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Anonymous (ID: AEaagfGG)

2014-12-03 21:04:26 Post No. 582897992

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2014-12-03 21:04:26 Post No. 582897992

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I don't think /b/ has solved this.

>>

50 percent.....

>>

>>582897992

Isn't it like 25%?

>>

50% you retarded fucks.

>>

50%

>>

one hundered percent

>>

Anonymous (ID: YFWta5UN)
2014-12-03 21:06:50
Post No.582898442

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>>582897992

It's still 50%, you faggot

>>

33%

>>

25%

>>

69% am I funny yet

>>

>>582897992

75% you degenerate fucks.

Add 50% then you have 50/2 thats 25% and add that to the 50

>>

>>582898717

what even

>>

i didn't even know /b/ was that stupid

>>

25% retards

>>

C-Crit N-normal hit

both things have 50%, there would be equal probability endings:

CC NN NC CN

due to at least 1 being crit the NN ending is out

so it can be one of the equal chance endings CC NC CN.

since equal chance and CN+NC+CC=1 it's 1/3 33.(3)%

>>

Anonymous (ID: Qi7ogWdz)
2014-12-03 21:13:16
Post No.582899626

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It's 75%

>>

>>582899626

This

>>

1/3

Easy

>>

first hit 50% roll

if crit, next roll is another 50%

if not crit, next hit is 100%.

50%-50% is the only possibility.

25% probability.

but of all possibilities, 1/3 because non-crit does not exist.

>>

>>582899459

gr8 b8 m8

>>

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

>>

Every hit is a 50% chance. It saying at least one hit is a crit is arbitrary information.

>>

>>582899459

Jesus thank you, somebody whose not retarded. It's pretty fucking embarassing when people can't even get the right answer when it's one you can actually map outcomes for

>>

50% is 1/2, 1/2 * 1/2 = 1/4 nad that iss 25%, learn math faggits

>>

>>582898442

>>582898308

>>582898184

>>582898103

It's not 50%... Take all possible outcomes...

1=crit 2=not crit

11 - 12 - 22

1/3

33.333%

>>

>>582901586

this shit is being tought in 5th grade, do u evern school /b/?

>>

Anonymous (ID: BmwTLTTW)
2014-12-03 21:27:14
Post No.582902151

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u cant die twice faggots, so 0percent

>>

Anonymous (ID: GmYt19yk)
2014-12-03 21:27:53
Post No.582902272

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>>582897992

Let hit number one be event A.

Let hit number two be event B.

Given that P(A) = P(B) = 0.5 and prior for A being 1.0, we use Bayes rule:

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

then applying the beta function for our newly assumed distribution, we arrive at P(B) = (0.5 * 0.1) / 0.3 = 0.1666

Source: I have a PHD in advanced mathematics and I have personally been awarded the highest recognition in math research by the government. Pic related is my alma mater.

>>

>>582901621

It says AT LEAST ONE of the hits is a crit.

So the only possible out come is either 1 crit or both crit.

>>

>>582902692

Then let me recant... you may be right.. I'm high, i missed that

>>

>>582897992

50% + 50% = 100% l0l i into math

>>

50% either it happens or not.

>>

Anonymous (ID: 6MSZaXZw)
2014-12-03 21:34:50
Post No.582903623

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>>582903053

It's all good. I'm waiting on Arrow to download and only read it 1,000 times until I got it.

One of us is spending our time better, and I'm pretty sure it's you.

>>

>>582902151

This.

If by critical hit you mean killing blow like D&D or something. You can't have both of two swings be critical hits in sequence if a critical hit eliminates the target.

>>

>>582903532

If it was really that intuitively obvious do you really think this would be posted?

>>

Anonymous (ID: BmwTLTTW)
2014-12-03 21:43:51
Post No.582905405

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i won?

>>

Anonymous (ID: gHpJPd/c)
2014-12-03 21:44:25
Post No.582905551

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>>582904383

>If it was really that intuitively obvious do you really think this would be posted?

>>

Given: There are two hits.

Given: One of the hits is a crit. It is not specified which.

Given: Each hit has a 50% chance to be a crit.

Normally, when making two hits with a 50% crit chance, there are four equally likely possibilities:

CC

CN

NC

NN

Per the problem, we know that NN is not the case, so we discard this possibility.

This leaves us with three equally likely outcomes:

CC

CN

NC

Out of all the remaining possible outcomes, what's the probability of CC?

1/3.

