@ Hypathia:
Some nice work there! I like the depth of the mathematics and the results give some reason to rethink some things.
But I would like to see the maths behind the "expected number of hits before n hits in a row" formula. I've been trying to work on something similar and I get different results: at least for n=2 and a=0.5 I get 4 hits as Expectancy value for the time until 2 hits in a row. Your formula gives 6 hits.
Either I'm doing something wrong, or you are including the 2 hits in a row in your number, or I'm taking the "expected" a bit too literally as the mathematical Expectancy Value, while you mean slightly different.
Could you give your derivation of that formula please?
Give a man a fire and he's warm for only a day.
Set him on fire and he's warm for the rest of his life!
I saw direct proof of what you were saying just the other night while fighting Nightbane.
4 crushing blows in a row. Our bad luck that it was at the very beginning of a fight, while establishing aggro. Taught me a lesson though a tough one. "A crushed can will wipe the party."
First of all /bow!
Great work!
The problem about this discussion I think is the fact of the Boss design in SWP.
The amount of HP to survive the "one-shot" abilities of the Bosses, such as Brutallus (Stomp + MH +OH) or Felmyst (Breath + MH) or Eredar Twins (MH + OH + Blow / + MH, she stimes even hits you after u get a blow before she switches targets) is only possible by stacking +15 Stamina Gems in Full T6 Gear. I just have Belt and Bracers out of SWP, but i dont think full SWP gear would change that, and also on the other hand, where is the need of that when you clear SWP.
We're currently working on Muru, havent gotten to Phase two, so I can't really say anything about the Avoidance/HP usefull for this fight. The Adds from the Wave though do some burst, three adds, a caster who can hit with 7,5k FB's and 2 dw melees who buff themself with flurry and have an insane attackrate with 4k mh and 1,5k oh, so you easily lose 20k in a matter of seconds if the hits come badly.
We'll see what Kil'jaeden does, but I would say, as much as i think it is an interessting discussion, you kinda have to stack up HP in the end. Cause those "one-shots" do happen due to the lack of 25% avoidance in SWP to often that you can cover that up with avoidance.
just my 2 cents
Rhea
Sorry about not saying anything for a while. Been distracted with work and AoC.
I am including the two hits—it's the "expected number of swings until you have just been hit for the nth time in a row", more or less.Originally Posted by Gwayne
The derivation:
T_H = 1 + (1-a) 0 + (a) T_H
T_H is our basic reaction time metric: The expected time until we're hit. Here it's expressed in terms of a recurrence. The number of swings until we're hit is one (the next swing), which can either be a hit with (1-a) chance (which means zero more swings are needed), or a miss with (a) chance (which means T_H more swings are needed.)
This recurrence technique is commonly used for algorithm analysis in computer science, because many algorithms may be described in terms of recursion, which this models pretty well. In this case, we're saying "In each recursive step we do one unit of 'work'" (this is the "1"), and then we describe the probability of the recursive sub-steps and their values.
So we take the above, and simplify out the zero term:
T_H = 1 + a T_H
And do some other algebra to get T_H alone on one side:
T_H - a T_H = 1
T_H = 1 / (1 - a)
And now we've derived the basic T_H metric using a recurrence, just to show how the technique works. Once I got this, I of course realized "Duh! Of course, the time between events of probability p is 1/p." But it's nice to show with a very simple example that this analysis technique works.
So, on to the real show:
T_B(1) = T_H
by definition: The time until we get a burst of length one is the same as the time until we're hit.
T_B(2) = T_H + 1 + (1 - a) 0 + a T_B(2)
The time until being hit by a burst of size two is: The time to get hit the first time (T_H), plus one more swing (1) which may either be a hit (zero more swings needed with probability (1-a)) or a miss (we have to start all over again and add in T_B(2) with probability (a)).
