If there comes a time where it is easier to ascertain the magic evasion of any mob of interest (probably never given the FFXI "team's" fetish for making basic game mechanics as opaque as possible, and lack of information sharing among the "playerbase"), perhaps it can be useful to quantify the difference in overall magic damage between a "high-resist" setup (with the purpose of maximizing magic accuracy) and a normal setup for resistant NMs and whatnot. But realistically this is just another parlor talk.
A long time ago, I argued that levels of resistance for a single "nuke" can be modeled by a one-parameter categorical distribution, with the parameter being the probability that a nuke is not resisted at all (full damage). This probability will be called "overall magic accuracy" for the remainder.
To reiterate, the distribution can be described as
no resist: π
1/2 resist: π(1-π)
1/4 resist: π(1-π)2
1/8 resist: (1-π)3
This assertion was based on prior observations by me and others that multinomial count data for nukes, categorized by level of resist, seemed to conform to such a pattern. (I will not discuss the speculated motivation for the programmers to use this model, assuming it is true.) If this is a reliable model, it seems reasonable to think about the effect of overall magic accuracy on magic damage in terms of expected value.
Ignoring rounding, let X be the actual damage of a nuke (subject to being resisted) with unresisted damage D. The expected value of X can be expressed as
E[X] = D[π+0.5π(1-π)+0.25π(1-π)2+0.125(1-π)3]
Based on this expression, overall magic accuracy can be thought of as attenuating the unresisted damage of a single nuke in the long run, multiplying that damage by some factor less than 1 that is a function of π. Therefore, in making some assessment of overall magic damage as a function of magic accuracy, we don't have to consider the actual distribution of resists given π, just as players calculating physical damage don't consider the distribution of pDIF given a ratio of attack to defense.
Just as magic accuracy attenuates unresisted magic damage by some factor less than 1, magic attack bonus (MAB) amplifies magic damage by a factor greater than 1. This is illustrated and summarized with the following graph plotting these factors described (for magic accuracy and magic attack):

As you can see, when "long run" magic damage is considered, there is decreasing return to overall magic accuracy, π (the expected value computed earlier is a third-order polynomial with respect to π), and constant return to MAB. The endpoints also make sense, too. If you happen to have 100 MAB, your overall damage is twice as high. If you happen to have 100% overall magic accuracy (recognizing that this is impossible in FFXI for nukes), then there is no attenuating of your potential magic damage.
Using the model for levels of resistance I described, it is possible and simple to estimate the percent change in long-run magic damage between two equipment setups of interest. Suppose you have a normal setup with +70 MAB such that you know will achieve 60% overall magic accuracy on some target of interest (this means in the long run 60% of your nukes will be unresisted) and you are interested in assessing whether utilizing your "high-resist" setup is worth the tradeoff in potential damage. Suppose your high-resist setup has +63 MAB and +26 more magic accuracy than the normal one.
At this point, there should be no need for quantifying the relative performance increase, but perhaps you want to quantify it anyway.
Since the magic damage "formula" is just multiplying various factors together, it is easy to calculate a percent difference that is independent of base damage, INT, weather effect, etc. (all of which could be considered constant). One needs merely to identify the multiplicative factors associated with MAB (+63 and +70) and m. acc (60% and 86%). Through direct calculation,
(1.63)(0.924757)/[(1.70)(0.752)] - 1 = 0.179
In the "long run," the overall damage using the "high resist" setup will be almost 18% higher than that using the normal one.
Again, is this useful or practical? Not really. But it could serve as a theoretical framework for "theorycrafting" (oh how I hate MMORPG-related jargon).
curious, in your categorial distribution, what did you do with the remaining (1-pi)^4 ? complete resist? (which, outside of magic-shielded mobs, i have seen only a handful of times, the one that comes to mind is casting cure spells on undead besieged mobs at level ~25. then again, healbombing could very well follow a different resist setup, like enfeebles). sorry if you addressed this when you first derived the model.
ReplyDeleteStealth_Duck
This probability model is based on the deduction of other people based on count data. The observed multinomial counts (from data collection) provide some support for the plausibility of this model. (See lodeguy's data sets in particular.) Unfortunately, statistical evidence is usually based on goodness-of-fit tests.
ReplyDeleteIt seems to me that a complete resist is more aptly characterized as immunity (perhaps 100% magic damage reduction) to a spell. Then again, I have not considered the behavior of Cure spells on Undead. (They outright resist Cures sometimes? I didn't know that.) If there is a fifth level (complete) to account for Cures, maybe it is lumped in with the eighth-resist, unobservable unless when using Cures.
I forgot to mention the proposed mechanism underlying the model. Under this model, the game supposedly flips a coin (metaphor) for a binary outcome. If the result is success, then that's no resist and no more flipping occurs.
ReplyDeleteIf the result is failure, then the coin is flipped again. If the result is success, then the result is a half-resist and the coin flipping stops.
If the result is again failure (consecutive failures), then the coin is flipped yet again. If the result is a success, then the result is a quarter-resist. If the result is a failure (three straight failures), then the result is an eighth-resist.
The probability expressions follow from conditional probability rules.
I got that it was multiple hit-tests, however the probabilities stated left a remaining probability of 4 'misses'. else 1/8 resist would have probability (1-pi)^3. (3 misses in a row). having pi*(1-pi)^3 leaves a remainder of (1-pi)^4 unaccounted for. So I was wondering where the remainder is accounted for.
ReplyDeleteYou're right. I will change that soon.
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