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[Biochem] 6.0 Kyrprax Medpac and Med Units - Artifact Schematics via deconstruction?

STAR WARS: The Old Republic > English > Crew Skills
[Biochem] 6.0 Kyrprax Medpac and Med Units - Artifact Schematics via deconstruction?

DawnAskham's Avatar


DawnAskham
12.16.2019 , 06:53 AM | #11
Quote: Originally Posted by MacCleoud View Post
I knew i could only learn schematics for my skills. What I was meaning was, can I only learn schematics from the items I craft personally. I bought 2 Medpacs to make a stack of 60, and my entire stack ended up showing "No research available". Even after separating out the stack, it stayed that way. I put that stack in the vault, and the next stack I crafted I could try to research with again.
Craft a new one and add it to the stack and it will allow you to RE for MK parts, which should also allow you to RE for a chance at schematic.

SnakeSD's Avatar


SnakeSD
03.25.2020 , 06:06 PM | #12
Is the "gold" medipac usable by just the Biochem character or can it be crafted and distributed across your legacy?

Void_Singer's Avatar


Void_Singer
03.25.2020 , 06:26 PM | #13
Quote: Originally Posted by SnakeSD View Post
Is the "gold" medipac usable by just the Biochem character or can it be crafted and distributed across your legacy?
requires Biochem (at the proper level) to use.
If it moves, shoot it.
if it stops moving... shoot it again.
- Anonymous Mercenary

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SteveTheCynic's Avatar


SteveTheCynic
03.26.2020 , 05:52 AM | #14
(Sorry for the necroism.)
Quote: Originally Posted by PatT View Post
Gyromaniac, that's not how chance works. 5% chance is 5% every time. Not stacked by the number of tries. Statistics
Actually, Gyromaniac is correct. To fail to have one after 30 tries, at 5% success = 95% failure, you must fail all 30 times, and the probability of that is (0.95) multiplied together 30 times (that is, 30 lots of 0.95 multiplied together, so 29 multiplications), which is 0.2146 (etc.) which is 21.46% chance of the accumulated failure.

The 30th try, however, is still 5% *that*time*.

((Why is it like that?))
The individual attempts are what's called "independent trials with replacement", meaning that no one trial affects the outcome of the next one, and nor is it affected by the previous one ("independent"), while the "with replacement" is more applicable to e.g. drawing cards from a deck where you do or do not put the card you drew back for the next draw. To finish the sequence with no success, you must fail *all* of them. The analysis works like this:

* Fail once = 95% chance.
* Fail twice = 95% chance *if and only if* you already failed once = (95% of 95%) chance from the beginning.
* Fail thrice = 95% chance if and only if you already failed twice = (95% of 95% of 95%) from the beginning.
...
* Fail 30 times = 95% chance if and only if you already failed 29 times = (95% of ... of 95%) from the beginning.
http://www.swtor.com/r/Hg3sV2
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Void_Singer's Avatar


Void_Singer
03.26.2020 , 11:11 PM | #15
Probably should have mentioned that calculation (95% ^ Attempts) only works for dominant odds over 50%, otherwise you have to calculate it as (1000% - [100% - non_dominant] ^ Attempts) which isn't very intuitive, but does give the right answer.
If it moves, shoot it.
if it stops moving... shoot it again.
- Anonymous Mercenary

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DougKiller's Avatar


DougKiller
06.22.2020 , 12:37 PM | #16
Is this broken? Already did a lot of artifacts medpacs, but never hit the 5% chance

Lodinn's Avatar


Lodinn
06.25.2020 , 02:44 AM | #17
Quote: Originally Posted by Void_Singer View Post
Probably should have mentioned that calculation (95% ^ Attempts) only works for dominant odds over 50%, otherwise you have to calculate it as (1000% - [100% - non_dominant] ^ Attempts) which isn't very intuitive, but does give the right answer.
I could only assume you actually meant 100% - [100% - non_dominant] ^ Attempts for success rate which is not only very intuitive, it's literally the same formula being used above. Get rid of the "dominant" and "non-dominant" notion altogether, it makes no difference at all because the calculation here is:
1) Find the probability of independently failing every time (since we're looking for 1 success only) - that's p_failure^attempts, or (1 - "chance to research the schematic")^deconstructions
2) Find the probability of this NOT happening = 1 - probability from the previous calculation.
No matter if the success rate is 1% or 99.9% - the formula stays the same, the numbers are changing.

For say 30 attempts that gives:
At 95% failure chance the calculation would be 0.95^30=21.5% chance of not obtaining a single success in 30 trials
At 5% failure chance it becomes 0.05^30=9.3e-40 chance of not obtaining a single success in 30 trials which, for all practical purposes, is a given success.

That's just statistics 101, the part that is normally taught on like the very first lecture.
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