Iksar Shaman - Solo Challenges?

[bcbrown]

Originally Posted by DeathsSilkyMist [You must be logged in to view images. Log in or Register.]

For Direct damage spells you do get twice the damage. For DoTs specifically you get the Direct Damage component of the DoT + X amount of ticks. That is why I reduced the damage by 50%, because on average it will proc halfway through the fight. You are getting half the DoT ticks on average.

[1 roll, 2 rolls, 3 rolls, 4 rolls, 5 rolls, 6 rolls, 7 rolls, 8 rolls , 9 rolls, 10 rolls, 11 rolls, 12 rolls] / 12 roll attempts = 6.5 rolls on average to get any specific number one time on a D12. This means you will get a proc halfway through the fight on average, as you will roll the specific number you want after 6.5 attempts on average.

We're trying to calculate an expected value, the expected damage done by scourge procs. Keeping a copy of https://en.wikipedia.org/wiki/Expected_value and https://www.stat.purdue.edu/~zhangha...1%20Sec3.3.pdf open will be helpful here.

The way you calculate an expected value is by summing up all the possible outcomes, each multiplied by it's probability weight. Our random variable, X, is in this case the event of a proc.

E[X] = sum over all x of E[x]

The probability weight is the proc rate, or 1/12. The expected number of procs with your 12 rolls is going to be the sum from 1 to 12 of 1/12, or 1. We are in agreement here.

But we're not just interested in the expected number of procs, we're interested in the expected damage dealt.

There's an identity, that for a linear function of x, the expected value of the function is the function of the expected value of X:
E[aX] = a * E[X]

This is the calculation you are doing. E[d(x)] = d(E[x]). However, the expected damage is a function of time, and it is also a function of whether or not there has been a prior proc.

Therefore, the linearity of expectation does not hold, and it is neccessary to calculate the sum over all the possible outcomes.

This is why you cannot simply multiply the damage by the average time for a proc to hit.

[DeathsSilkyMist]

Originally Posted by bcbrown [You must be logged in to view images. Log in or Register.]

The probability weight is the proc rate, or 1/12. The expected number of procs with your 12 rolls is going to be the sum from 1 to 12 of 1/12, or 1. We are in agreement here.

But we're not just interested in the expected number of procs, we're interested in the expected damage dealt.

Agreed!

Originally Posted by bcbrown [You must be logged in to view images. Log in or Register.]

There's an identity, that for a linear function of x, the expected value of the function is the function of the expected value of X:
E[aX] = a * E[X]

This is the calculation you are doing. E[d(x)] = d(E[x]). However, the expected damage is a function of time, and it is also a function of whether or not there has been a prior proc.

Therefore, the linearity of expectation does not hold, and it is neccessary to calculate the sum over all the possible outcomes.

This is why you cannot simply multiply the damage by the average time for a proc to hit.

This is where you are incorrect. You get 1 proc per fight on a 2 minute fight in the specific example I gave. On average that means you will get a proc halfway through the fight. That is an average of 9-10 ticks plus the DD. If you proc the weapon again during that time, you just get another DD proc (more damage), and the DoT continues to tick damage normally, as if the second proc never occured.

If the DoT is doing an average of 9-10 ticks, that is 10 x 24 (Damage per tick) + 40 (Direct Damage) = ~2.1 DPS per fight on average. This is not taking into account white damage, and the proc rate is actually 0.65, which means you are probably getting another tick or two on average.

Remember we are looking the the average over all possible fights, not an individual fight where you could get 0 procs, 10 procs, etc. Your average procs per minute is still 0.65.