020 Chameleon [100] v1.1.2

[What]

Like I said, I am fairly sure something strange is going on with the RNG at the moment. Long strings of evade and whatnot should only happen once a blue moon.

Chance for night elf to evade 1 shot: 30%
Chance to evade twice in a row: 9%
3 times: 2.7%
4 times: .81%
5 times: .24%
10 times: .00059%

I know that strings of evades at least this long happen with fairly common frequency. I'll test this out whenever I get a chance.

You have the right idea, but thats not how the percentages work

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You're conflating probability with the percentage chance

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first shot-30% chance to evade
second- 30% chance to evade
third- 30% chance to evade
etc.

If you want to look at the odds of this happening, that is fine, but the chance to evade does not decrease per-successive shot

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ie your fourth shot in a row still has a 30% chance of being evaded, but the odds or probability that this occurs is .81%

[CYBER]

You have the right idea, but thats not how the percentages work

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You're conflating probability with the percentage chance

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first shot-30% chance to evade
second- 30% chance to evade
third- 30% chance to evade
etc.

If you want to look at the odds of this happening, that is fine, but the chance to evade does not decrease per-successive shot

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ie your fourth shot in a row still has a 30% chance of being evaded, but the odds or probability that this occurs is .81%

U'RE BOTH right ... and wrong.

the CHANCE or PROBABILITY of each of those evades REMAINS a 30% EACH time...because each event IS "independant" from the other.

however, what pass is trying to convey is that the chain of evades can be represented by:
probability of the second evade to occur AND the first evade occured.
which roughly translates to P(A AND B) = P(A). P(B)

so the probability of the first independent evade to proc is 30%.
the probability of the SECOND independent evade to proc is 30%.
However, the probability of BOTH the first AND the second back to back is 30% x 30%= 9% ..

so when u have a guy evading 10 shots in a row, it's true that the probabilty of each evade was 30%, however the probability of that CHAIN of evades is determined by the multiplication of all the other events.

you can also apply the bernouli theorem, more specifically using the geometric distribution and call "evaded" as "fail" and then "not evaded" as "success"... (in the enemy's point of view),

so if we apply the geomtric distribution

q^(k-1) x p

where q is the probability of evading (30/100)
p the probability of NOT EVADING (70/100).
and let's just assume that we're looking at the "5th" back to back evade.

if we apply the formula, after 5 back to back evades, the probability of NOT evading (finally) is:
((0.3)^(5-1)) x 0.7 = 0.00567


... and that is PRETTY significantly low...

so it SHOULDNT be that low, so it SHOULDNT be proccing 5 times back to back, and the fact remains that for some reason, a lot of people are able to proc these evades back to back, as if it's nothing, when in fact the RNG might be messed up...


PS: i really hate probability and statistics, but i just googled all that shit to refresh my memory.

[Passarelli]

U'RE BOTH right ... and wrong.

the CHANCE or PROBABILITY of each of those evades REMAINS a 30% EACH time...because each event IS "independant" from the other.

however, what pass is trying to convey is that the chain of evades can be represented by:
probability of the second evade to occur AND the first evade occured.
which roughly translates to P(A AND B) = P(A). P(B)

so the probability of the first independent evade to proc is 30%.
the probability of the SECOND independent evade to proc is 30%.
However, the probability of BOTH the first AND the second back to back is 30% x 30%= 9% ..

so when u have a guy evading 10 shots in a row, it's true that the probabilty of each evade was 30%, however the probability of that CHAIN of evades is determined by the multiplication of all the other events.

you can also apply the bernouli theorem, more specifically using the geometric distribution and call "evaded" as "fail" and then "not evaded" as "success"... (in the enemy's point of view),

so if we apply the geomtric distribution

q^(k-1) x p

where q is the probability of evading (30/100)
p the probability of NOT EVADING (70/100).
and let's just assume that we're looking at the "5th" back to back evade.

if we apply the formula, after 5 back to back evades, the probability of NOT evading (finally) is:
((0.3)^(5-1)) x 0.7 = 0.00567


... and that is PRETTY significantly low...

so it SHOULDNT be that low, so it SHOULDNT be proccing 5 times back to back, and the fact remains that for some reason, a lot of people are able to proc these evades back to back, as if it's nothing, when in fact the RNG might be messed up...


PS: i really hate probability and statistics, but i just googled all that shit to refresh my memory.

Thank you for spelling it out Cyber.

What, based on my math, it should have been obvious I was calculating the probability of that string occurring from the first shot. Not to mention I've spelled this out in the last couple days myself.
All of my math was simple exponents, because each event had the same chance of evading. The formula for this is f(p) = p^k

If you flip a coin 100 times, it has ~ a 50% chance of falling heads up each time.
The odds of you flipping heads 100 times in a row would be .5^100 which is 7.89 x e-31

Cyber's method is a little different then mine, but both say essentially the same thing--something is going on with the RNG.