Does anyone know what the answer to this would be? I saw it on /v/ earlier and no one could seem to agree if it was 1/2...

It does matter because you're twice as likely to get one Crit one not Crit than two crits.

It's 1/2 OP. 50%.

There is no ambiguity here. /v/ is brainlet tier.

>dis guy saying hits are indistinguishable
>the absolute state of american education

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That is just wrong. One of them is guaranteed crit. The other crit is 50% chance. This is why the order is irrelevant here. You are not twice as likely to get "one crit one not" because the chances of the non guaranteed crit are 50%.

There is no reason to think order matters here. One of them is 100%, one of them is 50%. Whether the guaranteed crit comes first or second doesnt change this

you'll not get any crits because of game engine

Two hits.
Both have to crit.
1 always crits so you can ignore that one.
You're left with 1 hit with a 50% chance to crit.
EZ.

or the answer is always 50/50, it either happens or it doesn't.

Think about it as a coinflip experiment. Instead of crit or non-crit it's heads or tails.
Flip two coins a thousand times each. Now throw out all the ones where neither is heads. So you're left with all the ones where coin 1 is heads and coin 2 is tails, the ones where coin 1 is tails and coin two is heads, and the ones where both are heads. Are half of the remaining tosses going to be both heads? No, about 1/3 will be both heads and 2/3 will be mixed.

>Conditional probabilty: P(A|B) = P(AnB) / P(B)
>A: two crits
>B: at least one hit is a crit
>P(AnB): probability that both hits Crit and at least one is a Crit, P(A)is a subsection of P(B) therefore their intersection equals P(A), probability of two crits =1/4
>P(B), probability at least one hit is a Crit: CC, CN, NC, NN, 1/4 each, every outcome except NN, therefore equals 3/4
>then P(AnB)/P(B) = (1/4)/(3/4)=(1/4).(4/3)=1/3
1/3.

it's 25% bros

But you are missing the fact that you don't need to flip two coins. We are already being told what one of the coins is. It doesn't matter the probability of two coins being flipped.

It depends on what the rules are for at least one of the being a crit. If I dont crit the first time, is some magical being going to make sure that the second one does? Or has someone looked into the future and assured me that at least one is a crit?

Miss? the question says you hit twice.

The magical being sees two hits into the future, if your next hit doesn't Crit it forces the second hit to Crit. If it does Crit it doesn't intervene again and the second hit is up to chance.

They're retards. It's one of those simple questions that adds extra information solely to confuse you.

Well by those rules it's 1/2. But that isn't probability, it's divine intervention.

>into the future
Why are you looking at it like this? "at least one hit is a crit" implies that both hits already happened.

What? I dont understand what you mean

Jesus guys, it's a famous paradox. It's intentionally ambiguous.

en.wikipedia.org/wiki/Boy_or_Girl_paradox

Unless you say 1/4, then you're wrong.

If you assume it like that, then its 50%

If you replace the second sentence with "One of the hits is guaranteed to be a critical hit", I think that erases all ambiguity and puts it at 50%.

>"This extreme assumption is never included in the presentation of the two-child problem, however, and is surely not what people have in mind when they present it."
It's 1/3.

>at least one of two is X
>one of two is guaranteed to be X, the other may or not be X
It's the same thing.

how is the answer not 1/4? with a 50% crit chance, the chances of getting two is a row is 1/4. I dont understand how that changes because of the other information.

it can be cleared up by either
>at least one of the two WAS X
>at least one of the two WILL BE X

because it's worded in present tense, we don't know if it's this bullshit
or if it's an observation of what happened.

but only the second one makes sense in the real world. god probably isn't going to drop in on your game and promise you a critical hit.

The question is ambiguous, it depends on how you are told:

If you are being told "there were two hits, and at least one of them WAS a critical hit" this puts the answer at 1/2.

If you are being told: "if you hit an enemy twice, then at least one of them WILL be a critical hit", this puts the answer at 1/3.

youve arrived at the conclusion 1/4 by looking at the probability matrix:
nC nC
C nC
nC C
C C
and taking the number of desired outcomes (1) and dividing it by the number of ALL possible outcomes (4). However, the problem somehow forbids the possibility of nC nC, so the total number of possible outcomes is no longer 4

>no (you)s
I'm sorry, fellow mathanon.

This, but brainlets don't get it and get confused.

As with the Wikipedia article though, the assumption for 1/2 is bullshit. You're basically picking a situation at random then rerolling if you get two negatives. This makes NC and CN the same thing.

the only brainlet here is you
and most of this thread, actually

It's 1/3
Given that at least one of the hits is a crit and 50% crit chance, we have a sample space of three outcomes with equal probability
HM (hit miss)
HH
MH
One of these three outcomes has both crits. Since the outcomes have equal probability, the chance of having both crits is 1/3.

