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Originally Posted by Kreave
Math:
P = procrate
N = number of hunters
I assume that following a standard cycle you will be able to shoot 5shots/10sec = 1/2shot/sec
The number of shots needed in average before a proc is calculated the following way:
Number of shots before a proc( I will call this q): q = 1/(p/100)
The time between a proc to the next is therefore: q/(1/2shot/sec) = 200 / p
The proc last 7 seconds, this means that the proc will be on 7sec / ( 200/p) = (p*7)/200
I have entered my own equip and an mm build in the spreadsheet from lactose, and when I add 450 atk power, my dps increase by 80dps. Therefore the increase in dps is calculated the following way:
DPS increase = (p*7) / 200
This is on 1 hunter, now each hunter in the raid will gain a benefit from the buff. Which is the same, there the total increase in dps is:
DPS increase = (p*7*N) / 200
Of course two procs can happen just after each other, I will calculate this, by saying that if a shot procs just after the first, it gives no ekstra dps and if it hit the second shot after the initial proc, it will only give half the dps increase:
Chances of a proc happens on the first shot is p, and on the second is also p. Therefore the formula for calculating the dps becomes:
DPS increase – p*DPS increase – ½*p*DPS increase = DPS increase( 1 – 3/2p)
Ps. this only works for low values of p.
I haven’t been able to detirme a good value for p yet, but my results so far shows that p = 3,2%. If anyone have any good numbers for this, I would really appreciate it :).
But lets look at a raid with 5 hunters in. The DPS increase from wearing full DS is:
DPS increase:
P DPS
1 13,5
2 27,0
3 40,4
4 53,9
3,2 42,7
I tried to calculated the change in DPS from going from full tier 2 to 8/9 tier 3, according to Lactoses spreadsheet it gave a change of 57DPS ( if I entered it correctly and I added the 50 atk power from the bonus). This show that the change from tier 2 to tier 3 isn’t actually that great and the more hunters you have the less the change actually does.
And some random fact:
If you have 35 hunters, all with 8/8 DS in a raid. Each hunter will gain an addianial 299 DPS from the DS 8/8 effect.
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I tried to solve the problem by directly computing the probability that a single hunter would be under proc, and got a different answer for DPS contribution than you did. So, in the spirit of debate, here is my approach:
The weapon of the hunter is approximately 3.0 speed w/quiver. This means that if a shot procs, then the next two shots will gain the benefit of Expose Weakness since the proc lasts 7 seconds. (Of course, further shots may also gain if a re-proc occurs.)
From this, we can construct a Markov process with 3 states:
State 0= no proc up
State 1= proc up for the next shot
State 2=proc up for the next two shots.
Putting aside stacking damage contributions for the moment BUT recognizing that reprocs can occur, this completely describes the system if we can assign probabilities to the different states. Let p denote the probability of a proc, and bear in mind that this process works as follows:
from any state i there is a p probability of moving to state 2, and a (1-p) probability of moving to state \max(0,i-1). That is, you either proc (or reproc) and go to state 2, or you fall down a state.
Assuming the process has a well-defined ergodic distribution, which seems certain, let us define the following:
\gamma_0= (Long-run) Probability that you are in state 0,
\gamma_1= (Long-run) Probability that you are in state 1,
\gamma_2= (Long-run) Probability that you are in state 2.
to deduce these values, note that we have the following
\gamma_0=\gamma_0 (1-p) + \gamma_1 (1-p) ** (says "If you aren't proccing then it's either because you the proc wasn't up the shot before and also didn't proc on the most recent shot, OR you were proccing the shot before but it was the second proced shot and you didn't re-proc")
\gamma_1=\gamma_2 (1-p) ** (says "if you have one shot under proc to go, then the shot before you must have had 2 shots left under proc and didn't get a new proc on that shot")
\gamma_0+\gamma_1+\gamma_2=1 ** (say, "stuff adds to 1" :))
So, you can solve those. Doing so gives
\gamma_0= (1-p)^2 / [(1-p)^2+p(2-p)]
Hence the probability that the proc is up at any point in time is 1-\gamma_0 \approx 0.062 if p=0.032.
Assuming an 80 DPS gain from 450 RAP, the Expose Weakness gain for a single hunter is 80*(1-\gamma_0) \approx 5.03
This is substantially less the other approach suggests. Now, you might say "Well, you're not allowing stacking procs from the same hunter" and that's true but you can easily see that the probabity of two procs from the same hunter being up is very small and can't influence this result too much.
So, my approach seems more direct. I don't use this math often (ever) so I may have screwed something up. thoughts?
edit: typo fixed.