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Thread: Avoidance value of Expertise

  1. #41
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    So, I took some time to work through the math, and I figured out what's going on. I can now get my analytical result to match yours exactly, but both differ from the result I get with the iterative algorithm. Here's why:

    Let's start with your equations. I'll use the notation you used in your spreadsheet, with one minor modification:
    Wp = player base weapon speed before parry-haste effects
    Wp' = player base weapon speed after parry-haste effects (pWp in spreadsheet)
    pph = Wp/Wp' = player parry-haste correction factor
    Pp = player parry chance (~20%ish)
    A = Extra parry-able attack rate not affected by parry haste (HotR, for example)

    Wb = boss base weapon speed pre-parry-haste
    Wb' = boss base weapon speed after parry-haste (pWb in spreadsheet)
    bph = Wb/Wb' = boss parry-haste correction factor
    Pb = boss parry chance (7.5% nominally, i.e. at dodge soft-cap)

    Now, let's compare. Starting with your formula for player weapon speed:
    1/Wp' = 1/Wp + 0.24*Pp/Wb'

    multiplying through by Wp gives a formula for pph:
    Wp/Wp' = pph = 1 + 0.24*Wp*Pp/Wb'

    In my MATLAB code, the formula for pph is:
    pph = 1 + 0.24*Wp'*Pp/Wb'

    There's a similar discrepancy in the bph terms. In your analytical formalism:
    1/Wb' = 1/Wb + 0.24*Pb*(1/Wp'+A)
    bph = Wb/Wb' = 1 + 0.24*Wb*Pb*(1/Wp'+A)

    whereas in the matlab code I have (letting player_attack_speed = 1/Wp'+A):
    bph = 1 + 0.24*Wb'*Pb*(1/Wp'+A)
    I'll come back to the meaning of this discrepancy in a little bit. First, let's do a little math.

    If we make some variable substitutions to clean this up a bit, we can put your two equations in a more easily dissectable form:
    pph = 1 + Cp*bph
    bph = 1 + Cb*(pph + A*Wp)

    where Cp = 0.24*Wp*Pp/Wb and Cb = 0.24*Wb*Pb/Wp

    This system of equations is clearly solvable, giving the result you use in your spreadsheet. In condensed form, by plugging the second equation into the first and solving for pph, we get:
    pph = [ 1 + Cp*(1 + Cb*A*Wp) ] / [ 1 - Cp*Cb]


    Let's do the same for my two equations. Using the same substitution, I get these two equations:
    pph = 1 + Cp*bph/pph
    bph = 1 + Cb*(pph + A*Wp)/bph

    These equations are not analytically solvable - let A=0 for simplicity, and you'll see the equations are of the form:
    a = 1 + Cp*b/a
    b = 1 + Cb*a/b
    plugging one into the other doesn't eliminate a variable, and there's no simple way to eliminate either a or b from either equation. At best, you can substitute into the equation for a recursively:
    a = 1 + Cp*b/a = 1 + Cp/a*(1+Cb*a/b) = 1 + Cp/a + Cp*Cb/b
    = 1 + Cp/a + Cp*Cb/(1+Cb*a/b) = 1 + Cp/a + Cp*Cb/(1+Cb*a/(1+Cb*a/b))
    You could make some binomial-type approximations here if Cb*a/b is small, but barring that you can't really get anywhere with them. If you did make that sort of approximation, you'd get terms of alternating sign that increase in powers of (Cb*a). Since this is safely below 1, the series will converge eventually.

    On the other hand, these two equations are a recursion relation. By putting in a sample value, you can iterate to arrive at the exact value, provided the set of equations converges. In other words, starting with a0 and b0:

    a1 = 1 + Cp*b0/a0
    b1 = 1 + Cb*a0/b0

    a2 = 1 + Cp*b1/a1
    b2 = 1 + Cb*a1/b1

    and so on. This is what I do in the MATLAB code, and it's equivalent to extending the series I described out earlier to as many orders as there are iterations.

