WELL??

>25%
Maybe if you don't read
>Atleast 1 of the hits is a crit
Means that miss/miss is impossible. There are 3 possibilities hit/miss, miss/hit, hit/hit

33%

Doesn't matter. First crit 50%, second crit 50%. First hit has to crit or we can stop there. If it crits, we need another one. Every other possibillity is irrelevant to answer the question.

filtered

50% x 50% = 25%

as long as the chances are not 100% or 0% , the probability is always 50/50
this is just plain ol common sense
i have played enouogh Xcom

looks like everyone ITT is stochasticlets except OP but he is so shit at math he doesnt just do a proof, he rather uses empirics

en.wikipedia.org/wiki/Boy_or_Girl_paradox

however if you apply the power of faith , aka the law of assumption and
you believe you will hit it, you will hit it
if you can visualize it happening and feel its reality, it will happen , even if there is a 0% chance of it happening

It is 50%, because in every case you have to exclude one combination. You can not have both Hit-miss and miss-hit at the same time.

P(A | B) = P(A and B) / P(B)
Let C be the number of criticals landed
P(C = 2 | C > 0)
= P(C = 2 and C > 0) / P (C > 0)
= P(C = 2) / P(C > 0)
= P(C = 2) / ( 1 - P(C = 0))
= (0.5 * 0.5) / ( 1 - (0.25 * 0.25))
= 0.25 / 0.75
= 1 / 3

Oops, I meant 0.5 * 0.5, not 0.25* 0.25

dumb codefrog
intelligent combinatorics man

Wait, isn't she lying about the crit chance in this case? If the first crit misses, then the second will crit 100% of the time

This is the answer. You are all overthinking it.

Attached: ni.png (1181x657, 24.65K)

/thread

>it's an "i don't understand conditional probability" episode

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Fpbp

kill yourself

This is wrong. You are not given any information on What crit you are shown.
The person may only show you the second crit or the first crit consistently

>purposefully wording a mathematical question in a way where there is more than one correct answer
I fucking hate these Facebook-tier puzzles, they're such a waste of time and prey on people that are so conceited that they absolutely can't help themselves and MUST show other people how "smart" they are.

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It's 0%, obviously. If both hits are crits then the weapon doesn't have 50% crit chance and the whole question is invalidated. Suck it, math and code nerds!

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Glad someone sees it.

50%, you either get it or you don't :^)

>Using fancy syntax to still be wrong

With just the premise of hitting the enemy twice and having a crit chance, there are four possibilities. (C = crit, N = normal hit)

N,N
N,C
C,N
C,C

The question assumes at least one of the hits is a crit, so that limits us to three possibilities.

N,C
C,N
C,C

Now, if you're extremely Goddamn stupid, you might think "Hey, only one of those three possibilities is a double-crit, so the answer is one-third, or 33%!". You're the kind of dipshit moron who probably gets the Monty Hall problem wrong too. The question is NOT asking "Out of all the possible combinations where at least one hit is a crit, what portion of them involve two crits?", the question is asking, "If you already know that one of the hits is a crit, what is the probability that that the other hit is also a crit?". In other words, if you're in EITHER of these scenarios (U = Unknown), what are the chances that the U is a C instead of an N?

U,C
C,U

We already know that. In either case, we have a 50% crit chance, so we have a 50% chance of either scenario being a C,C (instead of N,C or C,N depending on which scenario we started with). For the answer to be 33%, we would need to be able to go from U,C to C,N (or from C,U to N,C) which isn't how the problem works.

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>Python
kill yourself, use C or C++ like a normal human

It is testament to how poorly people understand probability and conditional statements that every day we have threads like this. Not just this thread, but all of the near identical meme threads that have been around for over a decade.

Furthermore, people read these threads and still get confused. It plays out in every given situation too, whether choose your sword/buff/whatever threads that have some probability aspect, memes like this, or even stat discussion.

Not just Yea Forums either, it plays out here and there across boards.

Unironically this each event is independent

Use R or SAS you retarded mongoloid

0.75?
n=2
p=0.5
lower limit:1
upper limit:2

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Anons that say 50% assume you have one guaranteed (100%) crit. But this is not how it works, at least I don't understand it this way. The first hit has a 50% crit chance, and has to crit, and the second hit has again a 50% crit chance and has to crit again, giving you a 25% chance of two crits in a row.
I'm willing to accept 50% as an answer, because you could interpret the question this way. But I don't understand 33% fags.

>every four pair of attacks is a crit
no wonder fire emblem is so fucking easy

0 because in video games anything below a 99% chance to crit will never crit, and at 99% it has a 1% chance.

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What the fuck bros.

You attack two times.

One attack is always a crit.
I repeat, one attack is ALWAYS a crit.

So only the other attack to think about.

What's the chance the other attack also crits?

The dude literally says it. Scroll up.

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Bayes

>"people" still taking the "at least one has to be crit" bait

Here are your outcomes
Crit/no crit
No crit/crit
Crit/crit

Crit/crit makes up 1 of 3 outcomes, hence 1/3

Here are your outcomes
crit/crit
everything else is irrelevant

Eh, it's alright. If everybody was smart there would be nobody to flip my burgers.

Why did they put a computer in the retirement home? No one wants you online.

what about no crit/no crit

my mistake i think i calced the chance for no crit.

Ruled out bc states at least one crit.

Analogy is just coin flip. Fair coin, flip twice, at least one head. What's the chance of both heads?
TT ruled out
HT
TH
HH

HH is 1/3

Why are people saying the only options are
>Hit
>Crit
>No Crit
What about parries, dodges, resists? What happens if you crit but do 0 damage the same as a hit that does 0 damage, does that count as a crit? There simply isn't enough information to answer this question.

This would be correct if the questions was about the number of possibilities, but I don't see how you could interpret the question this way.

if you think like that in math class i don't want to know your grade. just follow what's written. it's never stated that your hit can be blocked or misses.

"At least one crit" implies three possible outcomes because it doesn't specify further:
Crit/no crit
No crit/crit
Crit/crit

Think of it like this. If the question was ok you landed one crit, crits are 50%, whats the likelihood of landing another crit? That would obviously be 50% assuming crits are independent.

Or if the question was what is the likelihood of landing two crits in a row? That would be 25%.

But the "at least one crit" yields three possible outcomes so its 1/3

It's not syntax that guys just using le generic game theory equation
So people are just arguing over which equation is correct

Confusing thread. Isn't every roll a 50% chance to crit and there's only one roll so its 50%

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This is not looking at percentages however, in the regular probability it would be
C/C 25%
C/M 25%
M/C 25%
M/M 25%
However, the conditions change M/M to 0% so many itt are saying the new chances become
C/C 33%
C/M 33%
M/C 33%
When it would actually become
C/C 25%
C/M 25%
M/C 50%
Because the second hit being a crit if the first is a miss is now 100%

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>le generic game theory equation
god damn I always forget how fucking stupid the people here are

strawpoll.com/polls/mpnbaxAlGy5

>people are reading this as a mathematical problem and not a word problem