Well, Yea Forums?

Possible outcomes
>Crit / Crit
>Crit / No Crit
>No Crit / Crit
1/3

Only 50% crit? 100% crit or go home.

>babby's first bayes

A = both are crits
B = at least one is a crit
P(A) = 0.5 * 0.5
P(B) = 0.5

P(A | B) = (P(B | A) * P(A)) / P(B)
P(A | B) = (1 * 0.5*0.5) / 0.5
P(A | B) = 0.5

These threads always make me feel dumb

Attached: 1564713247306.png (260x230, 105K)

This, now if the scenario didn't state you're guaranteed to crit at least once then it would be 1/4 of course for No Crit / No Crit. Well done.

its 50/50 since either it happens or it doesnt just like with everything.
There's a 50/50 chance I get married tomorrow

>That leaves two cases
>The probability is 1/3
you are fucking retarded

its either you crit or not so its 1/2 period

50%. one hit is a confirmed crit. all that is left is calculating the other hit

You tried too hard in your reply to the first user, 2/10 bait

Imagine thinking there's a point in differentiating between no crit/crit and crit/no crit as far as the question is concerned.
One of them crits, the order is irrelevant.

Reread You are retarded.
>two cases of one crit and one case of two crits.
Consider toaster bath

Attached: 1356827081589.jpg (379x353, 32K)

These threads always make me feel like everyone else is fucking retarded.

1/2 x 1 you tard

Well, if I'm playing Warframe it's about 0.1% unmodded.

crits are for faggots real men just get their atk as high as humanly possible and unga bunga the fuck out of crit dexfags in 1 (one) hit.

1/4

150 %

50%
It's either:
hit crit
Or
crit crit

It says one hit is guarateed to crit you fucking retard.
>at least one of the hits is a crit

Knowing that hit is confirmed to crit, the chances of the other crit is 50%.
There is a 50% chance that they will both crit.

You're all just pretending to be retarded right?

75%

Oh wait didnt read the first part. kek

"at least one of the hits is a crit" is insufficient information. How does this come into effect? If neither is a crit will it be retroactively applied? Is one of the hits randomly chosen to be a 100% crit? Without this knowledge, the question cannot be answered

1/3

50%?

It means either the first or second hit is a crit. It really doesn't matter which one crits because the answer is still the same.
With one hit guaranteed to crit, the other crit has a 50% chance of happening, meaning a 50% chance they will both be crits.

If the 100% guarantee is only applied after rolling no crits it's different than if it's chosen before hand. As written the question doesn't have an answer.

Well since there's only one variable here, and there's a 50% chance of that being a crit, there§'s a 50% chance of both critting.

12.5%

If both events are independent, it's p(A ∩ B) = p(A) · p(B)

p(Crit 1 ∩ Crit 2) = 1/2 · 1/2 = 1/4 or 25%

>ignoring possible outcomes

it's 1/3 retard

always switch doors

Ok so normally given two hits there's a 25% chance because 50% of cases will have the first hit be a crit and 50% of that 50% will have the second hit also crit so we get:

0.5 x 0.5 = 0.25 = 25%

However since we're given the information that one of the hits will always crit then we can make one of two assumptions that either;
1) We look at all cases where the first hit was a crit
2) We look at all cases where either hit was a crit

If the first scenario then the answer is 50% because we only need to consider the second hit's probability of critting.

If the second scenario then the answer is 33.33% (recurring) because we need to look at what amount of scenarios have at least one hit critting already, this results in 75% of cases remaining, 25% are 1C(Crit)-2N(Not Crit), 25% are 1N-2C and 25% are 1C-2C. So we have three amounts of 25% which when the total becomes 100% means we increase all of them by a factor of x1.33 giving us 33.33% or 1/3 probability of each occuring. Since the only one we're interesting in is 1C-2C then the answer in this scenario is 33.33%.

You're overthinking it way too much. You're trying to add information that isn't needed.

>at least one of the hits is a crit

I'm not, the answer to my question literally changes the outcome of the equation. You can't answer it as stated.

Distribute the remaining 75%

If you hit an enemy 45 times and crit chance is 30% what is the probability that 23 of the hits are crits

Attached: 1554324968717.jpg (990x990, 113K)

if you miss the first crit, that probability scenario gets thrown out from calculating the probably of both hits being crits, dumbass

It's not a possible outcome you goddamn fucking brainlet

AT
LEAST
ONE
OF
THE
HITS
IS
A
CRIT

YOU EITHER HAVE ONE CRIT AND ONE NOT CRIT
OR TWO CRITS

FIFTY
FUCKING
PERCENT

Hopefully suicide is a possible outcome for you, you should choose it.

Shit I glossed over that. So 50%, right?

The phrase "At least one of the hits is a crit." is too ambiguous and there are several possible implementations you can put in a game that fulfill this condition, but that have different probabilities for both hits to be critical.

>Crit / No Crit
>No Crit / Crit
>counting them as different outcomes

Its 50%, you're literally retarded if you say anything else

Attached: 4c9.png (645x729, 113K)

Attached: crit chance.png (940x790, 20K)

no crit/crit
crit/crit
crit/crit
crit/no crit
it's 50% you fucking retards, they don't specify which hit is the crit so it could be either hit 1 or hit 2, where the other hit could also be a crit, as shown by the two crit/crit instances. retards.

The probability is 25%, but we already know that one of the two hits is a crit, therefore the probability becomes 50%

>muh order
hits already happened genius. one of them is a crit. now what is probability that both hits are crits?

33.3% repeating of course

You forgot crit hit which is not the same as hit crit

there is only two scenarios, one guaranteed crit and two crits (one guaranteed and the othe by chance), there is NO 3rd scenario
cope

the probability that both were crits was 25%

however, we gained more information knowing that the first hit was a crit, the probability then increased to 50%

its 50%
You're forgetting the other Crit/Crit

sorry if i'm missing something, but what is the chance that the first hit was crit? and what is the chance the second hit was crit?

50%

It only raises the odds to 1/3, see

But user, you're thinking about this the wrong way. There's another 11 your not accounting for. 1 (100%) -> 1(50%) and 1(50%) -> 1(100%). So by your logic there are five possibilities. 00 01 01 11 11. Now it's 1 in 2 not 1 in 3.

It doesn't matter. The answer is the same in either scenario.

Exactly 23?
P(Crit)^23 * P(Normal)^(45-23)
0.3^23 * 0.7^22
= 3.6808 * 10^-16

Crit/Crit
Crit/Hit
Hit/Crit

1/3

You're overanalysing.

>guy literally crosses one of the scenarios.
That's not how you do that mate. You need to pick from all potential scenarios the number of scenarios possible fullfilling the criteria and divide it by all scenarios.

It's 50% you fucking mongoloids.

I knew Yea Forums had a lot of retards but this thread really puts things into perspective

It's never stated a set hit will 100% chance be a crit. The question is flawed because it doesn't state how the guarented works. For example it could be set up so that if the first hit doesn't crit the second will. This means the the whole always at least 1 crit thing can be ignored for answering the question and its 25%. If one of the two hits is set to always crit 100% of the time then it would be 50%. If it's a predetermined roll with 00 removed from a roll pool then you can get 1/3 but honestly that sounds like a stupid way to program the scenario l.

You're missing the other Crit/Crit. So it's actually 2/4 or 1/2 simplified.

100%

I'M FEELING LUCKY!

I can never tell if people are just fucking with each other in these thread or legit retarded.

25%
A probability of a roll with constant parameters doesn't change. Even if the question said that both of them are not crits, they're still wtf probable to be both crits.
Probability of an outcome is not the outcome. If you win a lottery that doesn't mean you have a 100% chance to win a lottery.

Crit/Hit and Hit/Crit are the same outcome.

How are this many retards not getting this? user put the answer right here. It is the most basic Bayes Theorem problem.

My method is perfectly valid within the constraints of the question as worded

by drawing a tree, you have one branch "hit" and one branch "miss" for the first shot.

on the branch hit you have two more branches "hit" and "miss" at 50% likelyhood each.

the miss branch just has a single branch of "hit" at a 100%.

Thus the chance to hit 1 should be 75 percent (100x50 + 50×50) and to hit twice 25% (50x50)

50%. We can ignore one of the two hits, since we already know that it's going to be critical.

It's the same thing with a question like-
>You have a multiple choice test with four answers per question. There are a hundred questions. You need 70% to pass. Guessing at random, what are the odds that you pass the test?
You can ignore 30 of those questions, since they don't matter for your problem.
>0.25^70

ok fine, but im asking you. what are the odds that the first hit was crit?

crit crit
crit crit
crit miss
miss crit

50/50

100%

user thats cheating you'll get banned.

This is why I play as a mage and don't talk to critfags.

Is 2+1 the same as 1+2?