    As a sanity check, your formula gives me pph = 1.0328, while mine gives me pph = 1.02914, a difference of around 0.35%. So as far as accuracy is concerned, either method is more accurate than we'd ever care to worry about.

    However, let's quickly address the meaning of the discrepancy, just so we know why the two methods are different. The two formulas for pph were:
    pph = 1 + 0.24*Wp*Pp/Wb' (spreadsheet)
    pph = 1 + 0.24*Wp'*Pp/Wb' (MATLAB)

    The second term here can be interpreted as follows: 24% is the amount of a swing you gain, on average, when you parry an attack. Thus we're multiplying 24% of our weapon speed Wp by the number of parries we get per second on average, as calculated using our parry chance Pp and the boss's parry-adjusted swing timer Wb'.

    This makes a lot of sense - we're taking attacks at a rate of 1/Wb', and we parry Pp percent of them, and each parry gives us 24% of a swing. The question is whether we gain 24% of our base (pre-parry-haste) swing speed Wp, or 24% of our parry-hasted swing speed Wp'.

    I would guess that the parry-hasted swing speed is the more accurate one, because the goal is to determine how many attacks we get on average during the course of a fight. On the other hand, any individual swing will be parry-hasted by 24% of our pre-parry-haste swing speed. So even I'm not sure which one is the correct answer.

    However, as I've shown, the difference comes out to be pretty trivial numerically. Whether we use one version or the other, the answers end up within 1% of each other.
    Last edited by theckhd; 10-27-2009 at 07:50 AM. Reason: fixed some text formatting
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  2. #42
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    Quote Originally Posted by drae View Post
    I'm eagerly anticipating that article Theckhd. I've got a couple of questions regarding what kind of impact different levels of ratings have on the final answer, but you'll probably answer them in your article. I'll have to hold my breath.
    Ask away, it's entirely possible you'll ask something that I hadn't intended to cover in the article (or hadn't even thought of yet). At worst I'll give you a brief summary of the answer and a "details will be in the article." At best, you might stumble upon something I've overlooked and make the article better for the effort.
    "Theck, Bringer of Numbers and Pounding Headaches," courtesy of Grehn|Skipjack.
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  3. #43
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    This makes a lot of sense - we're taking attacks at a rate of 1/Wb', and we parry Pp percent of them, and each parry gives us 24% of a swing. The question is whether we gain 24% of our base (pre-parry-haste) swing speed Wp, or 24% of our parry-hasted swing speed Wp'.

    I would guess that the parry-hasted swing speed is the more accurate one, because the goal is to determine how many attacks we get on average during the course of a fight. On the other hand, any individual swing will be parry-hasted by 24% of our pre-parry-haste swing speed. So even I'm not sure which one is the correct answer.
    If we take the following statement for granted:
    When you parry an attack, the remaining time on your current swing is reduced by 40% of your weapon speed, unless this would result in a reduction to less than 20% of your swing time remaining
    This is a rather easy probability-problem, if you draw this problem it would look like:

    Where the y-value symbolizes the reduce in weaponspeed and the x values the point in time when the parry occures. The Area under the curves is the average reduce of your swing. (we have a square 0.4*0.4 and a rectangular triangle 0.4*0.4/2 = 0.24) so the haste is based of your normal non-parry-hastened hits. (hopfully this is at least a bit understandibly it would be hard for me to explain it in my first language). If the Boss hits faster than 0.6 your weaponspeed the things might be a bit more complicated (or not) but it is definitly to late for me to think about that
    edit: ok the scenario: boss hits faster than 0.6 your weaponspeed wouldn't be that hard but I will have to do some further testing (deadmines I am coming)
    Last edited by kahalm; 10-26-2009 at 03:03 PM.

  4. #44
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    That's how you calculate the 0.24 in the formulas, but it doesn't answer the question I posed.

    Given the statement is correct, we all agree that on an individual swing, you gain up to 40% of your weapon speed Wp.

