You should be able to solve this

You should be able to solve this

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en.wikipedia.org/wiki/Boy_or_Girl_paradox
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50%

1 crit / 2 crit / no crit

33.3 %

75%

but 1 is 100% crit

>no crit
>at least one of them is a crit
Wew lad

its 50%

because you hit him twice.One of the hits are already known to be a crit for both of them to crit you just need to crit once meaning its 50%

question is made by someone who thinks he is a mathematical genius or something but in reality the question is fucking retarded

1/2 * 1/2 = 1/4 or 25% chance

oh then what said /thread
pack it up boys shows over

ATLEAST ONE HIT IS A CRIT, NO MATTER FUCKING WHAT. NOW WE ONLY HAVE THE OTHER HIT TO LOOK AT. SO 50%

1: 1st crit, 2nd crit
2: 1st crit, 2nd don't crit
3: 1st don't crit, 2nd crit
1/3 chance.

>At least one of the hits is a crit
First hit= crit 100%
>Assuming 50% crit chance
Second hit= crit 50%/no crit 50%
= 50%

50%, suck my dick, nigga

It's literally crit or not crit one of the times, so you can eliminate that time. After that, it's just 50/50

So much ameritard here.... 0.25 is the answer

this user gets it

Mathematican here,
Let A be the first hit and B the second. There are 4 possibilities:
A=crit B=crit
A=crit B=x
A=x B=crit
A=x B=x

We have the condition that at least one hit is crit so we only consider the first three cases. That means the possibility is
1/3= 33.333%

75%
100%(0.5)+50%(0.5)=75%

1 in 2

1 crit + depend on how many probabilities?
0.25

Right answer, wrong method. Draw a punnett square: there are 4 possible outcomes (ab, Ab, aB, AB). We know from the question that one of them (ab) is off the table and our solution is one of the other three. Of the other three, what percentage are double-crit? 33%.

4 different possible outcomes. 2 of them have 2 criticals. 50% chance.

This is the same as the coin problem. It's 1/3.

66%

can someone please legit answer this

...

⅓ you idiots

>at least one of them is a crit
>A= x B= x
pure autism

82%

A=x and B=x can't happen. And if you are counting all possible outcomes where we know one is critical there are 4 possible outcomes with 2 of them having 2 criticals.

It's obviously 50%. Next question.

its 50% mate

Yep easy. Make the problem harder. You hit the enemy 4 times the 1st and 3rd hits are crits. You got at least 3 crits. What is the probability all hits were crits?

73%

37.5% chance

1/2 * 1/2 = 1/4.

C = Crit, N = No crit

NN, CN, NC, CC

Only one instance of four possible outcomes is both crits.

25% chance.

67%

.3 repeating.

>35 replied
>25 posters
fuck off

>We have the condition that at least one hit is crit so we only consider the first three cases.

Do you autists even read?

the question is "what's the probality BOTH hits are crits?"
C:0.5 then C again=0.5 (or /C=0.5) so 0.5*0.5= 25%....
>>at least one of the hits is a crit
So fcking useless cause the question is BOTH are crits
fags

Still 50% assuming the hits are independent of each other. With one of the two trials having already been completed, we can look at the other hit isolated.

>A=x and B=x can't happen
read my post again
>here are 4 possible outcomes with 2 of them having 2 criticals.
No. If you don't understand probability theory than don't post sit.

If someone's interested: This problem is called the "Boy or Girl paradox", see
en.wikipedia.org/wiki/Boy_or_Girl_paradox

Except there aren't 4 instances because one is a guaranteed crit dumbass.

did you read the rest of my post. Please explain.

Mathematician here: 50%

Every scientific percentile is a 50/50 chance when devised by retards like you. Either the desired effect occurs or it does not.

...

No, see or the link in Your fault is that you assumed that A and B are independent. That's not the case.

P(crit n crit | crit) = (crit n crit n crit) / crit = (0.5 × 0.5 × 0.5) / 0.5 = 0.25

1/3 or 33% no arguing.

Wrong. You are not a mathematician.

50....

HC
CH
CC
All these are equally likely to happen since the chance to crit each time is 50%.