>mainhand hit is the same as an offhand hit
retard alert

You retards are all forgetting to factor in luck. LUCKCHADS hit crits everytime regardless of nerd % chance. Lucklets will not crit on 99% crit chance. This is not up for debate.

Yes. Unless you attach something on one of the sides.
2×2+1=/=2+1×2

I’m sure there’s a ten page paper on this somewhere

yeah I'll probably get a crit

Does the first statement infer that there's some kind of conditonal probability at play here?

Or I should say the second.

IT'S LITERALLY JUST THE MONTY HALL PROBLEM. THE ANSWER IS 1/3. HOLY SHIT THERE'S SO MUCH AUTISM IN THIS THREAD. PEOPLE HAVE ALREADY CORRECTLY EXPLAINED IT AND PEOPLE ARE STILL ARGUING OVER IT BECAUSE THEY DON'T UNDERSTAND BASIC STATISTICS

Attached: classicmontyhalloddstreediagram.jpg (432x324, 27K)

Your question was much more interesting when I thought it was the first one

Well for starters P(B) = 0,75

Then it would be (2+1)×2 = (1+2)×2. What you did doesn't make any sense in an equation.

People are arguing over it because the question is poorly worded

What if the game is X-com

your math is wrong, P(B) = 0.75
then we get (1 * 0.5*0.5) / 0.75
= 1/3

Hitting or not isnt the question though. It's critting or not. The question assume you will hit everytime.

>he thinks this is a monty hall problem

33%

if reading comprehension is really the problem here, it means the average iq of this board has to be in the low 60s

This is a traveling salesman problem you dumb shit.

>question is poorly worded
what part of the question you don't understand?

Easy question.
There are four possibilities
>No Crit on Hit 1, No Crit on Hit 2 (this is ruled out by the problem so we ignore it)
>Crit on Hit 1, No Crit on Hit 2
>No Crit on Hit 1, Crit on Hit 2
>Crit on Hit 1, Crit on Hit 2 (i.e. what we want)
These are all equal probability but the first one is explicitly ruled out by the problem. Thus there are 3 equally probably outcomes, one of which is good.
So it's 1/3.

fucking retard you are talking about something entirely different and irrelevant

So there's a higher chance of not even hitting the target.

It's not an issue of reading comprehension. The question is bad. There are multiple correct answers to it depending on how you implement "at least one of the hits is a crit"

What do you mean by "at least one of the hits is a crit"? That either hit 1 or hit 2 is 100% crit, and the other one is 50%? Then the answer is 50%.

>Crit on Hit 1, No Crit on Hit 2
>No Crit on Hit 1, Crit on Hit 2
its the same scenario dumbshit the odds are still 50%

>by drawing a tree
Here's your tree. It doesn't really matter whether shot A is the first shot or the second. What matters is that one shot will be critical, and the other will either be non-critical or critical.

Attached: bullet tree.jpg (961x432, 62K)

>there are multiple answers so that means the question is bad

user pls...

>at least one of the hits is a crit
So there's only a probability that the other hit is or isn't a crit. 50 percent, Alex.

Is it though? Since crits are a random chance each time how does the previous one affect the following hit? Its not like monty hall where there is either a donkey or a prize, this problem there can be 2 prizes. One behind each door (hit). Theres also no switching. Wait, what the fuck how is this like the monty problem at all?

this is it
B = {CN, NC, CC}
P(B) = 3/4

Depends on how the mechanic.
>"that's not how it works, probability says this"
Dumb of you to think programmers will make something like this 100% realistic.

Based on the poorly worded question.

I hit twice and at least one is a crit. (100% crit rate on one of the rolls)

The other attack as a 50% chance

50%SHITTERS BTFO FOREVER

Attached: retards.png (486x595, 46K)

It's not complicated but it requires some abstraction because normally you would have to consider the scenario where neither hit is a crit.

user, he didn't say the FIRST 23, just that exactly 23 of them were crits.
You need to multiply by 45 choose 23, to "select" the hits that will crit.

>IT'S LITERALLY JUST THE MONTY HALL PROBLEM
Monty Hall threads are the ultimate enemy of Yea Forums so that makes sense

0% this is an rng game now bois

You fucking dumbass this is exactly like the stupid fucking math problems that get posted. The problem is purposefully ambiguous so that two anons will post different answers which are both fucking correct depending on how you read the problem.

75%

it does though. In computer science that equation is legit, even if the intentions don't match, and the computer will still calculate using PMDAS.
2×2=4
+1=5
2×1=2
+2=4
5=/=4

huh?

Why distinguish the two hits though? They're the same thing in two scenarios....

What odds are there for it to be a glancing blow thus negating a crit completely?

I'm this case yes, that's exactly what it means

It's an axiom of real numbers.

If the first is a given, then the only variable is the other crit
50%

Crit, Crit
Crit, Not crit
Not crit, Crit

1/3

You strike a branch with a crit saying it's impossible because there needs to be a crit. Perhaps you should reconsider your way of thinking

0.333....

Since at least one hit is a crit, that only leaves the outcomes CC, NC and CN, i.e. we may disregard NN.
Now of those three equally likely outcomes exactly one fulfils the requirement.

It's 50%. OP is a faggot.

The first hit isn't a given though. The crit could be on the first hit, or it could be on the second hit. You have to account for both of those states.

33%

33% if order matters and it's a permutation
50% if order doesn't matter and it's a combination

Attached: 1577855691046.png (787x1025, 800K)

You mean

Crit, Crit
Crit, Crit
Crit, Not crit
Not crit, Crit

If the position of the non-crit matters then so do the crits.
It's 50%

no, because you can miss the first shot, in which case you get a guaranteed crit on the second. you cant just cut that part of the tree out

The equation that you end up with is syntactically correct, but appending "x2" to both sides is not a valid way of working with equations, you violated the semantic rules of working with equations.

The whole point of equations is that if you change them in certain ways the equation is guaranteed to hold true if it was true to begin with. If you break those rules you might as well just add +5000 to only one side.

1/3rders are so fucking retarded it just blows my mind you fucking retards there are only TWO, repeat with me TWO scenarios, crit and hit, and crit and crit
THERE IS NO THIRD "HIT AND CRIT"
IT DOESNT MATTER IF THE CRIT IS ON THE FIRST OR THE SECOND YOU RETARD THE % OF BOTH CRITTING IS STILL 50%

You're damn right I am

Attached: Im super gay.jpg (315x259, 33K)

no way to know, maybe there is pseudorandom crits or not, or some weapon effect that gives a guarenteed crit on first hit or crits on proc per minute limit

Can't argue with that

Attached: 1583275049965.jpg (1080x9242, 2.41M)

What the fuck?
>0,0
OP's condition states that one of the hits is always a crit, so you only need to calculate for the second one to hit 1.

Is this bait? I didn't know what there were nonwhites/women on Yea Forums

you retard this problem is about crit chance not some donkey shit

If no further information is given you always assume it's a perfect scenario. Otherwise we would have to estimate the uncertainty as well.

Attached: 1425070474704.jpg (490x327, 34K)

Probability theory doesn't work like you think it does.

If you feel like blowing your mind tonight:
en.wikipedia.org/wiki/Monty_Hall_problem

no, two of the options were dropped, not one. if the extra information had been "one of the hits was a crit", then yes, 3 options would be left, but the information was 'the first hit was a crit", so only 2 options are left.

>one of the hits is a crit
does not mean
>the first hit is a crit

Correct.

for me? 100%, im no lucklet.

Attached: akagi.jpg (572x564, 37K)

Here 1/3 tards, I also have a chart.

Attached: mechartmakemesmart.png (982x387, 16K)

It doesn't say "the first hit was a crit" it says "at least one of the hits is a crit"
Both answers are correct to this stupid question

50% assuming you cant miss because you are already did the first crit so next attack is a 50/50
33% if you mean the scenario as a whole

There's a problem with how the question is worded. Let's say the gun can hold two bullets at a time, does each individual hit have a 50% crit rate or is the probability based on a fully loaded gun? If the former, you already eliminated one of the variables, so the question becomes "if you have a 50% chance to crit, what's the probability you'll land a crit". If it's the latter, then the answer is 33.333...%.

33% because the scenario where you don't Crit will never happen

We can easily assume 100% of the hits won't be missed though.

Why are you drawing two separate trees?
Why are you throwing out an entire tree because it had one disallowed outcome?
Your diagram should look more like this

Attached: fixed tree.png (1095x631, 41K)

The crit chance is 50%. That's how it came into effect. You tossed a coin twice and it came up heads at least once.

Everyone always misses the wording that says “at least one will be a crit hit” which means that for one swing you are guaranteed a crit. Which means you are only concerned with the probability of one hit, and the question already tells you it’s 50%. So your answer would be 50% that both hits will be crits.