    But we're interested in aggregate statistics - many attacks over a long period of time. In that case, does the formula that uses Wp give a better estimate of the number of attacks, or does the formula with Wp' work better?

    Like I said, I can see both arguments. It's possible that the parry-haste term should just use Wp, since that's the amount you would see from a single swing. But that would seem to ignore the recursive form of the equations, in that when you parry, you swing faster, leading to more boss parries, leading to more chances for you to parry, and so forth.

    I need to think about it a little harder before I'd feel confident saying one way or the other.
    "Theck, Bringer of Numbers and Pounding Headaches," courtesy of Grehn|Skipjack.
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  5. #45
    Quote Originally Posted by theckhd View Post
    That's how you calculate the 0.24 in the formulas, but it doesn't answer the question I posed.

    Given the statement is correct, we all agree that on an individual swing, you gain up to 40% of your weapon speed Wp.

    But we're interested in aggregate statistics - many attacks over a long period of time. In that case, does the formula that uses Wp give a better estimate of the number of attacks, or does the formula with Wp' work better?

    Like I said, I can see both arguments. It's possible that the parry-haste term should just use Wp, since that's the amount you would see from a single swing. But that would seem to ignore the recursive form of the equations, in that when you parry, you swing faster, leading to more boss parries, leading to more chances for you to parry, and so forth.

    I need to think about it a little harder before I'd feel confident saying one way or the other.
    I'll start by saying that this does more or less go above my head in terms of math, but here's what I understand we're trying to do:

    [1] We're trying to find out how many melee swings we can prevent from happening.
    [2] We're trying to find out by how much we can reduce the burst damage taken.

    Now as I understand it, you're asking wether for the sake of our swing timer we should plug parry hasted numbers in because if you parry the boss, then your swing timer is reduced, now if the boss parries your parry then his swing timer is reduced, but we might parry that parry (...) etc.

    I think that for the sake of question 1, it makes sense to plug hasted numbers. After all it's how things could end up happening on a very rare occasion. And we're looking at how much damage we avoid over a longer period of time.

    For the sake of question 2 however, I don't think this is at all relevant. If I just parried a boss's melee swing, then I'm not really taking significant burst damage if my parry causes the boss to parry and last a parry hasted attack. Im not taking burst damage because I just parried the previous attack.

    Some other things that I think would matter is how synched up your melee swing is with the boss, and what your swing timer is vs the boss swing timer.

    Then again, I could be misunderstanding your dilemma entirely.

  6. #46
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    So, I spent some time last week to work this problem out from first principles. It turns out that both expressions are wrong, but for different reasons. Here's the derivation; afterwards I'll explain where kahalm's version of the deriviation went wrong.

    There are two ways to interpret the effect of parry-haste. One is to look at a fixed number of swings N. From this point of view, each parry-hasted attack reduces the total time T it takes to reach N swings. In other words, we get N swings in T-dT time, where dT is the total amount of time "saved" via parry-hasting. We can compute the parry-hasted swing rate as:
    R = N/(T-dT)
    And of course, our swing timer would be the inverse of this, 1/R = (T-dT)/N.

    The other way to interpret the effect of parry-haste is to look at a fixed amount of time T. In this formulation, each parry event gives us an additional "fractional swing," because now we're producing swings slightly faster than usual. In other words, we produce N+dN swings in time T, where dN is the sum of all the fractional swings we get from parry-hasting. The swing rate is then calculated as:
    R = (N+dN)/T,
    with the swing timer again being the inverse of this.

    It's important to note that in the long-time limit (T->infinity), (N+dN)/T must approach the same value as N/(T-dT), since they're both perfectly valid ways of calculating the average rate. It was while working through the algebra for these two approaches that I discovered what was happening.

    We're going to derive this for a generic system of two events A and B. The problem definition is very simple:
    We have events A and B.
    Event A happens with frequency Rao (in vacuo)
    Event B happens with frequency Rbo (in vacuo).
    Event B has a Pb percent chance to decrease the time until the next A by 0.24/Rao.
    Event A has a Pa percent chance to decrease the time until the next B by 0.24/Rbo.