By the same steps as before (after swapping the position of 1 and T_H for simplicity):
T_B(2) = 1 + T_H + a T_B(2)
T_B(2) - a T_B(2) = 1 + T_H
T_B(2) = (1 + T_H) / (1 - a)
Substituting in the value of T_H:
T_B(2) = (1 + 1 / (1 - a)) / (1 - a)
T_B(2) = 1 / (1 - a) + 1 / (1 - a)^2
These last two can be written instead:
T_B(2) = (1 + (1-a)^-1) * (1-a)^-1
T_B(2) = (1-a)^-1 + (1-a)^-2
At this point you may already have an inkling of where this is going. Let us make it explicit and look at T_B(3):
T_B(3) = 1 + T_B(2) + a T_B(3)
First, we have to get a burst of length two. Then we have one more swing. If it's a hit, we're done. If it's a miss (with probability a), we need to reset and start over again. Analysis is identical to T_B(2), mostly:
T_B(3) - a T_B(3) = 1 + T_B(2)
T_B(3) = (1 + T_B(2)) * (1 - a)^-1
T_B(3) = (1 + (1-a)^-1 + (1-a)^-2) * (1-a)^-1
T_B(3) = (1-a)^-1 + (1-a)^-2 + (1-a)^-3
And now we definitely see the pattern. Somebody with more math chops than I have could probably just write:
T_B(0) = 1
T_B(n) = (1 + T_B(n-1)) * (1-a)^-1
and go straight to the conclusion using something clever. Anyway, our pattern is this:
T_B(n) = \Sum_{k=1}^n r^k
where r is (1/1-a). This is a geometric series. There is a general formula for finite sums over such series (ASSUMING R IS NOT EQUAL TO ONE. Which in our case means that this equality does not hold if avoidance is zero):
\Sum_{k=0}^m r^k = (1 - r^(m+1)) / (1 - m)
In order to get our series into this form, we need to do this:
T_B(n) = r \Sum_{k=0}^{n-1} r^k
r^0 is always 1, so this is r*1 + r*r + r*r^2, or r + r^2 + r^3, which is what we want. Substituting in {n-1} for m:
T_B(n) = r \Sum_{k=0}^{n-1} r^k = r (1 - r^({n-1}+1)) / (1 - {n-1})
simplifying:
T_B(n) = r (1 - r^n) / (1 - r)
... lots of messy stuff left out as I tried to simplify it ...
T_B(n) = (1 - (1 / 1-a)n) (a / (1-a)^2)
This formula was quite unsatisfying to me, but I couldn't come up with any way to simplify it further. Or at least, I couldn't until I went further and tried to derive T_H(n) from it.
T_H(n) = 1 + a (T_B(n-1) + T_H(n))
That is to say, the time until we've just been hit for the nth time in a row, given that we've just been hit for n-1 times in a row, is equal to 1 swing, which may either be a hit (we're done), or a miss (probability a), in which case we must first be hit by a burst of size (n-1) and then hit another T_H(n) times. (i.e. Either it's one swing and we're hit and we're done, or it's one swing and we're missed and then we must be hit n-1 times from a state of being missed before we can add in T_H(n) again, since T_H(n) assumes we've just now been hit n times in a row.)
If we run through this, we get to:
T_H(n) = (1 + a T_B(n-1)) / (1 - a)
through exactly the same manipulations above. This doesn't simplify very well. However, when you plug numbers into it, you notice that it behaves exactly like 1 / (1-a)^n. And then (if you're me), you realize that you should've known that in the first place. OF COURSE, T_H(n) = 1 / (1-a)^n. That should have been obvious!
But that means we have two different formulas for T_H(n), and one of those is very very simple, and the other one is very very simple when given in terms of T_B(n-1). That means it gives a second way to approach T_B(n), one which allows us to simplify much more easily:
T_H(n) = 1 / (1 - a)^n = (1 + a T_B(n-1)) / (1 - a)
1/(1-a)^n = 1/(1 - a) * (1 + a T_B(n-1))
divide both sides by 1/(1-a):
1/(1-a)^(n-1) = 1 + a T_B(n-1)
flip it around:
1 + a T_B(n-1) = 1/(1-a)^(n-1)
Subtract one from both sides:
a T_B(n-1) = 1/(1-a)^(n-1) - 1
Divide both sides by a:
T_B(n-1) = (1/a) (1/(1-a)^(n-1) - 1)
And replace n-1 with n on both sides:
T_B(n) = (1/a) (1/(1-a)^n - 1)
And this is exactly the formula I gave above (and it does give the same answers as the more messy formula for T_B(n) I derived above, as well, unless I mis-stated something.)
All together, a terribly messy and slipshod piece of work. But it results in a fairly elegant little formula, so I am happy.
Learn to science and stop theorycrapping in its tracks.
/hides from hypatia's math.
As Hypatia explains this is for "non-memorized events" where as "memorized events" is mitigated by expertise (ie parry gibs) although the events (hits) are subjects of the theory Hypatia is presenting.
@ Rhea re-read the post, you are misunderstanding it.
I was just commenting the summary about EH vs avoidance with what ive experienced:
That even tho there is a pro and contra in both of them, you end up with EH cause of the way the endgame instance is designed.Over-emphasizing avoidance over Effective Health leads to fights in which the tank cannot survive bursts that inevitably occur. With less avoidance and more EH, these fights would be better survivable because the tank would live through unpreventable bursts. But at the same time, over-emphasizing Effective Health over avoidance leads to fights in which the tank is frequently faced with near-disastrous situations which cannot always be overcome. With less EH and more avoidance, these fights would be better survivable because the frequency of those dangerous situations would be reduced.