No. But what makes you think programmers won't just work it in such a way that the first attack is always a crit? I would, it makes my job easier and less retarded.

it doesn't, crit crit is the same as crit crit
once you crit once you enter a scenario with only two further possible outcomes
if you dont crit the first time you have only a single further possible outcome
3 outcomes
your second crit crit occurs nowhere

You forgot the part where the second hit on the bottom flowchart isn't a crit making it 1/3 retard.

It seems you're doing too many steps.
A = hit 1 is a crit (50%)
B = hit 2 is a crit ) (50%)
You KNOW one of the hit is a crit, so you have to calculate following the sentence
"If I know B happens, what is the chance of A happening" or viceversa, the result is the same.
This can be done using the formula P(A|B) = P(A*B)/P(B).

So the result is 0.5

It doesn't say anything about guarantees.

Attached: images (85).jpg (554x554, 29K)

>If the position of the non-crit matters then so do the crits.
Are you shitposting or retarded?
"Getting a crit, and then not getting a crit" is different from "Not getting a crit, then getting a crit"
"Getting a crit, then getting another crit" is exactly the same as "Getting a crit, then getting another crit"
How could you possibly suggest those are different?

100%

Attached: Untitled.png (63x61, 10K)

40-32/2
40-16
24
1*2*3*4
4!
I don't get it, what's the trick?

Because if the first hit is a crit it's not random.

That implies you don't care about the order then. so 0(U)1(G) and 1(G)0(U) would mix into one possibility as well since you mixed 1(G)1(U) and 1(U)1(G) together, so there would only 2 possibilities.

C/NC
C/C
NC/C
NC/NC

Given the condition, there are only 3 scenarios left
C/NC
C/C
NC/C

So 1/3
Also it doesnt say if the first hit has to be a crit.

It’s literally the correct answer. Don’t be mad you got fooled by a trick question you faggot zoomer.

The answer is 1/3 but it's only a coincidence because it's not the Monty Hall problem.

you guys are undersimplifying. ignore the first half, it is literally irrelevant. the actual question is, "what is the probability of getting a crit, when the crit chance is 50%?" the fist hit and crit are done with and guaranteed, they do not affect this variable at all.

1/3rders btfo

Attached: chucknorris.jpg (1023x1385, 220K)

crit/crit
crit/no crit
no crit/crit
no crit/no crit

1/4 so 25%

Yes it won't, but gamers will be expecting one to crit anyway because it is guaranteed so here's your guaranteed crit.
Now I only have to calculate for the second one.

go simulate two dice, eliminate all cases of double odds, see the ratio of double even compared to the rest
that's your guaranteed crit scenario and it is not 50/50

you're assuming that the first hit will always crit

yes, all we have to go off of is one of the hits will be a crit and there are three different ways to make that happen.

Now figure out the chance of getting a double crit if we assume each of the three systems for the problem have an equal chance of being used.

My best guess would be he was hoping there would be people fucking up PEMDAS, or people that have never heard of factorials.

We're looking at the probability of both attacks being crits not a single attack. The only stipulation is that one of the attacks is a crit so you can throw away the outcome where both attacks were non crits.

how the fuck is crit, crit different from crit, crit?

1. I am not assuming. This is a fact the question itself posits.
2. Wether it happens on the first or second is irrelevant. One hit always has 100% probability of crit. Therefore I am only concerned with one of them. Wether it’s the first hit or second hit, it’ll still be 50% for both hits to be crit.

It literally isn't though, look at There are two ways of getting one crit (01 and 10) and only one way of getting two crits (1,1)
It tells you that at least one hit was a crit, it says nothing about WHICH hit was the crit.

Probability theory is fucking retarded, miss me with that shit.

That formula is also correct, but as others have stated, P(B) = 0.75

P(A*B) = P(A) = 0.25

P(A|B) = 0.25/0.75 = 0.333

Because there are different type of critical attacks based on your crit chance at the time of the attack. Critical 100% and critical 50%.

.5 * .5 =.25

That logic is flawed though because OPs condition states there is one which will always be crit. You don’t even know what the fuck you’re linking you retard.

Is this the power of AI?

You decided
Crit/ no crit
And
No crit/crit
Are different but assume
Crit/crit
And
Crit/crit
Are the same it's 2/4 or 1/2

Okay retards
Get two coins
Get two fucking coins right now, I'm serious, actually physically do this
Heads = crit
Tails = no crit
Repeatedly flip them both
Record how many outcomes have 2 heads (2 crits) and how many outcomes only have 1 heads (1 crit)
Ignore outcomes where there are 0 heads (no crits), i.e. don't count them at all, because this outcome is impossible per the problem statement.
Report back.
I promise you you'll get double heads ~33% of the time.

>Everyone always misses the wording that says “at least one will be a crit hit”
Probably because the wording doesn't say that.

>learn programming language
>start every code with "scat retards"
Thanks, very gay

crit crit
crit miss
miss crit
miss miss

miss miss is discarded.
This leaves three equally possible options, one of which is the desired outcome.
Thus, 1/3.

wew lads.

Attached: tfwtointelligent.jpg (640x640, 33K)

>1. I am not assuming. This is a fact the question itself posits.
Quote on the problem where it says either the first hit or the second hit have a 100% unconditional chance of being a crit

>all we have to go off of is one of the hits will be a crit
No.

One of the hits IS a crit.

You throw out the entire tree because it's only possible if you accept an ambiguation which is not necessary to describe the problem as stated

Here you go my blind and seething friend who got tricked by a trick question.

Attached: BC56A3D7-A6E2-4303-A4A8-CF17C90D5AD8.jpg (750x174, 96K)

crit or miss?
I guess this never crit
huh?

Attached: bruh.jpg (480x360, 9K)

You are ignoring the fact the text says "At least one of the hits is a crit". That = One hit is a crit. So why the fuck would you get two coins? One coin's result is set, in the scenario described in OP.

Read what I'm replying to.

wow this shit got no replies, such a satisfactory read

33% or 50% depending on how the forced crit is applied

Read the OP retard also

Retard.

Attached: 1583272198312.png (390x345, 59K)

how can it be 1/3 if there are only two hits retard bitches

Read, faggot.
My experiment accounts for that by throwing out outcomes with no heads, i.e. essentially pretending they never happened.

"You hit an enemy twice. AT LEAST ONE OF THE HITS IS A CRIT".

It's in the formulation of the problem already. You managed to hit both attacks, and one IS A CRIT.

Bitch, if you're going to argue about wording, and the actual wording is not what you say it is, don't act like the problem is the wording. It's present tense. Not future tense. It describes a fact. Not something that is to come. You tricked yourself by reading something that wasn't there.

>That logic is flawed though because OPs condition states there is one which will always be crit.
Yes and it doesn't tell you which hit is the crit, which means it could be one of three equally probably outcomes
0 1
1 0
1 1
As you can clearly see, the probability of two crits is 1/3.

0. My luck is terrible and I'm lucky enough to have crit once

>trick question
It's actually a 101 tier question.
Not even in an exam, but rather in an exercise.

It’s hilarious how you reduced the whole question to a coin flip and you STILL fucked it up. It’s 50%. One hit being guaranteed a crit effectively removes it from the equation. Doesn’t matter which you pick. The leftover hit is a 50% of being the second of both crits.

If you changed the question to read: “You hit 100 times. 99 of these hits are guaranteed crits. What are the odds all 100 hits are crits if each hit is a 50% chance?” then it WOULD STILL BE FUCKING 50%.

It’s one of those questions where the teacher tells you to read the instructions and on the last page they tell you to underline your name on the front page and get 100% automatically.

Do this instead retard, take two quarters. Leave one down as heads (crit), then flip the second one over and over. Now, pick up both quarters. Flip one first and THEN place down the second one the same number of times as the first scenario. Now you tell me how many times you got two heads as a percentage.

read bottom left retard

After you do that flip one coin, if heads flip the second coin. If its tails don't flip the second coin and set it to heads. Repeat until you have a data set.

After you've done that flip the first coin and set the second coin to heads. Repeat until you have another data set. Notice all of these methods qualify for achieving OPs problem.

>but as others have stated, P(B) = 0.75
No.
It's always 50%. The chance of the hit being a crit is the same for both hits. What we know is that at least one of them must be a crit. This doesn't modify the original variables.

You idiots need to learn to read the problem better
All you 50% shitters are missing the phrase
>Assuming a 50% crit chance
If your argument hinges on the fact that "one of the crits is 100%" you're directly contradicting the problem statement.
The individual hits are STILL each individually 50%. What's 100% is "at least ONE of the hits being a crit"

>dude what you’re reading isn’t there!
Lmao seething instructionlet

Right, which means you are not concerned with the probability of two, but of one. See how that works my high school dropout friend?

Good read, thanks user.

The monty hall problem requires 3 choices. This only has 2.

