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KeyboardNinja
 11.02.2012 , 11:47 PM | #22 Quote Quote: Originally Posted by grallmate Take your time. I spent a while on it and as I said, I'm not convinced myself. Its much much more complicated and I think may require running a recursive algorithm to properly solve or perhaps just some rather advanced integral calculus. However I am certain that discreet groups of attacks do not provide an accurate answer. I've had time to think about it. :-) My original expression was wrong, but it was a closer estimate than your refinement. Here's the logic... Let x be the probability of a single event, where an "event" is a boss attack that slips past mitigation. What we're interested in is the following situation: P((x_1 \cap x_2 \cap x_3 \cap \ldots \cap x_10) \cup (x_2 \cap x_2 \cap x_3 \cap \ldots \cap x_11) \cup \ldots) That is to say, the probability of events 1-10, OR events 2-11, OR 3-12, etc, all the way up to 185-195. There is an immediate law of probability which can be applied here: P(A \cup B) = P(A) + P(B) - P(A \cap B) Thus, the probability of A or B is equal to the sum of the probabilities of A and B and the negation of the probability of A and B. For the first two strings, we can calculate this in a very straightforward manner: x^10 + x^10 - x^11 Thus, we have the probability of 10 consecutive mitigation fails, plus the probability of another 10 consecutive fails, minus the probability that all 11 were fails. Unfortunately, this sort of expression spirals out of control in a hurry. Getting a closed form is somewhat...involved. With a bit of algebraic munging in the first case, we can derive the following: x^10 + x^10 - x^11 = x^10(1 - x) + x^10 This gives us an inductive case for a recursive function that encodes our problem: f(i) = f(i - 1)(1 - x) + x^10 f(0) = x^10 Plugging this into Mathematica, we churn out the following closed form: f(i) = x^9(1 - (1 - x)^(i + 1)) (this bears a hilarious and entirely coincidental resemblance to the fundamental stat-scaling formulae which govern diminishing returns in TOR) Coming back to our original problem... x = 0.4595. Our maximum i = 185, since there are only 185 overlapping 10 x 2 second periods in a 195 x 2 second fight. The rest is arithmetic: f(185) = 0.000913208 That is to say, 0.0913208% odds. That's almost a 10th of what I originally calculated. So, my original estimate was *high*. Computer Programmer. Theory Crafter. Dilettante on The Ebon Hawk. Tam (shadow tank) Tov-ren (commando healer) Aveo (retired sentinel) Nimri (ruffian scoundrel) Averith (marksman sniper) Alish (lightning sorcerer) Aresham (vengeance jugg) Effek (pyro pt) December 13, 2011 to January 30, 2017