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Author100% magic proof?
Is it possible to be 100% magic proof, i know that you can get up to 70% percent with talents resistence and barrier along with the artifiacts _____ mitheral coif, cape of arcane protection, fullmitheral armor and mitheral boots. Is it possible to get the extra 30% from enchantements?
I do not believe it can be done.

Barb max resist = 68%
Barrier = 20%
Mith coil = 5%
Cape of ArcProc = 15%
Full Mith = 5%
Mith Boots = 5%

That's 100-(100-68)(100-20)(100-5)(100-15)(100-5)(100-5)= 82%

Given the formula, it's virtually impossible to reach 100%
Hmmm you missed 'Resistance' talent (fortune branch) offering 20% magic shield.

That's 100-(100-68)(100-20)(100-5)(100-15)(100-5)(100-5)(100-20)= 85% (approx)

Question: If a racial 9 barb adds 20% air shield to all the armour pieces : helmet,boots,armour,cape and shield - then how will the formula change?
100-(100-68)(100-20)(100-5)(100-15)(100-5)(100-5)(100-20)(100-20)(100-20)(100-20)(100-20 ) = 93%
maybe with enchants you could but it'd be pricy
93% is amazing!
Pity armour enchanting is underrated... and now getting a 20% in anything is ...
u could get a total of 40% protection from enchants
helm and armor
for gurumao: yeah and a babarian that has 93% of anti-magic proof that will be bad time for ambushing wizards he he he good point very pricy a bad time for wizards
Don't express the formula like that. It is technically incorrect, and would confuse a lot of people. It should be written as,

1 - (1-a)(1-b)(1-c) ....

where a, b, c are decimal numbers between 0 and 1.

Anyway, it's not just VIRTUALLy impossible to reach 100%. It IS impossible. The formula guarantees it. It doesn't matter how many sources you have. Even if you wear ten pieces of armor, each giving you 99% resistance, you still wouldn't get 100% (you might get 99.99...%).
The only way to get 100% resistance, is if you have a single source of resistance >=100%. If you have mulitple sources, each giving you <100%, you wouldn't get it.
This topic is long since last update and considered obsolete for further discussions.

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