See Here is, semantically, how to read the OP.
>You hit an enemy twice
"Hit" is either past or present tense. If it is present, it's a dramatic or historic present, using the present tense to describe an event in the past.
>At least one of the hits is a crit
Again, present tense. This is merely relating the outcome of your two hits to you.
>Assuming a 50% crit chance,
This is the only actual stipulation for critting. It's a coin flip.
>What is the probability both hits are crits?
You know the odds, you know the actions, you know the (partial) outcome. You have all the information you need. You attacked something twice, and had 50% chance to crit both times. You also know the coin flip was in your favour at least once. What are the odds it happened both times?

Aside from people that are completely retarded, the only reason there is an argument going on is because of 2 groups of people with different outlooks. You have the one group of people that are considering crit/hit and hit/crit as being 2 separate outcomes, then you have the people that are saying crit/hit and hit/crit are the same outcome, so you combine those outcomes because in either of those options it's the same outcome with 1 crit and 1 hit.

Using that image, the first group of people believe the picture 100%, the 2nd group of people are combining hit/crit+crit/hit. That's why some are saying 33%, others are saying 50% and everyone else is handicap.

So which is it Yea Forums? Does hit/crit + crit/hit count as two different outcomes because the order matters? Or are they the same outcome because order doesn't matter, either way it was 1 hit and 1 crit. Once you decide that, we all agree on an answer.

p(two crits | one crit) = p(two crits and one crit) / p(one crit) = p(two crits) / p(one crit) = 0.5 * 0.5 / 0.5 = 0.5

If we know for sure we'll get one crit, then the probability of getting two crits is 50%.

You are calculating "what is the chance for B to crit if A crit", rather than "what is the chance that both crit if at least one crit"

And there are two other methods for achieving that exactly worded condition.

Why do people always look at it as 50% chance to crit instead of 50% chance to not crit.

>Leave one down as heads (crit)
You're doing it wrong. You should have a friend flip both coins without showing you and tell you every time at least one is heads.

50% because yes

refer to
there is no
"0 1"
"1 0"
scenario faggot

50% stay seething retard

Nowhere in the problem does it tell you that one of the hits is guaranteed. Your flowchart assumes this instead of just checking the possible states after two swings and discarding the outcome that was impossible.

1/2, 1/3 AND 1/4 ARE ALL VALID ANSWERS DEPENDING ON HOW "FORCED CRIT" IS APPLIED
WE'VE HAD THIS SHIT THREAD NON STOP
STOP BEING RETARDS

42

as always you fucking idiots are aruging this based on probability math. which does not allow for 100% guaranteed anything

instead use VIDEO GAME MATH you fucking dipshits. and in VIDEO GAMES, which is the BOARD WE ARE ON, there are skills that allow a 100% chance to crit.

Using one of those skills on one attack either the first or the second will still result in 50% chance to crit. for the other attack.

this is how the math would work in a goddamn video game. probability is the worst tier of math, you can fuck right off with your "gotcha" moment of there are 3 different results. because in a goddamn game there are 2. fuck you.

25% crit, crit
25% crit, no crit
50% no crit, crit

>then you have the people that are saying crit/hit and hit/crit are the same outcome, so you combine those outcomes because in either of those options it's the same outcome with 1 crit and 1 hit.
No, the 50% crowd are imagining things about guaranteed crits.

>people that are considering crit/hit and hit/crit as being 2 separate outcomes,
The people that pass the exam.

>then you have the people that are saying crit/hit and hit/crit are the same outcome
The people that don't.

the equation if one of them wasn’t garunteed to be a crit would be 0.5/2 (50 percent chance over a sample size of 2)
But since one is going to be a crit it’s 1/1 + 0.5/1

The 50% crowd knows how to read and deduce facts, clearly, while the rest are seethe n copers doing mental gymnastics

I'm not seething, I am calmly and clearly explaining the error in your interpretation. The actual wording which you referred to doesn't mention anything about "will". It's describing a fact, something that happened, but wasn't guaranteed to happen. The very next line informs you in fact that it had a 50% chance of happening per hit.

depends if the weapon is a katar then it's 50%

B is "[between 2 attacks,] at least one is critical". B is then {CH, HC, CC} out of possible {HH, CH, HC, CC}.

There is no forced crit. There is in fact one correct interpretation. 1/3 is the right answer.

you roll the first attack at 50%, if it crits "at least one of the hits is a crit", so you have to roll the second one for anoter 50%

25%

If a particular attack is a forced crit then it's 50%
If the probability of no crits is just removed from the pool is 33.3%
If a crit is retroactively applied to one of the hits if both rolls fail then its 25%
All 3 interpretations are valid based on the obscure nature of the question

Right, good thing I already know one hit is guaranteed a crit. So that’s 50%! That was easy.

Hey Yea Forumssauce. user here, and today we're going to talk about the Monty Hall problem.

beleventeenone

Reminder that most people on Yea Forums are from America and they have an abysmal education system. Also the people here are mostly polmutts. That should explain to everyone why there’s so many people still saying anything other than 50%. Logic just isn’t their forte. Look at their president after all.

>The 50% crowd knows how to read and deduce facts, clearly
Quite the opposite, in fact. You make up for your deficiency in reading comprehension by filling the gaps in your understanding with leaps of logic. If you had understood it properly, then the actual fact you would deduce from the wording of the problem is that there are no guaranteed hits and that the problem is merely describing the outcome of your actions.

See

I genuinely believe that at least half the posters in this thread have not fully read the entire opening post.

>Does hit/crit + crit/hit count as two different outcomes
It literally doesn't matter if they're two different outcomes, they still make up 2/3 of the possibilities given one guaranteed crit.

Due to the obscure nature of the question, 1/3 is the most correct since it makes the least assumptions.

That says 1/3 is right retard.

You’re bringing tense semantics into a question that is all tense-consistent (i.e in the present). Just because the dude paraphrased doesn’t mean the interpretation was wrong.

READ NIGGER READ

It's 1/6th you absolute morons. What you have to do is if you're reading this right now you just lost the game you chucklefucker

This is why crit builds are fucking retarded.

Guaranteed crit/crit
Crit/guaranteed crit
Guaranteed crit/no crit
No crit/guaranteed crit

It's 50%
You're all stupid

>I literally cannot stop thinking about Americans

I have seen it, but still, based on the wording of the question, there is only one valid interpretation. It doesn't mention a guaranteed crit. Nor does it allow you to check the crits for yourself. It is, in fact, just telling you this information. That corresponds to the 1/3 answer.

>Just because the dude paraphrased doesn’t mean the interpretation was wrong.
It does if he paraphrases it wrong.

Seething retard. Maybe if you knew how to read properly you wouldn’t have to be replying to every post in the thread with paragraphs about how you fucked up your interpretation.

Let's do the guaranteed crit first.
If first hit is the guaranteed crit, then the results
0,0 1,0 0,1 ,1,1 become 1,0 1,0 1,1 1,1
If second hit is the guaranteed crit, then the results 0,0 1,0 0,1 1,1 become 0,1 1,1 0,1 1,1
4/8=50%

Attached: 1574865659358.jpg (497x544, 174K)

Sonofabitch

There is no most correct or least correct
Information is deliberately obscured to give all 3 answers a valid ground
The entire purpose of the question is that the information provided is valid for all 3 explanations

Please tell me you aren't white.

You don't know anything, you are wrongly assuming.

You fools, you’ve forgotten the legendary critical miss

It's 25%

The welfare crit only happens if you were not going to crit at all and prevents that outcome

>dude if the question tells you a hit is guaranteed to be a crit you’re assuming it will be a hit!

This is why you never passed 8th grade math

I am actually writing about how you fucked up your interpretation. If you couldn't even figure that out, your reading comprehension and/or deductive capabilities shouldn't be relied on for anything.

No, two of the answers require you to make assumptions to answer the question.

>Crit / No Crit and No Crit / Crit being different results
My god, you're just the biggest retarded dipshit

If it's in the context of a videogame you'll see the results yourself by the damage amounts
If the first hit always crits it's 50%
If either can crit it could be 33% or 25%
Stop being a fucking idiot

Said the retard still seething that he got tricked by a trick question that already gave him the answer.

The question does not, in fact, tell you a hit is guaranteed to be a crit.

>At least one of the hits is a crit
That is an innate implication that one hit is guaranteed to be a crit. The 1/3 crowd assumes that in a crit crit situation, the first crit was the "at least one". But I'm trying to recognize that "at least one" could be either the first or second crit. Allow me to rephrase, if the first hit IS a critical, what is the second hit chance to crit? Now think of the second situation, if the second IS the critical hit, what are the odds the first hit crit? The 1/3 crowd argue that they are both the same, but they also try to argue that the other two possibilities are separate. And by their own logic, that is not the case bring them back to 1/2.

Double dagger assassins are retarded.

if one hit is a crit, set it aside. P(1 crit) = 1. Question is P(2 crit) = ?