    Given this well-defined problem, what is the correct analytical answer, or at least the correct set of coupled equations we need to be solving?

    Event A will be player auto-attacks, and Event B will be boss auto-attacks. For weapon speed Wao, a player makes N auto-attacks in time T before parry-haste is considered. Thus,
    Code:
    Rao = N / T = 1 / Wao.
    The actual rate of player auto-attacks after parry-haste will be Ra.
    In a similar fashion, Wbo is the base boss weapon speed Rbo is the base boss attack rate, and Rb is the rate after parry-haste.

    (Note: The nomenclature here is partly blizzard's fault. They call it "weapon speed," but then give a value in seconds. Anyone with a basic science background knows that a speed is a rate, and should be in units of 1/s. Nonetheless, I'm going to stick with their convention of calling Wao and Wbo "speeds" but using units of seconds. Hopefully this isn't too confusing, since I'll be attempting to use Rao and Rbo rather than Wao or Wbo wherever possible.)

    We'll now calculate the average Ra with each approach to make sure they agree:

    In time period T, we'd have T.*Pb.*Rb parries. Each parry reduces the swing timer by 0.24*Wao seconds. We're sure this is correct, because on a successful parry the swing timer is reduced by 40% of the pre-parry-haste value Wao, not the post-parry-haste value Wa. In other words, whether or not your previous attack was parry-hasted has no effect on the reduction (in seconds) you get during your next swing. This is consistent with the old EJ parryhaste mechanics thread, and there will be further proof later in the derivation that using 0.24*Wa gives you an incorrect answer.

    This leads to a cumulative time reduction dT:
    Code:
    dT = 0.24*Wao*T*Pb*Rb
    Just to check that this makes sense, this is:
    Code:
    dT = (swing timer redux per parry)*(total time T)*(Boss Parry Chance)*(Boss Attack Rate)
    dT = (swing timer redux per parry) * (total number of parries in time T)
    I'm going to define a constant c = 0.24*Pb to simplify the rest of the derivation, such that
    Code:
    dT = c*T*Rb*Wao.


    First Approach
    :
    We have N attacks in time (T-dT), for a rate of:
    Code:
    Ra = N / (T-dT) = (T*Rao)/(T - c*T*Rb*Wao) = Rao/(1-c*Rb*Wao)
    Ra*(1-c*Rb*Wao) = Rao
    Ra = Rao + c*Rb*Ra/Rao
    Note that this differs from both the version we use in the MATLAB (which is Ra = Rao + c*Rb*Rao/Ra given that pph=Wao/Wa=Ra/Rao) as well as kahalm's version (Ra = Rao + c*Rb). I'll come back to that point in a second.

    Second Approach:
    The other approach is to say that in time T, we actually get N + dN attacks due to parry haste. N is still T*Rao, but since it only takes (T - dT) to the Nth attack, there's some time left over (dT) during which we can sneak in dN extra attacks. The average rate is then Ra=(N + dN)/T.

    This is a crucial juncture, because there's a "wrong way" and a "right way" to calculate dN. I will now demonstrate that by doing it the "wrong way" you get kahalm's answer, and by doing it the "right way" you get an answer consistent with the previous N/(T-dT) formulation that you suggested.

    Wrong Way: In time dT, you get dN = dT/Wao = dT*Rao extra attacks. Then
    Code:
    Ra = (N + dN)/T = (T/Wao + dT/Wao)/T 
       = (T*Rao + c*T*Rb*Wao/Wao)/T = Rao + c*Rb
    Ra = Rao + c*Rb
    This is exactly the form that kahlam's equations take.

