Iksar Shaman - Solo Challenges?

[bcbrown]

At 170 Dex you get 1.5 PPM. At 47 delay, that means a swing every 4.7 seconds, or 12.76 swings/minute. At 1.5 PPM, that's 1.5 procs per 12.76 swings, or a proc rate of 11.75% for each swing. I know very little about both weapon delay and proc calculations, so please correct me if I'm wrong here.

JBB has an 8 second cast, and with 132 seconds per fight, that's 16 casts. Let's assume you take a swing before first cast, and lets say you have excellent reflexes, and click JBB almost simultaneously with swinging. Note I see you've fiddled some more with the numbers since I did this calculation; I can update it if you'd like.

Let t=0 be when the mob has been pulled, slowed, and first engaged in melee. A proc here will last 132 seconds, or 22 ticks. A proc here will do 40 + 24 * 22 or 568 damage.
At t=8, second chance to proc. There's (132 - 8) / 6 = 20 (rounded down) ticks left. If there's been no prior proc, a proc is worth 40 + 24 * 20 = 520. If there's a prior proc, only the DD component counts: 40 damage.
So we can build a table:

Now, what are the probabilities of each possible outcome?
At t=0, it's simple: there's a 11.75% chance of a proc.
At t=8, the chance of a proc is 11.75%. There's a 11.75% chance there was a proc at t=0, and an 88.25% chance there was no prior proc.
At t=16, we get to the real meat of the problem. To calculate the chance of a prior proc, there's an 11.75% chance of a proc at t=0, and a 11.75% chance of a proc at t=8, but these are independent possibilities.

1) The chance of a proc at 0 and 8 is 11.75 * 11.75.
2) The chance of a proc at 0 and not 8 is 11.75 * 88.25.
3) The chance of no proc at 0 and a proc at 8 is 88.25 * 11.75
4) The chance of no proc at 0 and no proc at 8 is 88.25 * 88.25

This is a classic binomial distribution. You can also think of this as flipping a coin every 8 seconds, where the chance of heads is 11.75% and the chance of tails is 88.25%.

The relevant formula is the probability mass function: https://en.wikipedia.org/wiki/Binomi...on#Definitions

The vertical (n k) notation is spoken as "n choose k" or sometimes written as nCk and described in more detail here: https://en.wikipedia.org/wiki/Binomial_coefficient

To apply this to our problem, we can use this formula to calculate the probability of no prior proc, i.e. setting k=0

(n choose 0) * p^0 * (1-p)^(n-0)

(n choose 0) is always 1, so this simplifies to (1 - p)^n

So at each 8 second interval (the variable n), the damage of a proc will be
44 * (1 - (1 - p)^n) + (44 + 24 * (132 - 8*n) / 6) * (1 - p)^n

p is the chance of a proc, or 0.1175

Over a 132 second fight, there's 16 casts, so we need to sum this over n = 0 to 15 inclusive. Let's implement this in Python; perhaps that will be easier to follow:

p = 0.1175

def no_prior(n, fight_length = 132):
    """" 44 dd plus 24 per tick remaining in the fight """
    from math import floor
    ticks_remaining = floor((fight_length - 8 * n) / 6)
    return 44 + 24 * ticks_remaining

def prior(n):
    """" if there's been a prior proc, only the dd matters """
    return 44

expected_damage = 0
for n in range(0, 16):
    prior_probability = pow(1 - p, n)
    expected_damage_this_interval = prior(n) * prior_probability + no_prior(n) * (1 - prior_probability)
    expected_damage += expected_damage_this_interval

print(expected_damage)

[Duik]

I am right. EVERYBODY ELSE IS WRONG.
Especially at math, but especially at math.

[DeathsSilkyMist]

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

This is a classic binomial distribution. You can also think of this as flipping a coin every 8 seconds, where the chance of heads is 11.75% and the chance of tails is 88.25%.

Exactly your problem. You are using a single Binomial distribution in an equation where all other variables are not binomial distributions. It's an apples to oranges comparison. You are plugging in an incorrect variable into the equation.

The reason why people can use "PPM" as a metric is because when you have 1 PPM, you get 1 proc per minute on average. It's really that simple. You wouldn't be able to say that if the average never came out to be 1 proc per minute.

[bcbrown]

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

Exactly your problem. You are using a single Binomial distribution in an equation where all other variables are not binomial distributions. It's an apples to oranges comparison.

That... that makes no sense. I have no idea how to interpret this. Do you know what a probability distribution is?

Do we need to go through this more slowly?

[Troxx]

The assumption that this mob will do 10dps to you swinging a 2handed weapon instead of wearing a shield is also a farce. You will take more damage without the shield ac. You will also have to deal with some degree of ripostes.

I bet the JBB shaman would likely fair better by just equipping a shield and not trying to land 2hb procs to begin with. However much damage you think you’ll be taking … you’ll be taking more if you equip that hammer.

[Duik]

This is great. I dont actually know the answer. But I have clues.

[Troxx]

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

Do we need to go through this more slowly?

Yes. Pretend you’re talking to a junior high school kid.

[DeathsSilkyMist]

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

That... that makes no sense. I have no idea how to interpret this. Do you know what a probability distribution is?

Do we need to go through this more slowly?

I know you don't understand it. That is why you keep using averages incorrectly.

Let me go a bit more slowly for you.

In our example, you basically get 6 swings per minute with a 1/12 chance of proccing. It is like rolling a D12 six times in a row, hoping to get a specific number. Let's say a "success" is rolling twelve. Since the fight lasts 2 minutes, that means we get twelve chances to roll a d12.

We are not looking for a set of numbers to roll in a sequence (what are the odds of rolling a 6 and then a 3). This means each dice roll has a 1/12 chance of succeeding equally.

Worst case is you roll the twelve after 12 dice rolls. Best case is you roll the twelve on the first attempt. 1 roll + 12 rolls / 2 = 6.5 rolls on average to get your number. So on average you will proc it roughly half way through the fight.

[Toxigen]

jesus

titty

fkin

christ

[Troxx]

Look man he already spelled it out for you very comprehensively:

https://www.project1999.com/forums/s...&postcount=531

It is absolutely ok to admit it went over your head. I imagine that level of math would go over most college educated adults heads. There is no shame in that. I’ve got a doctorate and it have me a slight headache trying to follow him. But he is correct.

You aren’t going to win a math fight with a senior engineer bro