You just need P(1st crit)*P(2nd crit)
P(2nd crit) = 1/2

Answer = 1 * 1/2 = 1/2

And no, the order doesn't matter

That's the entire point of phrasing the question in the context of a video game you absolute cretin

Lmao nigga you serious?

Right, but the video game in question is entirely hypothetical so we don't have damage amounts to go by. We only have the description of the events to go on.
It's not even a trick question. Again, you're tricking yourself.

The portal conserves momentum thus the block should fly out instead of just falling out

I fucking love these threads.
You get to see so many people who are so OBVIOUSLY clueless talking with such authority.

>dude if you read the question and accept the facts you are tricking yourself! Ignore the question to win!

25%

It says at least one was a crit, not at least one was a guaranteed crit.

>That is an innate implication that one hit is guaranteed to be a crit.
no
>The 1/3 crowd assumes that in a crit crit situation, the first crit was the "at least one".
no

>Right, but the video game in question is entirely hypothetical so we don't have damage amounts to go by.

Literally the same thing. Seethe n cope.

Based and B-pilled

Dead serious. Point out to me where I made a mistake. You will instead only show where you made a mistake and require me to point it out to you.

The question is over whether both hits are a crit, the order doesn’t matter dumbass

Case in point:

It’s in the OP. We all know you will spend another 50 posts trying to convince us your retarded semantic interpretation means that there are no guaranteed crits even though the question outlined the facts for you.

100%

>Ignore the question to win!
That is what you're saying, isn't it? If you actually read the question and accept the facts and nothing but the facts as given in the question, the answer is unambiguously 1/3. Mathematicians have already shown how, depending on interpretation, there are three possible answers. Now, as a linguist, let me tell you there is only one possible interpretation.

Well if anyone needed proof that Americans are retarded, here is this thread.

The dress is blue and black
The best candy you can buy at a truck stop is reeses cups
B is the correct answer to the portal question
Thick crust is the best kind of pizza crust
The chance of both hits being a crit is 50 percent because it’s the chance of one hit being a crit because the other crit is already guaranteed

>I think I've proved quite enough times that 1/3 and 1/2 are each absolutely correct given the right assumption about how the stated information was obtained.

When it comes to “is” and “is not” those things are binary. There are no extra interpretations there. Sorry bud. You got tricked by an easy question.

I fail to see your point.
>It’s in the OP.
It's not. This isn't how you point something out. You don't just say "it's there somewhere". I have already gone over your mistake in detail. The fact that you can't do the same in return is telling in itself.

>doesn't post his stance
Coward

1/3 = 1/2

I think it's a mix of ignorance and retardation.

Seethe n cope, retard.

"At least one" is not, in fact, binary.

t. NEET drop out

Which is why I didn’t reference that but the “is” and “is not” you retarded non sequitir moron

>Getting an answer of 1/2 requires us to make an additonal assumption...
Did you stop reading right before the end?

It's 50%. Anyone who said otherwise is just exposing their poor vidya understanding.

As much as I wished this was only Americans, it's not.
Humans as a whole are fucking retarded.

We are ALL fucking stupid, and (You) and me are no exception.

You were doing so well.

You know you've lost when you're forced to write in all caps.

If we're basing this solely on vidya it's 0%. The game won't let you get two crits in a row because that would be overpowered.

aa | aA
Aa| AA
however, because there must always be one A in each quadrant,

Aa | aA
AA | AA

50%

>you called me dumb, that makes you dumber!
You don't even know what side i'm on here.

They're the same thing only you swap A and B.

Then you are ignoring the question.

No, Americans are especially retarded. They are intentionally taught wrong from birth to make them as stupid and malleable as possible. The entire world is sickened by the country’s constant setting of a new low in intelligence.

Yes.

100%
>not stacking whatever stat increases crit chance

Explain where I went wrong then, mr genius

>no double no crit as a possibility
end yourself

After you pick the door you want and one of the "shit doors" is revealed, you're back down to two choices again. The guaranteed crit is basically them revealing one of the doors

>both can be correct
Thank you common core

The end, if that wasn't clear.
The dress is black and blue
The answer to the portal question is B
But the chance of both hits being crits is 1/3

What is your stance on Monty Hall?

Thread should have ended here
Also here is a statistical demonstration that the probability is 1/3

If you say 50% you might be nonwhite or ESL.

Read the OP question retard

Good read.

I don't know what's more worrisome, the sheer amount of 50% fags or the failed attempts at unfolding Bayes theorem.

>common core
Lmao this must be why Americans are so bad at reading simple questions

Why would you buy chocolate from a truck stop? Guaranteed to be old and spoiled.

>natural 50% crit rate
[ Hit, Hit ]
[ Crit, Hit ]
[ Hit, Crit ]
[ Crit, Crit ]

>modded 100% crit rate for one hit
[ Hit, Crit ] [Crit, Hit]
[ Crit, Hit ] [Crit, Crit]
[ Hit, Crit ] [Crit, Crit]
[ Crit, Crit ] [Crit, Crit]

4/8 = 50% of both hits being crit with mod of 100% crit rate for one hit always being crit

Why did you turn aa into AA if only one A is required?
Your rationale would lead to 25% chance instead.

Though eliminating aa still makes more sense given that it's stating only that one hit is a crit, and not that one of the crits is guaranteed, so 1/3

>still seething that you can’t read a simple question

Spend the rest of the night coping for all I care. Entertainment at its finest!

It's 0% because if the crit chance is 50% and 1 has already crit then the other can't crit

Wrong, there must always be one crit in any scenario. So it’s crit/not not/crit crit/crit and crit/crit

>muh crit/hit and hit/crit are the same scenario!
Stupid dumb nigger faggots riddle me this, if we remove the one guaranteed crit and turn the question into a simple "given a 50% crit chance what are the odds of critting twice in a row", do you still answer 1/3 because muh crit/hit and hit/crit are the same?
Idiot subhumans why didn't your mother abort you

>seething
>coping
Might I suggest... projecting?

>Atleast 1 is crit
Not "Only 1 is crit"

monty hall threads are too hard for most boards to be fair

lmao this guy is as smart as 1/3fags and “multipleanswer”fags FYI

>taught wrong
You mean they start out with a set of blatantly false axioms?
I think you are opening a can of worms there, that ought be left untouched.

thats not what i said

starting from where this post fucked up:
If one attack is guaranteed to be a crit then we are either at 1- or 01. Since both attacks have already been made with 01 then our only valid starting condition is 1-
Which then brings us to only two equally likely events
10 or 11
thus it the probability of 11 occurring is 0.5

American here, 100% agree.
It's 50% by the way.

that's not how games work

fuck you 1/2fag you're objectively wrong and should kys
we all know its 0% really, your just deluding yourself

Based crab bucket escapee. Good luck to you.

>there must always be one crit in any scenario
It doesn't say that. It just says there is at least one in this scenario.

>100% crit rate for one hit
The probability is literally 50% it says this explicitly in the question.

Aa and aA become AA, aa becomes aA or Aa

Maybe you should be more worried about the fact that the discussion is based on an intentionally poorly worded question designed to make retards fight each other, and how you perfectly fell for the bait like the dumbass you are.

50% is always 100% for one

Don't pay attention to user, he's a retard. The whole reason Monty Hall works out the way it does is because Monty knows which door the prize is behind

Well?

Attached: hqdefault[1].jpg (480x360, 24K)

It's obviously 0% stop being retards

crits are independant from each other
its 0.25 retards

one hit critting doesnt affect the other

except that if you get [crit, hit], you don't need to mod the second hit to be a crit because you already got your guaranteed crit, and if you remove all duplicated that would leave you with
[Crit, Hit]
[Crit, Crit]
[ Hit, Crit ]

Only one is [crit,crit] because you can't count duplicates. the answer is 1/3

There is no guarantee. "At least 1" means you roll normally and then from the results, only examine the subsets that have at least 1 crit to look for doubles.

Now you’re just using semantics. For any two hit pair there must be one crit. Thus the only possibilities are hit/(hit changed to crit), (hit changed to crit)/hit, crit/(hit changed to crit), and (hit changed to crit)/crit

No it isn't. You can flip a coin twice and get heads twice in a row. It's just less likely. Given that 1 hit is a guaranteed crit you're only flipping the coin once. So 50% chance.

25%
50% chance of a critical hit = 0.5
P(critical and critical) = P(crit) x P(crit)
0.5 x 0.5 = 0.25
the answer is 25%

If there is no guaranteed crit and you just happen to get one: 25% (1/2 x 1/2)
If there is a guaranteed crit: 1/2

2/3

I'm not even gonna scroll through this thread. If one hit is guaranteed a crit, there is a 50% chance the second will be a crit. Because the first is guaranteed, its no longer part of the equation.
50%

Nigga read the question.