    Right Way: In time dT, you actually get dN = dT/Wa = dT*Ra extra attacks. The justification is very straightforward: There's nothing "special" about that period dT that would prevent us from seeing parry-hasting. Since dN represents the total number of attacks we generate in dT (including the effects of parry-haste), we have to use the rate Ra. This is no different from saying that over a 5-minute fight, we'd expect 5*60*Ra parry-able attacks. Thus:
    Code:
    Ra = (N + dN)/T = (T/Wao + c*T*Rb*Wao/Wa)/T 
       = Rao + c*Rb*Ra/Rao
    which agrees with the N/(T-dT) formulation.

    I'll note here that if you make an error calculating dT, specifically by using Wa instead of Wao, the N/(T-dT) formulation also gives you kahlam's answer, Ra = Rao + c*Rb. This is the proof I mentioned up when I defined dT. If you use Wa to define dT, the two averages don't come out in agreement, which they must if they are to have any physical meaning.

    To summarize, here are the three possible forms of Ra that we have:
    Code:
    Ra = Rao + c*Rb*Ra/Rao (derived from first principles two different ways)
    Ra = Rao + c*Rb  (kahlam's - caused by an error in the above derivation)
    Ra = Rao + c*Rb*Rao/Ra (MATLAB)
    The version I've derived here from first principles is correct.

    Kahlam's, while wrong, is the closest to correct. It's error is simply a factor of (Rao/Ra) in the coupling term, and it's arrived at by making a minor mistake in the derivation. It's also worth noting here that this is effectively a first-order approximation, since he's substituted Wao in a spot where Wa should have been.
    Note that the result of this is to slightly over-couple the equation, since (Rao/Ra > 1)

    The MATLAB version is the farthest off. It differs by a factor of (Rao/Ra)^2, thus over-coupling more than kahlam's version. Furthermore, I haven't found any sort of rational justification for this factor. The closest I can come is by making two consecutive mistakes in the derivation: first by calculating dT as c*T*Rb*Wa, followed by calculating dN as dT/Wao. Those two errors are mutually inconsistent, so there's no chance that this is actually a lucky string of errors that corrects themselves. My best guess is that when we were fooling around with the parry-haste code, I either made a typo or we were trying to do something unnecessarily complicated to account for higher-order parry effects (i.e. two parries within one swing timer).

    One can perform a similar calculation for Rb to get the complete system of equations that need to be solved. They are:

    Code:
    Ra = Rao + c1*Rb*(Ra/Rao)
    Rb = Rbo + c2*Raa*(Rb/Rbo)
    Raa = f(Ra) = A + B*Ra
    where A is the static term representing attacks that are independent of parry haste (33/6 for warriors, for example) and B is the contribution from attacks that benefit from parry-haste (1 for warriors).

    As a final thought, it should be noted that these equations are not as simple to solve as the ones kahalm came up, but they are solvable. Here's the derivation:

    Code:
    Starting from:
    Ra = Rao + c1*Rb*Ra/Rao
    Rb = Rbo + c2*Raa*Rb/Rbo
    Raa = A + B*Ra
    
    Solve Rb eqn for Rb:
    Rb(1-c2*Raa/Rbo)=Rbo
    Rb = Rbo/(1-c2*Raa/Rbo)
    
    Plug into Ra eqn:
    Ra - Rao = (c1*Ra/Rao)*Rbo/(1-c2*Raa/Rbo) = (c1*Ra/Rao)*Rbo^2/(Rbo - c2*Raa)
    (Ra - Rao)*(Rbo - c2*Raa) = c1*Ra*Rbo^2/Rao
    Ra*Rbo - c2*Ra*Raa - Rao*Rbo + c2*Raa*Rao - c1*Ra*Rbo^2/Rao = 0
    -Ra*Rbo + c2*Ra*(A+B*Ra) + Rao*Rbo - c2*Rao*(A+B*Ra) - c1*Ra*Rbo^2/Rao = 0
    c2*B*Ra^2 + (c2*A + c1*Rbo^2/Rao - Rbo - c2*B*Rao)*Ra + (Rao*Rbo-c2*A*Rao) = 0
    
    let:
    a=c2*B
    b=(c2*A + c1*Rbo^2/Rao - Rbo - c2*B*Rao)
    c=Rao*Rbo - c2*A*Rao
    
    and we have a standard quadratic equation:
    a*Ra^2 + b*Ra + c = 0
    
    The solution is then
    Ra=(-b - sqrt(b^2-4*a*c))/(2*a)
    where I've chosen the negative root of the square since it gives a realistic result
    (the positive root gives us something ridiculous like 13 attacks per second).
    Some quick testing shows that my iterative technique gives me the same result as the analytical solution out to 8 decimal places.


