
doublebear |
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Short-ish answer: to get DPR for an ability like shocking grasp that can only trigger once, and that triggers on any successful attack use the following formula:
(1-(m1)*(m2)*(m3)...)*D+(t1*f1+m1*t2*f2+m1*m2*t3*f3+m1*m2*m3*t3*f3...)*c*D
where:
m# = the chance that the #th attack will miss
D = the average damage that the ability will do
t# = the chance that the #th attack will be a critical threat
f# = the chance a critical threat on the #th attack will be confirmed
c = the number of bonus multiples you get from a crit, as it applies to this ability
I hope the sequences are clear with regards to how to change them for more or less attacks.
So for your particular circumstances:
m1 = .2
m2 = .2
m3 = .2
m4 = .45
m5 = .7
D = 35
t1 = .3
t2 = .3
t3 = .3
t4 = .3
t5 = .3
f1 = .8
f2 = .8
f3 = .8
f4 = .55
f5 = .3
c = 1
Plugging that in, you get a DPR of .99748*35+(.332493)*35 or 46.55 for just the shocking grasp ability.
Longer Answer:
The formula I provided consists of two terms, the first is the chance that any one of your attacks hits times the damage a hit will do, the second is the chance that the first successful attack you make is a critical hit times the bonus damage a crit will do.
To calculate the odds that an event will happen given a number of tries, you have to think about the odds that it will not happen, and subtract that value from 1, to get the chances that your desired event will not not happen.
Your second question significantly complicates things, because you cannot have two charges of shocking grasp active at once, and because you must make your attacks in order of decreasing base attack bonus, you must hit with the quickened shocking grasp in your first three attacks in order for it to be used before the second shocking grasp is triggered.
This leaves us with a probability chain that I don't know how to simplify, that is going to be a really long equation, so I'm only going to describe it.
You have to consider four different situations: that the first attack hits, that the first misses and the second hits, that the first and second hit miss and the third hits, and that the first three attacks all miss.
This equation will end up looking similar to the 4 copies of the equation I posted earlier, each with a different coefficient and number of attacks based on the probability of the preceding event happening.
Reading back through this post, it seems pretty hard to follow so if you have any questions I'll try to clarify.