie. a finite set of event outcomes each with varying chance, many of which satisfy the conditions. the total would be the sum of all these event chances.

(which probably ends up at 1/20 knowing my luck)

If my math is correct above, its 5.23%, which is basically 1/20 :)

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Mettle82 wrote:

{math} = 5.23% chance of rolling at least 2 green

Now, I need a statistics person to tell me where I'm wrong.

erm.. do you want to be wrong? :) to me it looks on target.
(oddly that result is close to 1/20th. pesky mathematics!)

contributing for science:

i got this copy from someone and it needed 357 chroms to get into these colours...


so if any one else needs RRGGGB on pure armor chest, be prepared ;)

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dolpiff wrote:

contributing for science:

i got this copy from someone and it needed 357 chroms to get into these colours...

so if any one else needs RRGGGB on pure armor chest, be prepared ;)

okey. so given this is a Strength chest piece, we take the off-colour chances to be 10% each and the primary colour (red) to be 80% likely for each socket.

so a combination that is 2*R 3*G 1*B = 0.8*0.8*0.1*0.1*0.1*0.1 = 0.000064

there are 15 possible positions where the two R sockets can be within the six sockets (5 where RR are not separated, 4 where RR are separated by one, and so on) and in the remaining 4 sockets there are 4 different locations where the B socket can be placed.

giving us a total of 60 (15*4) possible results that satisfy a 2R 3G 1B condition.

total chance therefore = 60 * 0.000064 = 0.00384 = 0.384% ~= 1/260

possibly unlucky that you needed 357 chromatics to get that result :( nice chest tho!

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ualac wrote:

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dolpiff wrote:

contributing for science:

i got this copy from someone and it needed 357 chroms to get into these colours...

so if any one else needs RRGGGB on pure armor chest, be prepared ;)

okey. so given this is a Strength chest piece, we take the off-colour chances to be 10% each and the primary colour (red) to be 80% likely for each socket.

so a combination that is 2*R 3*G 1*B = 0.8*0.8*0.1*0.1*0.1*0.1 = 0.000064

there are 15 possible positions where the two R sockets can be within the six sockets (5 where RR are not separated, 4 where RR are separated by one, and so on) and in the remaining 4 sockets there are 4 different locations where the B socket can be placed.

giving us a total of 60 (15*4) possible results that satisfy a 2R 3G 1B condition.

total chance therefore = 60 * 0.000064 = 0.00384 = 0.384% ~= 1/260

possibly unlucky that you needed 357 chromatics to get that result :( nice chest tho!

but not that uncommen .. you only have a 75% to got that combination at least once in those 357 tries.

I would do bad things for the chest piece.... perfect for my build

EDIT: My bad, misunderstood a posting.

exactly 2 greens out of 4 total sockets

.1*.1*.9*.9 = .0081

Now, that is just for green in slots 1 and 2. The total number of configurations is 6, so .0081*6 = .0486, or 4.86% chance.

Your math works for exactly 2 greens when there are only 2 sockets

I find it funny you have such a condescending attitude when you didn't read the initial post. Also, who added .9? its been multipled due to the chance of other sockets not being green is .9.

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Mettle82 wrote:

I find it funny you have such a condescending attitude when you didn't read the initial post. Also, who added .9? its been multipled due to the chance of other sockets not being green is .9.

Added in terms of "add it in the calculation".

But multiplying it just because the chance of other sockets not being green is .9 is COMPLETE nonsense.

The 0.1 is already the result of 1 (=100% chance) - the chance NOT to be green (0.9). This results in 0.1

Multiplying it again would only make sense if you wanted to know the chance to get a green socket and a socket in another colour (if, for instance, a socket could also become black with a 10% chance and it´s 80% chance to get red OR blue, THEN and only THEN it would make sense to multiply with 0.8)

It is 1/10 for each socket. Not more, not less. Except you want to know something else than "what´s the chance for 1 socket to turn green".
And it remains 1/100 for two because there is no chance for a socket NOT to be a coloured socket.