If they’re independent it would be 50

There's 3 possible options therefore 1/3
t. 1/3tard

>Now you’re just using semantics.
Yes. Because semantics are the only reason there is any disagreement here. You are also using semantics to support your wrong interpretation.

100% because I’m playing with power.

Attached: 8240198D-F17E-43FE-B98F-5516413830B0.jpg (799x1431, 260K)

Alright alright. Depending on whether you think a guaranteed crit already happened the answer can change from 1/2 to 1/3. It's a matter of whether you're estimating after flipping the first coin or before flipping either

People here seem to have forgotten basic statistics.
If you're looking at the probability of x event OR y event happening, then thats (probability of x) + (probability of y)
If you're looking at the probability of x and y occurring, then (probability of x) x (probability of y).
If you go more complex then you factor in shit like being mutually exclusive, etc.

Based!

No, the answer is 50%. Learn to read a question properly next time.

The probability is out of 4 because there are 4 possible occurrences. Two occurrences are functionally the same but aren’t the same in terms of calculation

thats the name of the file retard

Listen up, jizzheads. If you're having trouble, literally just picture it with a coin, because that's literally what it is.
>You flip a coin twice. At least on of the flips is heads. Assuming a 50% chance of heads, what is the probability both flips are heads?
Note that I stayed as close to the original wording as possible. Now, you wouldn't go and assume that the coin is weighted (but only some of the time), or that you're guaranteed to get heads once any time you flip two coins. That'd be retarded. So why can't you idiots see that it works the exact same way in the OP?

Uhh based department, hello?

[ (Hit, Crit), Crit ] .75
[ Crit, (Hit, Crit) ] .75
[ (Hit, Crit), Crit ] .75
[ Crit, Crit ] 1

3.25/4 = 81%

So which is the better game design Yea Forums?
>1/2fags: "Here's your guaranteed crit and now pray you crit the next."
>1/3rdfags: "You will always crit at least once."
>1/4thfags: "If the first hit didn't crit and the second it calculated to not to as well, the second one will crit."

Attached: 1580393955930.jpg (408x560, 44K)

Ah okay, so the "at least 1" is just a condition to only consider events with, well, at least 1 success and disregard events with less. Cool beans

They are different outcomes, because you need to represent that critting once is more likely than critting twice or not at all.

If you treat them as the same outcome you're saying it's just as likely you'll crit once from two attacks as twice/none.

>every 50 percent retard ITT using the word "guaranteed"
>the question never says that anything was guaranteed, only that at least one of the hits out of two ends up being a crit.
Learn to read niggers

Yeah, I'm thinking this is based.

What is the chance of a 50 percent thing happening if it happens twice but you only need it to happen once, retard?

There are only three possible occurrences. We know for a fact that (hit, hit) did not occur.

Of course Yea Forums can't make any games, they can't even work out statistics.

Go back to English class Apu

>1/3rdfags: "You will always crit at least once."
That's actually not what we're saying.

The context of a video game and a coin are not the same

Imagine making a flowchart and still not getting it.

if you are at the start before tossing the coin twice, yes it's 1/3. But if you already toss a coin and it is heads(crit) then the chance of getting another crit is 1/2 because of the next fair coin toss.

Pretty simple. Semantic argument indeed

The two relevant possibilities are natural crit/forced crit and forced crit/natural crit.
Hit/hit was never an option, read my posts retard

Attached: cards.jpg (2980x2528, 774K)

Even though i am a 1/3 fag here, if i was designing a game i'd go the 1/4 route.

It's like, fine you have shit luck here's your wellfare crit, fag.

They are identical when determining probability. The question itself tells you this. The odds are fifty percent. Just as in a coin flip.

It's obviously 0% I don't know why 50%fags are still like this

69, don't @ me ever again

Why are people acting like the order matters in this problem? One of the two answers is guaranteed.

WRONG

Power bloat is cancer so 25% is the best option in the context of bidya

What size is the deck? Standard size?
The probability of getting a 7 and a Queen is (1/4 * deck size) * (1/4 * deck size) I think

the problem with the monty hall problem is if you were there in person you could subconcsiously detect the goat's pheromones and your gut feeling would be to pick the non goat door
it only works in simulations

Fuck you m7, go grab your calculator and do it yourself.

You haven't played many vidya then.

There is no such thing as a "forced crit". I'd recommend reading the OP first.

>But if you already toss a coin and it is heads(crit)
The question doesn't say that the first hit is a crit though.

hint: not a probability question

It's NOT GUARANTEED, IT JUST DOESN'T HAPPEN.
There is a difference.

Because I removed one of the results after calculating it because it wasn’t relevant and also impossible given the conditions and failed to account for that

He never did give the answer, did he?

0% because I don't play little fairy gay boy builds that rely on getting crits.

0%

Oh you sweet zoomer child.

I know the order doesn't matter, you have to stoop down to 1/3 fag's level to explain it to them. The order matters for one crit and then suddenly the order doesn't matter for two crits.

I never did but someone found it out in the thread

>at least one of the hits is a crit
meaning one regular hit of the two will be turned into a crit. Hit/hit is logically impossible because you WILL crit, dumbass. Don’t lecture me when you don’t understand the conditions

*didn't
Obviously it could have, and probably does all the time.

>at least one of the hits is a crit
>not guaranteed at least half of the hits are crits

I mean, it just does happen i guess. Fuck.

zeromind

>not going for a 100% crit, crit multiply, heal on crit, crit on crits build
So you're a cuck?

100% if you abuse save states

because they think probability theory overrides real world practicality.

it does not, they are treating the math like a set of coin tosses and removing 1/4 total possibilities, when the real world application would not work that way in terms of video games.

they don't believe forced crits exist. as a person who has played video games i do.

Neither of them are guaranteed to be crits.
You know only that one of them is.

>meaning one regular hit of the two will be turned into a crit.
No, it does not mean that. It means that at least one of them was never a regular hit in the first place. What were the odds of that? Fifty percent, as it turns out. The problem tells you this. It's not a condition. It's a result. I will lecture you on poor reading comprehension all night long.

>this is what STR fags think
Projecting much when you're compensating for your small dick with high STR?

>I roll a dice, and get a 6
Would you say based on this wording that I was guaranteed to get a 6 and that the dice will always roll 6?

>they don't believe forced crits exist. as a person who has played video games i do.
They do, in some games. Not the one described here though, as it clearly states the chance of critting is 50% every time.

Sorry you need to rely on luck and trinkets to win a fight noodle arms

So you’re telling me you’re ignoring the actual text of the problem, replacing it with your own bullshit, and then lecturing me for following the problem instead of your bullshit?

Better game design would be forcing probability IMO. In this example misses do not matter, it's 100% chance to hit. For example if you have 25% crit chance and you hit 3 times, your 4th swing 100% should be a crit. On the other end, if your first swing was a crit, the next 3 should be hits. Basically, I don't like getting fucked over by RNG. By forcing the crit or hit, it makes damage calculations a given, no more luck involved, which is how I prefer it.

I completely understand why people would hate this, because it completely removes luck from the calculation, but in my experience luck tends to lead more to frustration than it does satisfaction. People are more likely to notice they died or lost because they got assfucked by RNG than they are to notice they won because of it.

Two independent events so P(A)*P(B) = P(A and B) if one is guaranteed then P(A) = 1/1 and P(B)=1/2 (nothing about order mentioned in OP) so (1/1)*(1/2) = (1/2).

Probability is 50 percent.

>100%
>luck

>It happened
>Therefore it was guaranteed to happen
Yeah, hard determinism is real fun from a philosophical perspective but kind of useless from a probability perspective.

Hit/hit is an impossible result dumbass. Whether you believe the hit is replaced or it never happened in the first place there can be no two hits, only one crit or two.

>theory overrides real world practicality
You can program crits however you want.

I still have enough faith in humanity to believe you idiots saying 1/3 are just trolling. I have just enough hope in the future to think you're not all really that fucking retarded.

protip: read the question, it's worded very specifically, you're confusing it with the monty hall problem which is worded intentionally to trick you, this is not the monty hall problem, not at all

for the sake of a question then yes. those are the rules of that question.

There's a difference between one crit being guaranteed and one crit having happened.

Let's put it this way. The enemy has 5 HP. You deal 2 damage per hit, but you have a 50% chance to deal a 4 damage critical hit. You hit the enemy twice before it goes down. Because you'd have only dealt 4 damage total with normal hits, you know you crit at least once. But what are the chances that you crit on both hits?

I want to fucking strangle you

The exact opposite, in fact. I am telling you to stick to the actual text of the problem instead of replacing it with your own bullshit and the lecturing me for following the problem instead of your bullshit. Man, you people have poor reading comprehension.
>Hit/hit is an impossible result dumbass
Yes. I know. Hence why it's 1/3 instead of 1/4.