The probability of hitting both crits=1/3

Every other answer is wrong.

50 percent u retard

That's not how it works.

1/3 nigga

forget to write text...

the chance that both are crits is 0,25
we can't just ignore the possibility that both won't crit

That's very far from the answer. It's 1/3

the probability of the pool is not affected by the base chance. so the probability that the other hit is crit is just 50%. it isnt affected by the other one.

Yes but you sound like an asshole when you say mathematician here.

This all things in life boil down to it will or it won't.

1/3

there are only 3 cases possible:
First hit crit, the second don't crit.
First hit crit, second crit
First hit don't crit, the second crit.

so there are 33% chance of a double crit.

the correct answer

We can, it's literally stated in the problem that at least one of the attacks will crit.

a voice of reason in a sea of filth

...

Yes we can. That's what called "conditional probability".

well you calculate each one indenepndently, its not like well the first one was a crit so 1 is 50% of 2 and it was a crit so 2 wont be a crit..

no its each on its own, at each hit its 50-50 crit or not. In theory you could hit 1000 times and non of them could be a crit,, or all of them could be, if you were lucky. Altho 50% is pretty large so thats unlikely but,, not impossible.

so one is a crit no matter what, then its 50%, since its either the other is a crit or not, no other option.

Yes we can. Yes we do.

Dear God...

The world is heading into dark places.

I object.
The proof is kinda complicated, but regarding the theory of conditional theoretical chances, the answer is 0.5

50%
since the crit chance doesnt drop after a crit is made each attack has an independent 50% to crit

25% chance. ezpz.

Well, at the moment I'm working on my master thesis in math so I would say I'm a mathematician.

thats my thought as well. the second hit isnt dependent on the first and the result of the first hit has already been established.

>mfw niggers really arguing the order in which the crit is going to occur and making it a probability in outcome percentage

its 50%

Thank this guy
>youtube.com/user/MindYourDecisions

His videos taught me a lot about probability and logic

>math is hard

Jesus Christ, it's fucking 33%, if you're really that stupid just use Google.

No you can't treat them as separate events one the question is about multiple events happening.

explain

The people who don't say 1/3 are the same people who don't know the difference between your and you're

itx 33%, any other answer is mentally fucking retarded and cant do simple math

25% obviously what the fuck is wrong with you dumb mongs

Fine you can look at it as 1 crit or two cits, just 1 crit is twice as likely so you know, still 33%

Anyone who says something other than 1/3 is clearly underage, as anyone who's completed high school should be able to calculate such basic conditional probabilities.

Wrong, it's 1/3. Rather famous question actually.
Three possible outcomes, one of them has two critical hits. Answer is therefore 1/3.

1/3ed

normally there are 4 outcomes
nc/nc
c/nc
nc/c
c/c

The question removes nc/nc as an answer

This leaves us with
c/nc
nc/c
c/c

so it's 1/3

A lot of peopel say 1/2 because they make the incorrect assumption that nc/c and c/nc are the same.

They are retarded.

Wrong, it's 1/3. Rather famous question actually.
Three possible outcomes, one of them has two critical hits. Answer is therefore 1/3.

On a table I have an apple and a coin. If the apple is an apple and the coin is a fair coin, what's the chance that a coin flip will result in heads and me also having an apple on the table?

50%

Wrong, it's 1/3. Rather famous question actually.
Three possible outcomes, one of them has two critical hits. Answer is therefore 1/3.

one result has already been abrogated by assuming that we already have a crit. so the answer is .5

ding ding

No, we do not fucking ignore the possibility of neither hit being a critical. If that was the case, then would we not also ignore the possibility of both hits being crits?

Say the question was 'what is the possibility both hits aren't crits?' Would that be impossible to work out?

This

Wrong, it's 1/3. Rather famous question actually.
Three possible outcomes, one of them has two critical hits. Answer is therefore 1/3.

you are wrong, but its ok to be retarded

everyone is just over complicating it when the question is literally whats the chance a hit will crit when its a 50-50 chance to crit.
theres no chance of missing since question never gives us a hit chance thus its guaranteed to hit