>>

Neither crit affects each other, they are two seperate events.

One hit has 50% chance, it crits

Next hit has 50% chance, it may or may not crit but it is still a 50% chance.

>>

>>582897992

0.5x0.5x100=25

>>

>>582905665

Fuck I gotta learn to read first...

>>

>>582897992

Let C denote "Critical hit" and N denote "Non-critical hit".

With initial condition p=0.5, there are four possible outcomes:

NN, with P(NN) = 0.5^2 = 0.25

NC, with P(NC) = 0.5^2 = 0.25

CN, with P(CN) = 0.5^2 = 0.25

CC, with P(CC) = 0.5^2 = 0.25

(All have the same probability, because the probability of N and C are the same. P(N) = P(C) = 0.5)

Now, we know we have at least one hit, so we can rule out NN.

The question then becomes how likely each of the possible situations NC, CN and CC are.

They are equally likely, so the likelyhood of each one is 1/3.

In other words: The answer to the question is 33,3333...%

tl:dr: You are all newfags for not having been on Tor and seeing loads of CP in your lives.

>>

>>582906236

It doesn't say the first one is a crit. This cannot be assumed.

>>

>>582905551

I agree 0.5 seems the obvious answer. I've a degree in probability and statistics and if it wasn't for the fact that there's clearly some sort of trick in this question, I probably would have overlooked it.

As said a few times already, but poorly worded, we have

P(first shot is critical, second isn't)

P(first shot isn't critical, second one is)

P(both shots are critical)

= 1/3

I know intuitively its hard to picture

>>

>>582897992

problem is poorly written, 50% of each hit being critical independently?

if so then the probability of the both hits being critical is zero

P(A or B)=P(A)+P(B)-P(A and B)

1.0=0.5+0.5-P(A and B)

P(A and B) = 0

>>582902272

this is bullshit

>>

>>582902692

>>582901621

You were close. You had 11, 12, and 22, but forgot 21.

>>

>>582907235

Not sure what your point is there. There are two separate events, each with two equally-likely possible outcomes. Independent of each other.

>>

>>582907235

ITT everyone forgetting elementary set theory

P(A or B) = P(A) + P(B) - P(A and B)

where A=first hit is critical, B=second hit is critical

>>

>>582897992

(1/2) x (1/2) = (1/4), so 25%

If you don't understand this you're a waste to society and/or underageb&

>>

It's 50%.

One is already guaranteed, period.

All your checking to see if the other one, just one hit, is a critical or not.

Since it's a 50% chance of it being a critical, and since we are only monitoring one hit since the other hit is guaranteed, the result is 50%

>>

>>582907761

No they aren't independent because the probability that at least one hit is critical is one, hence

P(A|not B) = 1

conditional probability =/= unconditional

pleb

>>

>>582897992

33%

>>

>>582897992

50% as one hit is done and as we have no evidence to make us believe that a random number generator would consider prior hits we must then assume that every new hit is its own instance of a hit and therefore the information regarding one hit being a critical is not relevant for the question.

So this leaves us 1 variable where we are told to assume "a 50% crit chance"

therefore 50%

However if we were in a situation where hits in proximity may affect other hits then we would have to have a better understanding of what type of system we are operating under

>>

1/3

>>

>>582908068

>No they aren't independent because the probability that at least one hit is critical is one

The original problem states that each hit had a 50% chance to be a crit. This is not conditional.

>>

>>582902272

Number 1 -- I don't believe the PhD or the Harvard bullshit.

Number 2 -- on the very off chance that that's true, you're an idiot.

It's one third.

>>

>>582907851

This guy gets it

>>

>>582901621

At least one hit is critical, so it is down to a 50/50 chance now isn't it genius?

>>

>>582908664

>50% as one hit is done

No, we cannot assume that the first is a crit. We are told only that one of the hits is a crit, without knowing which, and are then asked to determine what may have happened from that information.

>>

>>582906957

Ok well then lets assume the second on had crit

The first one still had a 50% chance to crit.

Again two seperate events

>>

I'm just going to say it again since I know people

are ignoring me, but I happen to be right.

The answer is zero.

The people who say 1/3 are wrong because we

have no indication that the three possible outcomes are equally probable.

The people who say 1/4 are wrong because the

critical hits are not independent, as I explain here

>>582908068

The people who say 1/2 are wrong for the same

reason the answer to the monty hall problem is

not 1/2

I gave the correct answer here

>>582907235

>>

>>582897992

50% probability BOTH hits are crits. One hit is fixed, the other hit is variable. Only the variable hit matters, and in this case the hit has a 50% chance of being a crit. Some people are thinking way too much.

