Aggravation

[Penril]

Of course, I could be talking out of my ass here.

You? Noooooo never

I think I pretty much end all my opinions with "I could be wrong", so I have no clue what you meant with that post.

[dansari]

Of course, I could be talking out of my ass here.

You? Noooooo never

I think I pretty much end all my opinions with "I could be wrong", so I have no clue what you meant with that post.

lol same here mate, just giving you ****

[bwdaWAR]

My guess is that, when you attack something, you simply roll twice for the proc (both at 10% in this example).

Your scenarios are the following:

1) Proc and Proc (0.1 and 0.1)
2) Proc and No Proc (0.1 and 0.9)
3) No Proc and Proc (0.9 and 0.1)
4) No Proc and No Proc (0.9 and 0.9)

From the scenarios up there, there are 3 where you WILL get the proc (1, 2 and 3). So the overall probability of getting a proc would be:

(0.1 x 0.1) + (0.1 x 0.9) + (0.9 x 0.1) = 0.11 = 11%

In other words, I believe having 2 pieces with a 10% chance to proc the same effect simply translates to an overall 11% chance. Of course, I could be talking out of my ass here.

Admittedly it's been a while since I had anything to do with probability/math, but that looks like 19% to me.
0.1*0.1+0.1*0.9+0.9*0.1=0.01+0.09+0.09 or 1-0.9*0.9=1-0.81

[xtc1999]

My guess is that, when you attack something, you simply roll twice for the proc (both at 10% in this example).

Your scenarios are the following:

1) Proc and Proc (0.1 and 0.1)
2) Proc and No Proc (0.1 and 0.9)
3) No Proc and Proc (0.9 and 0.1)
4) No Proc and No Proc (0.9 and 0.9)

From the scenarios up there, there are 3 where you WILL get the proc (1, 2 and 3). So the overall probability of getting a proc would be:

(0.1 x 0.1) + (0.1 x 0.9) + (0.9 x 0.1) = 0.11 = 11%

In other words, I believe having 2 pieces with a 10% chance to proc the same effect simply translates to an overall 11% chance. Of course, I could be talking out of my ass here.

Admittedly it's been a while since I had anything to do with probability/math, but that looks like 19% to me.
0.1*0.1+0.1*0.9+0.9*0.1=0.01+0.09+0.09 or 1-0.9*0.9=1-0.81

10%*10%+10%=11%

[Egoish]

Unfortunately i do not have a source but this was extensively tested during live, unless the RoR devs have changed something procs with an identical name do not stack as the game only checks each named proc once.

I believe the tests were carried out with a witch elf attacking a tank being healed and combat log parsed for several thousand hits, with 2 concusive daggers the dd proced at exactly 1 procs rate over time however with a concusive and a jolt the proc rate doubled.

[Penril]

My guess is that, when you attack something, you simply roll twice for the proc (both at 10% in this example).

Your scenarios are the following:

1) Proc and Proc (0.1 and 0.1)
2) Proc and No Proc (0.1 and 0.9)
3) No Proc and Proc (0.9 and 0.1)
4) No Proc and No Proc (0.9 and 0.9)

From the scenarios up there, there are 3 where you WILL get the proc (1, 2 and 3). So the overall probability of getting a proc would be:

(0.1 x 0.1) + (0.1 x 0.9) + (0.9 x 0.1) = 0.11 = 11%

In other words, I believe having 2 pieces with a 10% chance to proc the same effect simply translates to an overall 11% chance. Of course, I could be talking out of my ass here.

Admittedly it's been a while since I had anything to do with probability/math, but that looks like 19% to me.
0.1*0.1+0.1*0.9+0.9*0.1=0.01+0.09+0.09 or 1-0.9*0.9=1-0.81

You are correct, i failed at the last part ha! Assuming it calculates both individually, the probability would be 19%. Should be pretty easy to test.

[Anomanderake]

Ok, i will test out. If the proc is 20% it’s like having a permataunt anyway, or something similar, since i can obtain +60% attack speedrate on a 2.4 weapon + skills. Quite good.