Which movies have the best fight scenes?

It's an impossible question, we don't know when the free crit is applied.

Boy or girl paradox of conditional probability. It's both 50% and 33% depending on how you interpret the question.

The Raid 1 and 2

Ones with close-range weaponry and few combatants. There has literally never been a good gunfight scene with many fighters.

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This.

wtf is a crit?

ITT people posting the % if whiteness in every American LOL

It can also be 33% though, it's one of those stupid paradoxical problems.

If you look it from a statistical perspective, it's 33%. There are 3 equally likely outcomes. If 0 and 1 represent non crit and crit respectively, you have 00,01,10,11. Since you know you're getting at least 1 crit, you rule out 00, leaving 3 equally likely outcomes

It's what you can't find when going down on a girl.

have sex

100%. Has some of the best choreographed fights ever committed to film.

Also a director that understands fight scenes simply need well choreographed fighting and wide long takes, not 50 jump cuts every few seconds.

Yeah its just bait for high school math dropouts to shit out their poor understanding of probability to try and seem intelligent

It's actually just a stupid paradoxical problem.

en.wikipedia.org/wiki/Boy_or_Girl_paradox

This has multiple answers and is one of many trappings the people who write questions for tests fall into.

But if one is guaranteed a crit, then the chances of both criting is just dependent on the non-guaranteed one, which is 50%.

but there are three possible outcomes if there is an auto crit, it's a stupid conditional probability fake question see:

They Live

Which one is the “non guaranteed one”? It’s unknown. If the question was that the first was a crit, then it’d be 50%. We only know that one was, not which one.

It doesn't matter, because it's not asking for a specific order.

50%

25%

2 hits, 50% chance of crit. 2 x 2 = 4.

100% / 4 = 25%

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it does matter though, let 1= crit and 0= non crit

possibilities: 00, 01, 10, 11

possibilities in frame: 01, 10, 11

11 is double crits so it's 1/3 if you view it in this frame. Just read the fucking wikipedia link the user shared.

No. It's a trick question, you don't need to even do math just read carefully

You're overthinking it. You can have one crit, no crit, or two, but one is given so its's just the 50% on the second attack, it's not asking for specific outcomes like first hit crit

It does not matter, since its already given and the events are independent of eachother

Ong Bak 2 [2008]

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And you are under-thinking it and haven't taken a stat course. It's a loaded question and both 33 and 50 are correct depending on interpretation.

100% you fucking lucklet

Its only a parafox if you are retarded, its 50%

Okay, retard.

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>going luck
Hows those re-rolls and retries doing for ya?

The problem is that the question is ambiguous.

If you were asked the probability before knowing that one got was critical, you would answer 25%. If you were asked after, you would answer 33%. Different people make different assumptions about how the problem is framed, leading to disagreement.

A universal narrator tells us in the present tense that something happens, then tells us the probability of the event happening. What is the probability of the event happening? Is it 100%, because the universal narrator strictly told us that must happens in this hypothetical scenario, or is it whatever probability the event normally has, ignoring the present tense narration claiming that the outcome of this event is already set? Only YOU can answer this question. Everybody else is wrong. It is very important that you argue that this is purely a math problem and not a problem that pertains to english language comprehension in the face of unclear syntax.

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And the interpretation is that the order doesn't matter since it didn't ask for order.

50%

the Warriors

75%

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that one with the pistol kungfu, equilibrium
fite me

>At least one of the hits is a crit
This is the sentence so many posters in this thread are ignoring.
>Assuming a 50% crit chance
Given that "at least one of the hits" is going to be a crit, that means that this 50% chance only applies to the other "hit".

So 50% is the answer

It doesn't specify which one retard

>tfw you have a 50% crit chance
>char gives 10 straight non-crit hits.

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75% chance. At least one hit is confirmed a crit (100%) which increases your chances for both hits to be crits, so what would've been only a 25% chance for both without the 100% confirm, the chance now increases to 75% chance that both hits are crits.

Possible outcomes are
{crit, not crit}
{crit, crit}
{not crit, not crit} (not possible, we're told at least one was a crit).
{not crit, crit}
Therefore we are looking for one of the three possible outcomes, so the probability is 1/3.

>the next attack is a guaranteed critical hit
>misses

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50%

There are literally only 2 outcomes.

2/2 Crits or 1/2 Crits

excellent bait

50%

and?

MILK
TRUCK

I drink my luck potion so 100%

>{crit, crit}

you need to take that into consideration twice, not once

Because at least one of the hits is a crit then the possibilities are:

Crit Normal, Normal Crit, Crit Crit.

So 33%

*honk, honk*

It's 25%

There are FOUR possibilites and CRIT, CRIT happences ONCE
Ergo, it's 1/4 or 25/100 or 25% is percentage form

nope

01 and 10 are the same though, the one crit is guaranteed and the order doesn't matter. So it is just the one hit differing between them, at that hit has 50% chance

This, with the one guarantee crit the odds change

There're 4 possible outcomes:
Crit Crit - 1/4
Crit Hit - 1/4
Hit Crit - 1/4
Hit Hit - 1/4

P(A|B) = P(B|A) * P(A) / P(B)
P(A|B) is the probability of event A, given that event B occurs
P(B|A) is the probability of event B, given that event A occurs
Assume P(A)= two crits = 1/4
P(B) = at least one crits = 3/4
P(B|A) = you get at least one crit = 1

Therefore P(A/B) = 1/3

wtf is a crit?

All of you are retarded. It’s 33.33 percent, repeating of course.

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NN
NC
CC
CN

Literally that fucking simple, you uneducated third worlders.

50%

The events, first hit and second hit, are independent. That mean that the probability of the first strike being a crit and the second being a crit is independent of whether previous hits were crits or not. Therefore, assuming one of the hits was crit, the probability that the other hit is critical is 50%

I'd tell the people saying 50% to write a program and prove themselves wrong but they're not smart enough to

There are four possibilities

Guaranteed crit -> non crit
Guaranteed crit -> crit
crit -> Guaranteed crit
non crit -> Guaranteed crit

50% chance of 2 crits

how is it 33% you dicksuckers

In terms of logical Induction this question leads to oscillating probabilities.

/sci/ here.
the answer is trivially 50% applying the gaussian-wildberger distribution and integrating thrice

Brainlets. One hit is a guarenteed crit. It's not asking if the first or second one is the guaranteed crit. It's asking if both hits will crit and one already has so it can be removed from the equation which leaves us one variable, will the other 50% hit crit. There is a 50% chance both hits will crit.

Since one of the crits is guaranteed, only the second one is left to chance, and the second one has a 50% chance of being a crit.
Therefore, 50%.

/sci/ here. two chances at 50% is mathematically the same thing as one chance at 100%. so the anwser is 100%

this only works in finite vector spaces

tryhaaaaaaaaaaaaaaaawrd

and you're wroooooooooooooong

shut the fuck up and go back in your hole!

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There are 3 outcomes.

Crit, normal
Normal, crit
Crit,crit

Answer is 33%

why dont you come over here and make me, you fucking pleb

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read that again
the first hit was a crit, the order matters.