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# Black Widow 3% Evade? Cool Story Bro

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## Comments

7,196★★★★★Thanks for replying. Just amused that you use the sun as an example. I was thinking if the sun rises ... wait the Sun is in a 'fixed' position?

Anyway, my view (or understanding) is that over large data, the 'averages' is x%. So, if there is 'inconsistent' occurrences, then something else is in the game mechanic (a bug?)? Is this correct?

17,603GuardianGeneral relativity says the laws of physics honor no preferred frame of reference in the universe.

That depends on your definition of "inconsistent." Almost by definition, "randomness" implies inconsistency. For example, if you're observing something like a coin flip that is supposed to have a 50% chance to land heads or tails, then the sequence H T H T H T H T H T H T H T H T H T H T H T H T has a 50% distribution, but is completely non-random. A genuine random generator will tend to generate almost any sequence with equal likelihood relative to other sequences of equal length. That means, counter-intuitively for most people, that H H H H T T T T happens just as often as H H T H T H T T. The second "looks more random" but it isn't. It is just that H H T H T H T T looks to human eyeballs similar to H T H H T H T T so we tend to lump those sequences all together, and because there are a lot of them they collectively come up more often.

This isn't strictly speaking correct, but for our purposes, randomness implies unpredictability. If we saw that every single set of 33 attacks against Black Widow included an evade, that level of consistently would actually be proof that the game wasn't properly triggering BW's evade randomly, and had been rigged to generate that rate of evade, using something like a streak breaker or a quota bucket. Genuinely random means sometimes you'll see a lot less evades than the averages predict, and sometimes you'll see a lot more. But the distribution of how often you see these deviations is itself something that is statistically predictable.

527★★17,603GuardianThe statistical margin for error for such a test would be about the square root of the expected result. In other words, to have a reasonable confidence level (statistical measurements are never "certain") that you were roughly within 10% of the correct value, you'd need to perform a test where the expected result was about 100 evades (which would then have an estimated margin for error of about 10), which would require a test of about 3000 attacks.

357★★When BW evades once every 5 hits and she’s got only a 3% chance 🙄

44★91★286★★1,958★★★★★Don't mind me.

1,696★★★★Edit: just realised this was posted almost 2 years ago my bad

322★★901★★★369★★★493★★★My mind was thinking if it's 3% out of 100% then that means there should be a 97% chance she does not evade, but that's not how any of this works?

2,421★★★★369★★★369★★★1,958★★★★★Imagine that you toss that same coin 10 times. How many times would you expect it to land on heads? You might say, 50% of the time, or half of the 10 times. So you would expect it to land on heads 5 times.

This is the theoretical probability.The

theoretical probabilityis what you expect to happen, but it isn't always what actually happens.Let's say you toss a coin 10 times and get following data:

HEADS: 7

TAILS: 3

The

experimental probabilityof landing on heads in this case is 70%It actually landed on heads more times than we expected.

Now, we continue to toss the same coin for 50 total tosses.

HEADS: 27

TAILS: 23

The

experimental probabilityof landing on heads is 54%The probability is still slightly higher than expected, but as more trials were conducted, the

experimental probabilitybecame closer to thetheoretical probability.So, theoretical probability is what we expect to happen, where experimental probability is what actually happens when we try it out. As more trials are conducted, the experimental probability generally gets closer to the theoretical probability.

369★★★2,968★★★★369★★★8,005Administrator∙Moderator ›