0%, crit is instant death.
No second swing if the first is instant kill.

It’s not 1/3 because the 4th you are removing was never part of the equation you fourth grader

>read the question, it's worded very specifically, you're confusing it with the monty hall problem which is worded intentionally to trick you, this is not the monty hall problem, not at all
I am really curious as to what's going on in your head at this moment. Could you clarify this statement at all? What exactly do you think our calculations are?

remember that if you don't think it's 50% you don't know basic math

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0% because crits have an internal cooldown that stops them from occuring more than once per attack set

Change doors

>"you hit an enemy twice"
>"you hit"
>WE HAVE TO ACCOUNT FOR MISSES, MUH MONTY HALL, MUH 1/3!

Jesus fucking christ, drink bleach, all of you

I am so braindead...

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>Two independent events
>one is guaranteed
Which makes them dependent

I have never seen someone with less reading comprehension than you
Get laid, read a book, and get off the internet

>It’s not 1/3 because the 4th you are removing was never part of the equation you fourth grader
That is what makes it 1/3. Instead of 1/4. You can't remove it twice.

That's not even a percentage. That's just making every fourth attack a critical hit.

No, good design would be nerfing the bad probability and buffing the good. So if you have a 90% chance to hit (10% chance to miss) you would actually never miss. And if you only have a 25% chance to crit you get poverty crits as well as being able to luck into multiple crits in a row.

I think there are 3 outcomes but they are not equally likely.
It's a 50% chance the first hit doesn't crit. Since this only leads to [Hit, Crit], that outcome has a 50% chance.
The first hit being a crit leads to a 50% chance of the second hit being a crit. Being 50% of 50% means the chance for the [Crit, Crit] is 25%.

There's a 50% chance each hits are crits. The first hit has a 50% chance to crit. It was a crit. The second hit still has a 50% chance to crit. It will either not be a crit (50%) or be a crit (50%). Therefore, at this point of the calculation, after the first hit critted but before we know if the second will, there is a 50% chance that both hits are crits.

50%. Final answer.

>f3
>retard
>60 hits
kek

How can you apply Bayes theorem here? The two events have no impact on each other. One of them is a crit. For both to be a crit the remaining one must be a crit. So it's 50% chance. It's the same as how you can't say it rained today, so it's unlikely to rain tomorrow.

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You can’t remove it at all it’s not there
The possibilities are 1 hit 1 crit and 2 crits

I have the best reading comprehension out of anyone in this entire thread and I have proved it already. You, on the other hand, keep misreading my posts, let alone the problem in the OP.

protip, shut the fuck up and read this

You miss an enemy twice.
One of those misses could have been a crit.
Which one?

You can crit the dead body and cause it to explode, causing AoE damage to surrounding foes (also with independent chances to crit).

this is the only correct answer

we know one is a crit, which means the other is still left to chance. since the odds of a crit are 50% so the odds of two crits is 50%

>1 hit is guaranteed a crit
so it's 50%. The first hit literally doesn't matter since it's a guaranteed crit, remove it from the text as it's redundant.

What’s the other 25% user.

>The first hit has a 50% chance to crit. It was a crit.
Not what it says, read more carefully.

You're mistaken it's just that
>it happened
The question stated that it must have happened THEN in that one was a critical hit. While it may not have been guaranteed, you know that in fact, one of the two critted. That's why we use the word guaranteed. Otherwise, you admit that the order of the crits doesn't matter and therefore the answer is still 50 percent.

This is a flat out troll image, the question is ambigiously worded in a way that makes it impossible to answer and I really thought more people would call it out instead of trying to "solve" it

How many of them were crits?

But there is a third option where you git at least one crit: the first hit didn't crit, but the second still did. Out of the three possible scenarios where you get at least 1 crit, there is only one scenario where you get 2 crits. 1/3.

5% because crits only happen on a 20

Me use big word will not pass you your stat exams, fren

remember that if you avatar fag and think that it's 50% you don't know how probability works

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>One of them is a crit
Which one?

Because the fact that it could be one or the other is the issue.

But you don't know which hit was the crit.

Step back and think about what you’re saying
You’re saying that the thing you are accounting for by removing one possibility, changing it from 4ths to 3rds, is impossible and thus shouldn’t even be in the statistic in the first place. You’re removing something that isn’t there

See

This guy is smart and knows his shit.

Read:

It doesn’t matter, the relevant result is when they’re both crits.

Impossible to answer isnt quite right
There are 3 solutions that are possible based on the information we have, there's just not enough info to narrow it down further than that

>newfag
We get those threads way too often.

Don't worry these threads are just baits baiting baits anyway like pic related

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thank you for finally making me understand what all this retards are thinking when they say it's 1/3

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It's not that hard to understand the two views on the problem if you think about the underlying code.

Team 1/3 considers two independent fair draws, you can build a simple tree (in which non crit/non crit is a thing) and then calculate the probability of having crit/crit considering at least one crit. It's kinda annoying for the programmer though since he can't enforce the 'at least one crit' rule with that kind of fair draws. He would need to toss away noncrit/noncrit draws till he gets one that has a crit. It's just like user's simulation in this thread, the player would see that 1/3 of the attacks are double crits.

Team 1/2 just considers the whole thing is fucking rigged from the beginning and noncrit/noncrit isn't even a possible outcome, you don't even need to calculate conditional probability in that case since it's always true. The programmer will just randomly choose one of the two attacks (this is why some anons here say crit/crit and crit/crit may actually be different things) , give it a guaranteed crit and then proceed drawing the other attack outcome with p=0.5. The player would see double crits half of the times. It's kinda boring but that's what the developper would choose I guess.
(If double crits happen 1/4, developper kun is a faggot and fucked up)

>The first hit has a 50% chance to crit. It was a crit.
Did you even read your own post? How can you say the first hit was a crit for sure? If you calculate the chance after the first hit it's obviously 50%, but it's like calculating a 1/100 after 98 outcomes are already known.
If you don't have info about the first outcome the answer is 1/3.

I'm assuming people are thinking 10 and 01 are two different outcomes and the answer should reflect such, despite the question never specifying the order being relevant(it just says at least one is a crit, it doesn't matter whether your first hit or your second hit is a crit, thus the two outcomes can't be counted seperately)
11 00 10 01
00 is not possible, per the question description
01 and 10 count for the same percentage of probability, because it just says at least one is a crit, it doesn't say chance that the first is a crit or chance that the second is a crit
11 is the other outcome
the answer is 1/2, anyone saying 1/3 is either retarded, trolling, or is skimming over it and thinking they're a genius because they think they see the monty hall problem
the key difference here is in the monty hall problem the order of 0s matters, so the number of possible outcomes is larger

Assuming actual randomness, and assuming the first hit already crit, the only variable here is the second hit.

The answer is in the description: 50%

Well, then it's not your interpretation that's wrong, just your math, because under that interpretation, the mathematically correct answer has been adequately demonstrated to be 1/3.

>cheese is not removed from the last burger

step up your game brah

>retards
So you admit that you understand the evidence in front of you, but you refuse to admit that you were wrong? I hope you're at least a 1/2fag and not a 1/4fag

I still contend that the interpretation that yields 1/3 is the only valid one because it takes every statement as straightforwardly as possible.

You’re adding aspects to the problem that weren’t there in the original.
We hit twice. We don’t know if we killed anything. We know we have a 50 percent chance of critting normally and that one hit is going to be a crit. What is the chance that the other hit is a crit too

Nowhere does it say that the first hit is a crit. It says at least one of the hits is a crit.

Crit followed by Hit.
>[Hit, Crit] - 50%
>[Crit, Crit] - 25%
>[Crit, Hit] - 25%

I thought the amount of fries was usually different

>1 hit is guaranteed a crit
>hurr it must mean the first hit is the crit one
I think 1/2fags are trolling at this point

that's baby mode

>over 500 replies
Thanks for all the (you)s, Yea Forums. ;)

Here, this is the empirical answer

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That interpretation only works if you consider all of them as valid but 1/3 as the "most correct", not if only 1/3 is valid

Are you even trying anymore?
Of course it fucking matters.

i understand what they are thinking but that's not how probability works

But you already said crit/hit and hit/crit are identical and make up 50% of the statistic

>You’re saying that the thing you are accounting for by removing one possibility, changing it from 4ths to 3rds, is impossible and thus shouldn’t even be in the statistic in the first place. You’re removing something that isn’t there
Normally when you hit twice there's a 1/4 chance to get two crits, given a 50% crit chance. But we're not in the normal situation because we've been given extra information. So we take (hit, hit) out of the equation. Whether you want to phrase that as taking it out or never taking it into account in the first place, now there is an actually useless semantic distinction.