There's not enough information

>t. autist that needs to go back to school
If the first isn't a crit the second will be
If the first is a crit, the second has a 50% chance to crit, since each hit has a 50% chance to crit
You're not smart

25% chance. There's not a 50% chance you flip heads on a coin 20000 times. Formula for this is (percent chance of something happening as a fraction)^(the amount of times you want it to happen). So in this case it is (1/2)^2 which is .25 or 25%

It's okay, I only ever post to satisfy my own autism.

But this is relying on the assumption that both hits have not already occurred. If they did already occur, there is no reason to have both HM and MH

>brainlets
brainlet spotted lmfao. The reason it isn't 50% is because it could be (NC, C) or (C, NC). The total possible outcomes is 3, not 2, and we want to chose 1 of them. Therefore, 1/3.

3 possibilities:
first is crit, second is not
both are crits
first is not crit, second is crit
they all have equal probability (0.5*0.5 originally, then disregard situation with no crits)
1 in 3 times, you will have two crits
answer is 1/3

Except 1/3 is impossible, fucking brainlet.

Both of you are brainlets
en.wikipedia.org/wiki/Boy_or_Girl_paradox

I was assuming that at least one hit was a crit. That leaves the 3 options with at least one crit.
However, 2/3 of options with at least one crit have only one crit, so the chance of two crits is 1/3

Mutebloxxors

it's 50%
the order doesn't matter. you're only calculating the crit chance of one hit.

You're the brainlet here
Check out who solves it with Bayes' theorem

but there are not 3 options. there are two options
1. both hits are crits
2. one hit is a crit, and one hit is normal
thus, 50%. in option 2, the order doesnt matter

What is a crit?

you're forgetting to count both possibilities with one critical hit
- first crit, second not
- first not, second crit
- first crit, second crit
these are all equally likely, so it's 1/3

If you actually bother to read any of the analysis the steps required to get 1/2 are complete bullshit. You have to pick a family at random that doesn't have two girls, but you don't know that until you select the family. So you automatically assign one to being a boy regardless of whether it actually is.

but the probability would only be 1/4 each if NN was possible. since its not possible, it cant be 1/4 each.

There are 3 options which have at least one critical hit. They are all equally likely.
1. first is critical, second is not
2. first is critical, second is critical
3. first is not, second is critical
One of these three options has two critical hits, therefore the probability of two is 1/3
You could also divide it into one critical hit vs. two critical hits, but you're ignoring that getting one critical hit is twice as likely as getting two.

>since it's not possible
It's not possible at the conditional step, P(B) is done before then.

the order doesn't matter. how hard is that to understand.

why the fuck are all these retards pretending order matters in this thread

That theorem involves both conditional probabilities and absolute probabilities. The absolute probability of each outcome is 1/4. With the condition that there is at least one critical hit, you have three outcomes with conditional probability 1/3.

How is 1/2 bullshit? Its a completely reasonable scenario. If I tell you that I have two children, and one of them is name Samantha, what are the odds that both are girls? How can you possibly say that its 1/3?

If order didn't matter, if you flipped two coins, two heads, two tails and heads and tails would all be equally likely. Pro tip, they aren't.

the assumption that you can ignore the 100% chance crit and just look at a single hit with 50% chance is wrong
there are two cases where there is only 1 crit:
first hit crit + second hit does not crit
or
first hit does not crit + second hit crits
and only one case where both hits crit:
first hit crit + second hit crit

3 potential cases by the terms of the setup, means that each case has a 1/3 chance meaning that both hits crit is 1/3 chance

you are missing the fact that one of the crits is 100% and this actually leads to 4 outcomes.
-first 100%, second 50% chance of crit
-first 100%, second 50% chance of not crit
-first 50% chance of crit, second 100%
-first 50% chance of not crit, second 100%

4 possible outcomes, and in 2, both are crits. so 1/2

The order doesn't matter, which is why both orders count as having one critical hit and one not critical. You're twice as likely to have one critical hit as you are to have two.
It's counterintuitive, but if you sit and think long enough it will make sense to you.

>the order doesn't matter
>it's 1/2
I get what you're saying, but in order to make that claim you would have to say that, for instance, the crit occurs on the first hit. Because that statement wasn't in the prompt, we have no idea when the crit will occur. 2 hits are made sequentially so there are 3 possible outcomes: (NC, C), (C, NC), and (C, C).

but we already know the outcome of ONE of the flips, which changes the equation.

>assuming 50% chance of crit
>one of the crits is 100%
no... both are still 50%, but you can disregard any scenario in which neither one hits that 50%. that isn't the same as one of them being 100%.

the order doesn't matter, so there are only two outcomes. either they both crit, or one crits. and you only need to calculate one crit, with a crit chance of 50%. so the chance of both critting is 50%. its not hard to understand.