>>

>>582898491

>>582900395

>>582901621

>>582908674

There lads here get it. Anyone who said anything else but one third or 33% needs to check their iq privilege.

>>

Really take 10 sec to read the question again.

What is the chance BOTH hits are critical?

0% u can't kill someone twice

all of u should be ashamed of yourselves

>>

>>582909073

No. You're failing to consider the fact that either the first or the second may be a crit, and that these are independent states.

>>582909323

>assume

That's where you're getting stuck. We cannot infer from the original problem which hit was a crit, and thus cannot assume one or the other. All valid possibilities must be considered.

CC

CN

NC

1/3

>>

Logically, it should be obvious who's right here.

1. its clearly some sort of trick question.

2. In trick questions, the majority are almost always wrong.

3. the majority of people are saying 1/2 as the answer, with very little reasoning behind it.

4. The people who are saying 1/3 as an answer all seem to be giving pretty good arguments, as well as sounding like they know what theyre talking about.

even if i had no math knowledge at all I would be pretty confident the people saying 1/3 are right

(however, I of course already know that they are right)

>>

>>582909249

that doesn't matter we know one will be true regardless

so all we have is 1 and 50% leaving us a 50% chance for the second to be true

>>

>>582909532

there are only 2 possible outcomes, cause one hit is a guaranteed crit. the possibilities are:

1 crit, 1 hit = 50%

1 crit, 1 crit = 50%

1 hit, 1 hit doesn exist, because the first hit will crit 100%

>>

>>582909360

You are wrong.

We do know the answer is 1/3.

Check out my solution:

>>582906656

I teach maths at a university level, and this is not a hard problem for me.

And even if you were unable to determine the probability, it wouldn't be 0.

>>

Anyone saying 1/3 are trolling or don't know logic. First hit, second hit, doesn't matter. There are 2 undefined 'hits' one is known to be a crit, the other has a 50% chance to be a crit. The OP asks what the probably both hits are crit, which is dependent entirely on the one unknown hit, in this case 50%.

>>

>>582909360

You are not correct.

We know that the three possible outcomes are equally probable because we know the probability of each event. It's given to us in the problem itself.

>>582909563

>implying one crit will always kill the target

>>582909837

>its clearly some sort of trick question.

True. The problem has tricky wording on purpose. but it is possible to answer the question as it is posed.

>>582909982

It does matter.

First hit, 50% crit chance: Crit or Normal, equally likely.

Second hit, 50% crit chance: Crit or Normal, equally likely.

Four possible outcomes, equally likely:

CC

CN

NC

NN

We know from the problem there's at least one crit, so we discard that possibility. We know that didn't happen.

What do we have left? One desired outcome from three equally likely possibilities.

One Third.

>>

>>582910260

you're gonna need a bigger bait

>>

Anonymous (ID: GjYLfzcl)
2014-12-03 22:11:33
Post No.582910469

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>>582909669

omfg you retarded neckbeard. Just because we don't know WHICH of the two hits is a guaranteed crit, doesn't change the fact that ONE of the two is GUARANTEED to be a crit, ERGO, the only one that matters when calculating the odds, is the remaining one, which has a 50% chance.

You morons are either trolling or incrediby stupid.

>>

>>582897992

50%. if each individual hit has a 50% chance to critically hit, it doesn't matter how many times you have landed a critical hit, each hit will have the same probability of it being critical.

>>

>>582897992

p(both critical hits | at least one)

= Pr(both critical n at least one critical) / Pr(at least one critical)

= Pr(both critical) / Pr(at least one) = ans

Pr(both critical) = 0.25 = "a"

Pr(at least one) = 0.5 + 0.5 - .25 = 0.75 = "b"

hence ans = "a" / "b" = 0.25 / 0.75 = 1/3

Answer: 1/3

>>

>>582910260

Information changes everything mang

http://en.wikipedia.org/wiki/Monty_Hall_problem

>>

>>582897992

so the possible outcomes are

1st hit - 2nd hit

a) crit - crit

b) crit - no crit

c) no crit - crit

d) no crit - no crit

each with an equal probability of 25%

>at least one hit is a crit

this rules out d)

we're left with

1st hit - 2nd hit

a') crit - crit

b') crit - no crit

c') no crit - crit

each with an equal probability of 33 1/3 %

>what is the probability both hits are crits?

