Math equation for variance
 

So I had this random thought the other day while I cam camping Verina Tomb. That specific mob spawns between a random 2-7 day variance. I started thinking if you could mathematically calculate the probability for any mob to spawn within their window but I couldn't create the formula on my own. So here's a thought experiment :)

Lets say a given mob's variance is 16 hours for the pop to occur.
Evenly distributed each hour in the window comes out to a 6.25% chance each hour that the mob could pop.

It is known that if the mob does not spawn by the 15th hour, there is a 100% chance the mob will spawn in the 16th hour...but how would you calculate each change mathematically as the hours progress?

For example, A mob's window is open and 3 hours pass but no pop or event has occurred. I am guessing you just take the remaining hours left and divide by 1. So in this example, on the 3rd hour of the window, the next hour becomes 7.69% , 8.33% on the 4th hour, etc.

Is my math behind this accurate? There is no way you can predict the random variable of when the mob occurs, but this might be a safe bet. Any opinions?

I dont know anything about the raid scene here, but would have to start with whether or not "Variance" is time before and after a previous kill for mob to pop..

Say its 10 hours..is that a 20 hour window? 10 before previous kill to 10 after?

When I think of variance I usually refer to the end result of a +-8 hours, which comes out to 16 hours. Most raid mobs are on a 7 day +-8 hour variance, which comes out to a 16 hour window.

I would think the only variable that changes is 100% during the last hour if the mob hasnt popped during the hours before...since each hour is its own variable with a yes/no not a % based on number of hours left..

That's what I was thinking too, I feel like something is missing in the calculation but I couldn't put my finger on it...AND were doing it by diving by hours of the window, you could get really into it once you start considering the minutes and quartiles of each hour for 15 minute increments.

is there a certain "tick" the mob can spawn in or is it on the hour during each hour in the variance window? if thats the case the mob could spawn in any 6 second interval during a "tick" right?

I think the spawn is based on in-game /time, but if it were based on server-side ticks, then the working variable grows exponentially:

16 hours = 57,600 seconds.

Converted 57,600 seconds into increments of 6 seconds per chance event = 9,600 instances a mob can pop within a 16 hour window.

If you take it to the extreme "per tick" limit, and work the opposite way, like cheating on a maze...

With one tick left on a mobs 16 hour window, there is a 100% chance of it spawning on that final tick, if it hasn't spawned yet.

With two ticks left, the mob has a 50% chance to spawn on the second to last tick, and a 50% chance to spawn on the last tick.

With three ticks left, the mob has a 33% chance to spawn on each of the ticks...

With four ticks left... 25%

With five ticks left... 20%

I think this is what you might be getting at? If this is the case, with only a minute left on a 16 hour window, you'd still only be at a 10% chance of the mob spawning on that tick. Any individual tick more than a minute out would be very unlikely to spawn the dragon. Practically, it makes more sense to start camping something closer to the end of its window, but of course then you run the risk of missing an earlier pop. Where that line in the sand is drawn most likely will come down to how important of a target it is, and how much time you're willing to waste in an elf sim :P Sounds like a fun math problem, if you could put some weight on those last two variables.

I'm pretty sure when the mob dies there's just a rng to set the exact value of the countdown timer until it repops. GMs can see the value of this timer.
E.g. Repop_timer = 7 days - 8 hours + ran(0 9600)*6 sec

Every time point within the window is equally probable.

When possible spawn times within the window have passed, the remaining times are still all equally probable, just with the time that has passed excluded (you could say their probability is redistributed among all remaining times).

1.0 / (TotalTimePointsInWindow - TimePointsThatHavePassed) = ProbabilityPerRemainingTimePoint

No real advantage to looking at it this way.