1/3 only seems that way because you're breaking it down as if the order matters. We KNOW one is a crit. The chances are 50% that both are crits, since the other has a 50% chance.

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>you are missing the fact that one of the crits is 100% and this actually leads to 4 outcomes.
That's not how conditional probability works.
To find the probability given that there is at least one crit, you first take all outcomes from the initial sample space where there is at least one crit, and then find what the probability of two crits is compared to not having two crits.
We take three equally likely possibilities:
1. CN
2. CC
3. NC
As you can see, one of the three has two crits. Therefore, the probability is 1/3.
This is because the absolute probability of one critical hit is 0.5, compared to 0.25 for two critical hits. So given that you have at least one critical hit, the probability of two is 0.25/(0.5+0.25), which equals 1/3

Because preselecting like that is just dishonesty designed to confuse people like the article said. You're twice as likely without preselecting to have one child of each sex than two girls.

>a wikipedia article describing the ambiguity of this question was posted
>retards still arguing about an intentionally paradoxical problem
retards

You present 2 outcomes, not 3. Fucking brainlet.
1 crit and 1 not crit = 1 not crit and 1 crit

>the order doesn't matter, so there are only two outcomes. either they both crit, or one crits.
Let's start this off from the beginning. Four possibilities, all equally likely.
1. NN
2. CN
3. CC
4. NC
The condition is that there must be at least one critical hit, so we discard 1. This leaves us with three outcomes, still with equal probability.
2. CN
3. CC
4. NC
The probability of two critical hits given these options is 1/3, since they are all equally likely.
I would recommend looking up Bayes' theorem and conditional probability.

we know one is a crit, but which one, user? There are 3 total possible outcomes, and you cannot just disregard one of them even if both satisfy your condition. Doing so wouldn't make any sense.

>not getting the point
We're all right in that it's too ambiguous. That's the point of this arguing. We can all smugly call each other dumb for not getting it and still be right.

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Yeah the but the way the ambiguity is generated is bullshit.

this user is correct
only retards who don't understand probability think it's 50%

There are three outcomes in which there is at least one crit.
CN, CC, and NC
You're attempting to group CN and NC into one possibility. If you group them as "one crit", that's fine, but you must realize the chance of one crit is twice the chance of two crits.

the order doesn't matter, so why are you treating NC and CN as two different outcomes.

Its 50% wtf is wrong with this dumbass 50iq board no wonder you are fucking robots

Well shit, we better make it 2/4 because NN and NN are totally different because of the order.
Let's simplify that a little...

Ah, the chances are 1/2

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the only point in arguing this is to enlighten the retards in this thread who want a neat and tidy shitty english language. im smart and know it's ambiguous so i refuse to help others while i gain nothing. slaves should stay slaves and masters should stay masters

Order is the way of describing the "one of each" outcome as being twice as likely as two crits.

>You're twice as likely to have one critical hit as you are to have two.
Not if one of the hits is guaranteed to be critical

1/3
Fucking idiots

Look at this , fucking brainlets.

Here for the intuitive explanation.

>but you must realize the chance of one crit is twice the chance of two crits.
False. There's a 50% chance (even) of one crit or two crits.

>tails tails is just as likely as heads tails and tails heads together
Lmao

>To find the probability given that there is at least one crit
But we already know this. Its 100%.

Its not ambiguous. The answer is 50%. There is no way you can seriously tell me that the question implies the order actually matters. There is absolutely nothing in the wording of the question, that indicates order should be taken into account.

Because they are two different outcomes
1/3 chance of NC
1/3 chance of CN
1/3 chance of CC

If I flip a coin, my chances of getting one heads and one tails is 0.5, since the chance of first heads second tails is 0.25 and the chance of first tails second heads is 0.25. Do you understand that both those outcomes are distinct, and both involve getting one heads and one tails?
This question is asking, given that you have at least one heads, what is the chance you have two heads?
So you take the chance of two heads, 0.25, and divide it by the chance of at least one heads, which is (0.5+0.25), or 0.75.
0.25/0.75 is 1/3
Do you get that if you flip a coin twice, the most likely outcome is getting one of each?

You are missing that fact that "guaranteed crit" and "50% crit" are actually two different variables.

Anyone saying its anything other than 50% is a brainlet

the problem is the 100% can't be one or the other randomly, otherwise it wouldn't be 100%. it has to be the first or second 100% of the time.
you're dual wielding two swords, the left sword has 100% crit chance, the other has 50% crit chance. therefore it's crit + not crit / crit + crit (or not crit + crit / crit + crit)

Its still 50/50 if both are crits u dense retard