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Rune of Invention
Old 03-15-2013, 04:06 PM   #1
SilverLink
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Default Rune of Invention

Currently if you have both Rune of Invention 5+++ and Rune of Invention 6+ equipped the effects from them stack allowing you to craft at a skill of +22.

I believe that runes of the same type aren't supposed to be able to stack like this but I suppose it could be intended.

I have tested and am able to make much higher tier planks than I would usually be able to.
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Old 03-15-2013, 04:13 PM   #2
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Ugh this is true... I just equipt 2 of them and you posted this!

so yes glitch this is true
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Old 03-15-2013, 04:17 PM   #3
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omg someone try make a +42 stone! or 43 is it?
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Old 03-15-2013, 04:18 PM   #4
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I hope this gets to glitch before its exploited too much
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Old 03-15-2013, 05:14 PM   #5
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It'll be fixed with this reload.
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Computing the probability that at least one of the following events will occur:
P(a or b ... or z) = 1 - P(!a and !b ... and !z)
Probability
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Old 03-15-2013, 05:16 PM   #6
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20 TC reward to Silverlink.
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Computing the probability that at least one of the following events will occur:
P(a or b ... or z) = 1 - P(!a and !b ... and !z)
Probability
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Old 03-15-2013, 06:29 PM   #7
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Quote:
Originally Posted by Glitchless View Post
It'll be fixed with this reload.
Many are wondering what measures you will specifically take to fix this, and deter/prevent such sort of exploits from happening in the future again

Don't forget the impact on Arena what with the top teams and all
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Old 03-15-2013, 06:58 PM   #8
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He did say in chat after the reload that all such stones would be removed as well as any enchanted armour having the enchant wiped.
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Old 03-15-2013, 07:30 PM   #9
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Did the reload only fix the problem with invention rune or was there a check made to make sure that all rank 6 runes will not stack with the others? A large portion of the other runes would be far less noticable if they stacked but would technically also be exploits.
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Old 03-15-2013, 07:45 PM   #10
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It was a fix for all rank 6+ runes to prevent equipping 2 of the same type.
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Computing the probability that at least one of the following events will occur:
P(a or b ... or z) = 1 - P(!a and !b ... and !z)
Probability
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Old 03-15-2013, 08:38 PM   #11
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We are reloading to wipe all of the exploited enchanting stones and stats of items that were modified with them.
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Computing the probability that at least one of the following events will occur:
P(a or b ... or z) = 1 - P(!a and !b ... and !z)
Probability
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Old 03-15-2013, 09:21 PM   #12
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Quote:
Originally Posted by Glitchless View Post
We are reloading to wipe all of the exploited enchanting stones and stats of items that were modified with them.
+1 for also being merciful, since exploiting is against ToU.
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