    I've already updated my local version of the code to reflect this model. I'm in the process of nailing down the particular values of A and B for each class (and possibly another constant C for DK's courtesy of the way Rune Strike interacts with boss attacks). Once that's done, I should easily be able to calculate efficiency values for each of the four classes, and thus get around to completing the article.
    "Theck, Bringer of Numbers and Pounding Headaches," courtesy of Grehn|Skipjack.
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  7. #47
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    That, Theckhd, would be a superb achievement. Thanks for your efforts so far!
    The DK tank site: pwnwear.com.

  8. #48
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    Just wanted to return to this topic to let you know that despite being busy, I had not forgotten about you. I've finally found time to finish the calculation for all 4 classes.

    The results are here, with a TLDR summary at the beginning:
    Quote Originally Posted by Theck
    • For Icecrown Citadel, expertise is roughly 91% as effective as dodge rating for reducing incoming damage, assuming a boss swing speed of 1.4 seconds (after JotJ). It varies significantly with boss attack speed though, from 78% @ 1.2 to 154% @ 2.4 speed.
      • For Warriors, this range narrows to 52-102%, with an "average" of 60% at 1.4-speed.
      • For Druids, it's 49-98%, with an "average" of 60%
      • For Death Knights, the range becomes 36-71%, with an "average" of 46%.
    • For bosses outside of Icecrown, we would use an average swing speed of around 2.0, giving average values of 87% for paladins (53%-104% range), 57% for warriors (30%-69%), 56% for druids (28%-68%), and 29%-45% for Death Knights (28%-52%).

      All of these can be seen in the plots in the "results" section.
    Though I highly recommend going to the post to see the pretty plots. :P
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  9. #49
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    Thanks for posting this.
    Quote Originally Posted by Tengenstein View Post
    just don't let them melee you up the bum.

  10. #50
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    Great article Theck. I am happy to see some final results for warriors (and others).

    A couple of questions.

    1. How does the tanks weapon speed effect the avoidance value of expertise? Does a slower weapon like a Quel cause the avoidance value of expertise to go up or down? Is the impact substantial between say a 1.6 vs 2.6s swing speeds?

    2. Are the avoidance gains the same for each point of expertise? Like if I have 6.5% expertise in case A, and 11% expertise in case B, do I gain the same amount of avoidance by adding .25%?

    Sorry if this is painfully evident in the maths, but I haven't done calculus since college and I can't follow your calcs

    Thanks!

    Be a Champion, not a hero.
    Drae

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  11. #51
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    Quote Originally Posted by drae View Post
    1. How does the tanks weapon speed effect the avoidance value of expertise? Does a slower weapon like a Quel cause the avoidance value of expertise to go up or down? Is the impact substantial between say a 1.6 vs 2.6s swing speeds?
    Yes, the results will depend on weapon speed quite a bit, because you are creating more parry opportunities with a fast weapon than a slow weapon. This is the primary reason DKs have a much worse efficiency ratio than the other classes. Note that these simulations are for 2H tanking, I need to do a little work before I can generate values for DW tanking.

    I can actually generate plots that equate weapon speed to dodge rating, I just didn't find a good place to put them in the article itself. But here they are, since you asked (generated for a paladin though):
    http://sites.google.com/site/theckhd...3_exp_ps_N.png
    http://sites.google.com/site/theckhd..._exp_ps_Np.png
    http://sites.google.com/site/theckhd...ps_eqdodge.png

    and, perhaps more importantly



    So for a paladin, changing from a 1.6 speed weapon to a 2.6 speed weapon is roughly equivalent to adding 39 dodge rating.