So it checks if at least one is a crit
So
Crit no
Crit crit
No crit
So 33% if actions are recorded in order
If it is recorded by how many crits and not order it would be 50%

It's 0% for fuck's sake why are you retards still arguing over this shit and posting bullshit explanations to justify your "muh 1/3, muh 1/2" faggotry
Actual fucking brainlets

The question is asking what the chance is of them both being crits. a non-crit isn’t relevant. All we care about is crit/crit.

The question isn't "What are the odds of this scenario happening?", it's "What are the odds both are crits?". The only thing standing in the way of getting two crits is the 50% chance on one of the hits, because the other one is confirmed to be a crit and you have a 50% to crit no matter what. You're trying to argue it from a place of not having any knowledge of how the hits will land when the question gives you that knowledge.

its 50 percent because the condition that the first one was a crit was already given (such that)
t. took probability and statistics in uni

>print(twocrits / onecrit)

oh shit nigger what are you doing

You must have failed both
t. passed probability and statistics in uni

This. I find it amusing so many Anons, are missing this important little fact.

no, it's not, this counts 10 and 01 as two separate outcomes, the wording in op doesn't count them as two separate outcomes, because it doesn't matter if the first is a crit or the second is a crit, it just matters whether one is or they both are

IT depends on what you mean when you say that at least one is a crit
if hit 1 is always a crit it's 50% for double crit
if hit 2 is always a crit it's 50% for double crit
if you remove from the roll pool the possibility of no crits than it's a 33% for double crit
if you make it that the 2nd roll is always a crit if the first roll wasn't one than you have a 25% for double crit

i passed ;)

Yes, hit/hit is impossible. But that doesn’t make it a 1/3 probability because that possibility is replaced by occurrences of hit/crit and crit/hit. It’s not out of 3. It’s 4 possible things.

>01 and 10 count for the same percentage of probability, because it just says at least one is a crit, it doesn't say chance that the first is a crit or chance that the second is a crit
There's your mistake. Because it's 50% crit chance for both, I think you will find that the combined chance of getting either 10 or 01 is twice the chance of getting 11.

see

>The question isn't "What are the odds of this scenario happening?"
>it's "What are the odds both are crits?"
user, "both are crits" is the scenario for which it's asking the odds...

There is no "forced" crit. You're looking for solutions that have at least one crit.

Obviously it's 0%
t. im smarter than you

I beg your pardon?

>01 and 10 count for the same percentage of probability
Each. EACH. You have two different 25% outcomes but you lump them together in a single 25% because? Without the "at least one is a crit" part how do you split the remaining 75% between 11 and 00?

P(B|A) = P(A and B)/P(A)

P(A and B) = .5*.5 = .25

P(A) = .5

.25/.5=.5

Probs is 50 percent.

you're thinking about it too much
we're told one is a crit, that rules out 00 as an outcome

>We don’t know if we killed anything.
Entirely irrelevant. It's purely for the purposes of the explanation. The damage and HP are likewise entirely arbitrary and made up for the example.
>one hit is going to be a crit
No. One was a crit. In retrospect.

A probability tells you the chance of something happening.
If the chances of not getting the desired outcome could vary between two and a million, you'd think that WOULD matter.

Noncrit-noncrit is literally an impossible outcome.

>At least one of the hits isa crit

With this in play, the only possible outcomes are nc-c and c-c. The only variable is the hit that is not a guaranteed crit. The description clearly states the crit chance for a single hot is 50%.

Therefore, the answer is 50%

I'm smarter than you.

If you don’t get a crit the first time you will crit the second. 50% failure 100% success
If you get a crit the first time you have a 50 percent chance of it being a normal crit or a non-natural crit. 50% success 100% success or 100% success 50% success

It's not replaced with anything. Those occurrences are present either way.

because 00 is not a possible outcome, it says that in the problem very plainly

are all you 1/3 idiots just ESLs? is that what's going on here?

OP's question is worded so you can either asume the attacks are about to happen (aka the crit is going to be "forced") or it already happened and you are looking for the solutions like you said. This is not a mathematical problem, nor is a reading problem, it's an question to make Anons run in circles.

You aren't incrementing numberofruns if you get a 0,0 since you have a while loop, and are either being disingenuous or retarded. If change that you will get 1/2.

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25%, 33% and 50% are equally correct
What this means you can argue amongst yourselves

THE ROLLS ARENT INDEPENDENT MORONS STOP ASSUMING THAT

What’s the difference between it being in retrospect and being a given, functionally

Cope 50%er

If you acknowledge that one hit must come before the other, then you can't help but treat them as separately.

>B-BUT IF YOU CODE BADLY ON PURPOSE ITS 50%!
sneed's cope and seethe

Because there must always be one crit, any hit/hit will become a hit/crit or crit/hit.
It’s overwritten. They’re replaced not removed.

>two separate attacks
>their crit chance is not independent

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Doubtful.

If you say 1/3, you paradoxically say that 01 and 10 don't have any guaranteed 1s but you also throw out 00 because there MUST be at least one 1. If there are no guarantees, then why is it that 00 isn't viable? Why does the order matter for 10 and 01 but the order doesn't matter if both hits were in fact 1s? You may as well reduce it to 2 1 0 and then rule out 0 to find that in fact that the chance is 1/2.

Ok, but I said 33% is just as viable, because it really depends, but sure, keep that "us vs them" attitude my man.

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Yes, move the goalposts so you can be partially correct instead of completely incorrect.

Given hit/hit is impossible, explain why it’s 1/3 and not 50%
What happens to the 4th

You're not really understanding my point though I admit it looks silly, but I went into it afterwards. Splitting it into the possible different scenarios doesn't have any effect on the actual crit chance which is the ultimate arbiter on whether you actually get both crits or not.

You're fucking stupid.

Yeeeees, and that leaves 10, 01, and 11. And your mistake, as I just pointed out, is treating 10 and 01 as the same outcome, when the odds of having EITHER 10 OR 01 is twice the odds of having 11.

yes I can, because the question isn't asking the probability that one is a crit, it's asking the probability that both are crits
I think this is the key disconnect that's happening here, 1/3 people are solving the wrong problem because they see this and they immediately think monty hall and as a result they "solve" that instead of solving this

Retard. You made the same mistake as the codefag

Well, if you're clairvoyant, nothing. But if you're not, then knowing the outcome beforehand would require the odds to be calculated differently.

It's 37.5% because of pseudo-random distribution modifiers

It's not that they're removed, they're just not a possible outcome in the scenario presented, so we disconsider it.

Irrelevant. It does not specifythat order matters, therefore we still have one guaranteed crit, and one variable no matter how you look at it.

My original post was me saying that both can be correct depending on how the "one of the hits is a crit" part of the questions is interpreted, but sure, go ahead, revisionist user. You are so smart. Want a cookie?

It's literally in my first stats course in uni course and that's how they word all of the questions. Get fucked.

Completely wrong.
P(A|B) = (P(B|A) * P(A))/P(B)
= (1 * 1/4)/(3/4)
=1/3

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Good God. No. You are entirely wrong about everything you just said.

No, your "order doesn't matter" argument has nothing to do with 00 being impossible. You still merge 01 and 10, two distinct 25% outcomes, into a single one just because an unrelated 00 is not possible, and therefore you have 50/50 between 11 and your 0110 monstrosity.

I'll ask again, let's assume 00 is possible, are 10 and 01 still the same? So 11 is 1/3 instead of 1/4?

they are the same outcome, because the question isn't asking the chance that one is a crit, it's the chance that both are crits, the order doesn't matter, what matters is whether you get 11 or some combination of 1 and 0

We’re looking at this from a mathematics perspective not a personal perspective user

Low IQ:
25%, 33.3%, 50%
High IQ:
0%, 12.5%, 20%, 37.5%, 75%

It's actually worded in such a way as to make it clear it already happened, 50%fags are just really keen on assuming things anyway.

>50%
If I crit at least once, what's the probability of the second hit being a crit.

>33%
If I attack twice, what's the probability of getting a crit.

The possible results are hitting and critting or critting twice

If you differentiate based on order you have to differentiate based on which crit was not a result of the conditions

cool, except that's the solution for monty hall, and this isn't monty hall, because the order of 1s and 0s doesn't matter

THERE I S ONLY ONE VARIABLE YOU RETARDS

Why are you calculating a sample size of 2 out of 3 retard

Yeah, my mom!

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The probability of getting a crit is 1 tho

if you want to change the problem and for some reason say 00 is possible despite being explicitly stated as being not possible, then yes, it would be 1/3

It's 100% because my build has 100% crit chance.

no crit/no crit

Look, man, it's like this. 00, 10, 01, and 11 are all equally likely. 00 didn't happen. So one of these three equally likely things did happen. The order of the hits is the same in all those outcomes. First the first hit, then the second. It is precisely because we have two separate hits that we can count 10 and 01 as distinct outcomes. There is only one way to get 10, and there is only one way to get 01. But that makes for two ways to get one 1 and one 0.