Sry if the other post was a little confuse, even tho i lived in the us for a year, my written english still suxx^^
I am not oppose to avoidance gear tanking, I actually use it a lot, but the probability you are referring to is an average over time probability, but each instantaneous hit has its own check.
You flip a coin 9 times and it all landed head and you are about to flip again, with average over time probability that is 1 in 1024 chance that the coin will be head again, but for that instantaneous probability, that coin still has 50% to be head.
It really has been about 15 years since I did probability and statistic, can you address this concern?
IMO, gearing a tank should be, research the fight for the EH requirement for the encounter of a worse case scenario and then stack avoidance much as you can after you meet the EH. Can you critique this idea?
I don't know why people have asked about 'max hits' instead of 'average hits'. If you are worried about 'max hits' just choose a higher 'average hit' to look at. Its not a problem with the model, you can choose to use it anyway you want.
You're missing the point, we asked about max hits because these scenarios are drawn out for worst case scenarios. We understand that you can interpret or choose to use it how we want to, we just wanted to ask Hypatia, the originator for his take on it.
And relax in the attitude, you're starting to sound condescending, this ain't the WoW Forums.
Hypatia, I don't know if you know this, but the forums can't display TeX
My assumption is that anybody who actually wants to read that math can read TeX, however.
Actually, if you use the basis of "given that you've just gotten heads eight times in a row", there is no average "over time probability" or anything. The unconditional probability of flipping up a heads is 1/2. The conditional probability of flipping heads nine times in a row given you've already flipped heads eight times in a row is... 1/2.
Without that given, though, things are different.
Look at it the other way around a little bit. At any given point in time, you're right, the unconditional probability that you're going to get hit is 50%. What the 1/1024 chance is, is the chance that the current swing will be a hit and that the last eight swings were also all hits. That is, it is the probability that any given swing will (unconditionally) be a hit that was preceded by (at least) eight hits. We're still talking probability here, with no averages involved at all. This is the exact probability of that event occurring. (NB: Be sure you understand which event I am referring to.)
Taking this sort of measurement and then switching over to the expected time side of things is then just looking at the average frequency of hits. And that is not a probability, but an average, it's true.
Last edited by Hypatia; 06-10-2008 at 11:27 AM.
Learn to science and stop theorycrapping in its tracks.
Oh, and since I didn't say it earlier, as a probabilist and a tank, great article.
There are fights where you cannot possibly meet the EH requirement.
Yeah, like when Sargeras hits you for 200 000 damage? Wait thats not till World of Warcraft IVThere are fights where you cannot possibly meet the EH requirement.
Although I'd like to see a list of unattainable EH minimums for bosses (I'm guessing brutallus) but you gotta cool it foolishness. Cat wasn't giving anyone attitude, so why do you have to respond in such a harsh manner.
I'm not being harsh, thats just my sense of humor. Either that or I'm a miserable angry man. I could start making completely ridiculous statements like Roana if you would rather?
I just thought it was kind of silly. If an EH minimum is the minimum EH needed to survive effectively on an encounter, than if you were not able to reach it then you it would be impossible for anyone to complete the encounter. Ergo.
What are you gonna do if you cant meet the EH minimum? Hope you get lucky enough with the dodges every time you attempt?
Edit: My idea of EH is probably wrong, apologies to roana
Last edited by Foolishness; 06-11-2008 at 10:07 AM.
I think that's not what Roana meant, I think she meant that some encounters (especially when doing progression type fights) you won't be able to survive the burst of a bosses attacks/skills with just pure EH, and you'll want some more avoidance in that case to try to avoid any potential 3-shots or whatever. I remember when I first tanked prince his goddamn thrash would 1 shot me, but putting on some avoidance I'd survive it because i'd get lucky and dodge 1 of the hits.
I get what you're saying but it just seemed a little too directed towards a member. I think you should've just stated what you stated in this post immediately after your remark. Then it wouldn't have seemed so "flame" like. But carry on.
My understanding was that the minimum EH to an encounter was to survive 3 seconds of burst, or some such? Brutallus is an encounter where you get hit for 15K every two seconds with bursts up to 28K?
It's fair to say that nobody meets the EH requirement for brutallus simply because the HP requirement is so high that you can't meet it, so your healers just hook their mana bar up to your HP and pump it in as fast as they can.
So this is how brutallus is won? I'm still not convinced but ill take it on board.Hope you get lucky enough with the dodges every time you attempt?
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