>what is the probability of outcome a')?

the probability of outcome a') is 33 1/3 %

>>

>>582910469

>guaranteed

Neither was guaranteed when the test was run. It was done, the outcome was observed, and some of the information was relayed to us.

The problem does not state - and thus we cannot assume - that any sort of probability modification was done to ensure one crit. The probability for each hit was 50%, and this is what we must use to do our calculations.

1/3.

>>

>>582897992

what if its pseudo-random

>>

>>582910469

Obviously incredibly stupid. Trolling makes no sense because it won't change the fact the answer is 50%

>>

>>582910181

sorry im dumb,

forgot:

1 hit, 1 crit

>>

>>582910469

You forget there are two different possible outcomes with only one crit... It could be the first, or the second.

See

>>582906656

>>

>>582910238

i think people are misreading this.....

"at least one is a critical hit"

that puts a constraint on one of the variables.

one of them IS a critical hit.

so the only odds needed are for the other hit. which is 50% .

you only had to look at the odds for one of the hits.

>>

>>582908990

It's true. I also have several other PHDs in a variety of topics and I can guarantee with 100% certainty that my post contained the correct results. For further support I have contacted Johnson Bayes, son of the great mathematician and a prodigy in his own right. He has verified my results.

>>

>>582910181

>1 crit, 1 hit = 50%

There are two ways that can happen.

This problem is the same as the "if i flip 2 coins and one is heads, what is the probability that both are heads?" and the indisputable answer is 1/3. Look it up.

Source: I'm an Actuary

>>

>>582910765

>Neither was guaranteed when the test was run.

OP posted "At least one of the hits is a crit".

Trolling or stupid, only options. Pick one.

>>

>>582910596

Except we know at least one is a critical hit. If "N" is not a critical hit and "C" is a critical hit, then all the possible outcomes are:

NN

NC

CN

CC

We know that one of the hits was critical so the first combination is not possible. That leaves three possible combinations:

NC

CN

CC

Out of these three, only one of them has two critical hits.

Hence the probability is: event / number of possible events = 1 / 3

Answer: 1/3

lrn2BayesTheorem

>>

Anonymous (ID: w5OKaOAh)
2014-12-03 22:16:34
Post No.582911286

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>>582897992

It's always 50%.

It doesn't matter how many times it crits, there is always a 50% chance. You've established that.

>>

Anonymous (ID: aiAfxlJA)
2014-12-03 22:16:55
Post No.582911341

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>>582897992

lets make a tree

0. Start

| \

1. X 0

| \ | \

2. X 0 X 0

^ ^ ^ ^

| | | |

( first 3 count)( last one doesnt count)

1/3 have double crit

33%

>>

>>582910682

Monty hall theory doesn't apply. We already know there are only 2 doors, know a crit is behind one of them, and there is a 50% chance of a crit behind the other. The chance of having a crit behind both doors is 50% no matter how much you want to deny it.

>>

4 options

HC

HH>HC

CC

CH

HH will be turned into HC afterwards because of magic., therefore its 25% chance.

DEFINE THE RULES OP

>>

>>582911037

It does not put a constraint on one specific variable. Still three possible situations, all equally likely: NC, CN and CC.

>>

>>582911237

fucking hate the bayes theorem, but you are right

1/3 is the only correct answer

>>

>>582910409

i understand your argument and it is fine however we simply do not care which one is a critical or not when we know one will be

So what we have is as you put

00

10

01

11

We can stirke out 00 giving us

10

01

11

We know one will be a critical, it is irrelevant which one as one will be we care more about which one the other one shall be therefore the only stat we have is

10/01

11/11

two results either one will be critical or both will.

The 1/3 argument does matter if we exclude one always being critical or care more about which one is critical however all we care about in this is the final result

>>

>>582910931

The question was not how many possible permutations there are. The question was simply what chance is there of both hits being crits. That chance is 50%. The end.

>>

>>582907809

>set theory

If you are actually using set theory correctly, then this result will give you 3 elements in the set.

However, you are using probability, and you are defining XOR here, not or. Here you found the probability that A=true, B=true, and both A and B are false. Obviously the probability is zero

Consider set theory however. Let us break this down into outcomes. There are two sets, one in which the first hit crits, and one where the second hit crits.