    Quote Originally Posted by drae View Post
    2. Are the avoidance gains the same for each point of expertise? Like if I have 6.5% expertise in case A, and 11% expertise in case B, do I gain the same amount of avoidance by adding .25%?
    Not exactly, but very very close. Boss parryhaste is linear in expertise, but the efficiency ends up being very weakly quadratic thanks to the recursive nature of parry-haste. However, the efficiency parameter varies by less than 0.2% going from 7.5% parry to 0% parry (26 to 56 expertise), so it's safe to treat it as constant.

    In other words, if expertise is 78.0% as good as dodge at 26 expertise, it'll still be at least 77.8% as good as dodge at 55 expertise. It also means that if 10 expertise rating is equivalent to 7.8 dodge rating at 26 expertise, it'll be at least as good as 7.78 dodge rating at 55.
    "Theck, Bringer of Numbers and Pounding Headaches," courtesy of Grehn|Skipjack.
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  12. #52
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    Thanks Theck! Your awesome, now if only we could convince you to re-roll warrior so all your writings are based on them!

    Seriously tho, thanks a lot for this work; really appreciate it.

    I would imagine the DK efficiency to sky-rocket when dual wielding.

    Be a Champion, not a hero.
    Drae

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  13. #53
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    70
    Quote Originally Posted by drae View Post
    Thanks Theck! Your awesome, now if only we could convince you to re-roll warrior so all your writings are based on them!
    To get warrior values, I think you can just divide by a rough estimate of the efficiency ratios for the two classes.

    For example, if you look at the data I posted, a paladin's efficiency is 78%, while a warrior's is 52%. That means we get almost exactly 1.5x as much "effective" dodge rating from a point of expertise as a warrior does. So just divide the numbers on the plots I posted by 1.5 to figure out what you'd get. If I get 39 dodge rating from switching weapons, you'd get 26.

    Also, for the record, those values were generated using a 1.2-speed boss with Chill of the Throne active.

    Quote Originally Posted by drae View Post
    I would imagine the DK efficiency to sky-rocket when dual wielding.
    Yeah, it should increase quite dramatically for DW. Maybe not as high as paladins, but almost certainly ahead of warriors.
    "Theck, Bringer of Numbers and Pounding Headaches," courtesy of Grehn|Skipjack.
    MATLAB TPS 4.0, Talent Spec & Glyph Guide 4.0

  14. #54
    Join Date
    Oct 2008
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    4,930
    Mmmmm, Theck-math.

    Lemme get a spoon, I'm hungry. =)

    Thank you, sir.
    The (Old) Book on Death Knight Tanking
    The New Testament on Death Knight Tanking
    -----------------------------------------
    Quote Originally Posted by Horacio View Post
    Who f-ing divided by zero?!?

  15. #55
    Join Date
    Jan 2009
    Location
    United Kingdom
    Posts
    1,644
    Nice numbers.

    However, in regard to the Warrior figures it is getting increasingly uncommon for Warriors to actually maintain Thunder Clap and Demo Shout as the majority of encounters require 2 tanks and other tanks will often apply said debuffs without a rotational loss.

    Additionally, Shockwave is rarely used every 20s on the dot as it is often inserted into a specific rotational point to avoid conflicting with SnB procs. Due to the conflict, I would suggest that Warriors only typically use a rough average of 1.5 Shockwaves per 60 second period. Even in maximum Warrior DPS/TPS logs, this theory is almost supported by the Shockwave execute count present over the fight duration.

    So, those points being the case, the worst-case (and increasingly common) Warrior rotational scenario involves having only 1.5 unparryable GCDs out of the 40 present in a 60 second period.
    Maintainer of Rawr.ProtWarr theorycrafting tool. Feel free to PM suggestions or feature requests!

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