knaughty wrote:For a 6 minute "stand and tank" fight, given my assumptions:
• What's the chance the buff doesn't fall off?
• 10 minutes?
• Full 264 gear, assuming same itemisation proportions as T-9?
Well, the basics are pretty easy:
For 45% avoidance, you have 55% chance to be hit and a 60% chance that those hits proc the trinket. Thus, the probability of a proc is
q = 0.55*0.6 = 0.33
The probability of not proccing is p=1-q, or
p = 0.67
Yes I know I'm using p and q in an odd fashion here, we'll see later why.
The probability of r attacks in a row not proccing the trinket is p^r. For this, we're interested in r=4, because following a successful proc there are 4 chances for it to be refreshed before it expires. Thus, the probability of 4 non-proc attacks in a row is
p^4 = 0.67^4 = 0.2015
Note that this is also the downtime of the proc, because this is (chance of no proc)^(number of chances to refresh). The uptime is then simply
uptime = 1 - downtime = 1 - p^4 = 0.7985
In other words, we only expect about 80% uptime on the proc, making it good, but probably not reliable enough to count as effective health. Also note that this is only 80% uptime of the buff, it says nothing of how many
stacks of the buff, which means that it's average worth isn't simply 0.8*10*27. More on that later.
Calculating the chance for the buff to fall off is... a pain in the ass basically. While calculating uptime and non-overlapping proc bonuses (like ICD trinkets) are easy, calculating the probability that you have at least one string of k consecutive successes or failures (aka a "run") out of N trials is particularly difficult. There's a brief description of the topic
here:
Dr. Math wrote:We must be careful in defining our terms here. If we want a success
run of length r at the nth trial, then a further success at the
(n+1)th trial could undo the run completed at the nth trial. Equally,
if we stipulate AT LEAST r successes, then any run can be extended
indefinitely and a run does not reestablish the initial situation.
Either of the above definitions makes analysis very messy.
I won't go through the details of the math (and neither does he really). But he does provide an approximation we can use:
Dr. Math wrote:With the above definitions and a lot of hand-waving explanation, the
probability of no run of length r in n trials is approximately
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(1 - px) 1
q(n) = --------- . -------
(r+1-rx)q x^(n+1)
where x = 1 + q.p^r + (r+1)(q.p^r)^2 + ....
Note that in our caes, "no run of length r" means no cases where we have r attacks in a row that don't proc the trinket. It's now clear why I chose p and q as I did - we can plug them directly into this formula, and q(n) will represent the chance of having the buff up the full time. Conversely, 1-q(n) is the probability that the buff will fall off at some point during a fight of length n.
For p=0.67, q=0.33, and r=4, we get:
x = 1.0886 (out to two terms)
for a N minute fight, n=floor(N*60/2.4).
30 seconds -> n=12
1 minute -> n=25
3 minutes -> n=75
6 minutes -> n=150
10 minutes -> n=250
q(12) = 0.3895
q(25) = 0.1150
q(75) = 0.0011
q(150) = 9.2E-7
q(250) = 7.7E-9
In other words, it's almost guaranteed that the buff will drop off for a fight of any appreciable length. For anything past 3 minutes, we have a 99% chance that the buff will drop off at some point during the fight.
It's likely to be worse than that as well - this is for the best case scenario where you're being attacked full-time. In most encounters, there are periods where the tank isn't being directly attacked (tank swaps, phase changes, submerge phases, etc) which guarantee the proc will drop and need to be refreshed. Since it will take
at best 24 seconds (for a 2.4 swing boss) to build up the stack from nothing, that will count for a lot of downtime.
In fact, since each attack is an independent event, they follow binomial statistics and we can easily calculate ramp-up time. The binomial probability distribution function
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binopdf(k,n,p) = nchoosek(n,k) * p^k * (1-p)^(n-k)
tells us the probability of getting k successes in n attempts with probability of success p. Reversing our old notation a bit, the probability of a success in this case is a trinket proc, which has probability p=0.33. If we calculate binopdf(10,n,0.33), this gives us the probability distribution for the number of attacks n we need to take to get 10 procs. The mean of this is simply
sum(n.*binopdf(10,n,0.33))/sum(binopdf(10,n,0.33))
Evaluating this for n=0:100 gives me a mean of 32.33. That means it takes on average 32.33 attacks to build the 10-stack, which is 32.33*2.4=77.6 seconds for the slow-swinging boss.
So the trinket takes over a minute to charge up to full strength, and has a 20% chance of falling off, forcing you to rebuild the stack from scratch (taking another minute). In other words, this trinket is crap.
Calculating the effective value of this trinket would also be a bit of a pain to do analytically. It's just complicated enough that it's easier for me to write a quick script to simulate it. I'm starting work on that now, once I have some results I'll post them.