A={CN, CC}

B={NC, CC}

A U B = {CN, CC, NC}

of these any outcome is equaly likely

>>

>>582911209

so we're taking into effect that even though one of the odds is absoluteley guarenteed, eliminating the odds, we still have to take into it;s theoretical effect on the chance?

because i'm not getting how people are getting .33

they're taking a factor that has been made absolute and saying it contributes to the odds.

>>

>>582902272

>beta function

>prior for A being 1.0

This is bait.

>>

>>582910682

but this is not the monty hall problem, in the monty hall problem the odds change as your options decrease. in this problem, the chance of a hit being critical is 50%, it is always 50% no matter how many hits are made and no matter how many are critical. if you make a hit, your odds of a critical hit are 50%, the next hits odds are 50% as are the odds of the next hit and the next hit.

>>

>>582902272

>beta function

>prior for A being 1.0

>highest recognition in math research by the government

This is bait.

>>

>>582901004

there should be more people replying to this

>>

>>582911436

even though one result is guarenteed? that there is absolutl--oooh fuck i just figured it out. because it cannot be guarenteed which hit is the crit.

goddamn i hate being wrong.

>>

62.5%.

One is 100% and the other is 50%

50/2=25

25+100=125

125/2=62.5

>>

>>582911853

Trolls gonna troll bro. It's 50% and no theoretical magical variable changes that fact.

>>

25%

0.5 * 0.5

The sets are not mutually exclusive.

>>

>>582911237

but thats wrong. it has been established that a hit, any hit, every hit has a 50% probability of being critical.

>>

P(X=2|y=1) Sample points. (0,0) (1,0)(0,1)(1,1) where 0 is no crit and 1 is crit. What is probability of digit 2 as crit if one is crit. That is one in 3 or 33.3%

>>

>>582911218

Yes, but it does not state which.

Let's play storytime.

The Crazed Buttstabber tells you he met two guys on his way over.

He flipped a coin when he met each guy, and would stab their butt on a heads.

He also tells you he stabbed at least one of the guys, but doesn't tell you which one or exactly how many he stabbed.

What are the odds he stabbed them both, given that they each had a 50/50 chance, and at least one of them got stabbed?

>Hint: it's 1/3

>>

>>582911237

lrn2ReadTheQuestion

>>

50% (whatever happend next is irrelevant)

/

100

\ -25%crit

crit 50 % - 25%hit

>>

>>582912183

Doesn't matter which hit is the guaranteed crit, one is guaranteed, so shelf that hit. Only look at the other hit, in which the probability of being a crit is 50%

>>

>>582897992

50%

when it comes to "Chance" things in the past doesn't affect the future.

You can hit a enemy twice having a 1%

>>

>>582902272

You don't have that write. Bayes isn't written like that.

>>

>>582899459

The problem is you assume you can knock out NN from the set, but the question says nothing of the sort.

out of two hits, what are the odds that you get two criticals if the crit % is 50%.

It is entirely possible to NOT crit each time.

So with you math:

CC (1) / CC CN NC NN (4) = 1/4 = 25%

>>

>>582912183

:)

Yeah, you are not guaranteed that it will be the first, or the second. Just that it will be at least one of them. So you have to look at all possible outcomes and see how likely they are compared to one another. NC, CN and CC, all equally likely since the odds of crit/noncrit is 50/50.

>>

>>582911592

You're so close!

It's fine if you want to combine 01 and 10 into one result, but you must also account for the fact that 01+10 is then twice as likely as 11 is.

>>

holy shit people, this requires no math at all. OPs problem states that hits have a 50% chance of being critical. 50% is a constant, that is the answer.

>>

You guys are thinking into this too much.

>protip: Its 50%.

>>

>>582912703

Except it's not possible since the OP states that one hit is a guaranteed crit. You absolutely no option get one of two crits regardless of any other factors.

>>

(0.5)(1.00)... P=0.5

one of the two swings is 100% chance to crit, the other is 50%

>>

Four possibilities: 1h2h, 1H2h, 1h2H and 1H2H.

1h2h is impossible, therefore it's between 1H2h, 1h2H and 1H2H.

1H2H is 1/3 of the possibilities, therefore the likelyhood is 33%.

>>

Fucking Lel.

Did any of you graduate?

It's 50 percent.

One hit has to be a crit.

Therefore the second hit has a 50 percent chance of also being a crit.

Therefore there is a 50 percent chance of landing two crits.

Fucking dumbasses.

Answer is 50 percent

>>

>>582899626

Yes.

>>

>>582897992

4 possible outcomes:

1 hit, 1 hit

1 hit, 1 crit

1 crit, 1 hit

1 crit, 1 crit

Answer: 25% or 1 in 4. Problem solved.

Don't over think it faggots.

>>

It's 50% either they both crit or they don't

>>

>>582912548

>Doesn't matter which hit is the guaranteed crit

Neither is guaranteed.

Source: wording of the question.

>>582912703

>The problem is you assume you can knock out NN from the set, but the question says nothing of the sort.

Actually, it says that at least one is a crit, which is how we know NN can be ignored here.

>>

>>582905665

there's other ways to interpret the wording

for instance let's say that your outcome pairs are

CC

CN

NC

but both outcomes are hidden

you reveal one outcome and it's a C

what are the chances the other outcome is a C?

50%

it only works out to 33% if you're worried about the probability of both dice being a crit

but you don't have to worry about that, because it's a given that one die is always a crit

you could rethink this problem as follows

you have two dice

you place one die on the table as a crit

you roll the other die and it will either be normal or crit

what are the chances both dice will be crits?

50%

you could interpret the original problem both ways

>>

>>582913217

>One hit has to be a crit.

No, it does not. One of them is a crit, but neither of them had to be.

>>582913310

But we know that "1 hit, 1 hit" is not the case based on the wording of the problem. This reduces it to 1 in 3.

>>

Anonymous (ID: BmwTLTTW)
2014-12-03 22:30:10
Post No.582913564

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0 %

0 %

0 %

you people overcomplicate things

no wonder you don't get laid

read the question and gtfo

>>

>>582913339

Thank you.

Someone who actually knows things

>>

>>582913332

>at least one of the hits IS A CRIT

How does this not guarantee 100% that one hit is a crit?

>>

>>582913310

I correct myself. Missed the line "At least one hit is a crit"

so the answer is 1 in 3 or 33%. Outcomes are

1 hit 1 crit

1 crit 1 hit

1 crit 1 crit

>>

>>582913113

You are assuming that the second hit (or the first hit, however you thought) will always be a crit. Not the case.

Each shot is a paranthesis in your calculation. Neither of the shots have a 100% crit chance, even though you know at least one hit crits.

See this:

>>582906656

>>

>>582913339

>but both outcomes are hidden

>you reveal one outcome and it's a C

>what are the chances the other outcome is a C?

If you interpret the problem like this, your answer may be correct. However, that is not what's being asked.

>>

>>582913310

You fucking tool. One hit is always a crit so there is no "1 hit, 1 hit" outcome.

>>

>>582913515

You are correct. Missed that line.

>>

>>582913586

It doesn't guarantee either hit to be a crit, it's simply an observation on the current state. Per the question, when the test was done, each hit had a 50% chance to crit.

>>

>>582905665

Finally someone who doesn't sound like a smart ass explaining the answer in plain english.

>>

Event A = First hit is crit

Event B = Second hit is crit

We'll assume the first hit (A) is the guaranteed crit. So P(A) = 1, P(B) = 0.5

So now using this equation for compounding probabilities:

P(A^B) = P(A) * P(B)

P(A^B) = 1 * 0.5 = 0.5

Choosing which hit is the guaranteed crit is irrelevant. (i.e. swapping P(A) for P(B) has no effect on the outcome)

>>

>>582913515

Exactly.

>at least one is a crit.

Therefore there's

CC

NC or CN (same thing)

There has to be a crit in there. And we want to know what the possibility of two is.

>>

33 tards :

1hit 1crit is 2x25%

Think like this. You NEED to crit first timel 50%, and 2nd time , 25%

>>

>>582913857

It does guarantee one of the hits is a crit, it's not a variable. At least one hit crits. It makes no difference if that hit had a 1%, 50%, 99% chance of being a crit, you know that it crit. The variables for that hit can be disregarded. The only variable that matters is the 50% chance of the other hit being a crit.

>>

>>582897992

Let C be crit, N be not a crit

There are 4 possibilities:

CC

CN

NC

NN

we know al least one crit so not NN.

so the probability is (1/4)/(3/4) which is 1/3 same as coin problem

>>

Anonymous (ID: X6X23OWi)
2014-12-03 22:35:28
Post No.582914394

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[iqdb]
[SauceNao]
[Google]

794KB, 500x281px

>>582897992

1 in 3. we've fucking done this one.

>>

Anonymous (ID: ninLQK/a)
2014-12-03 22:35:58
Post No.582914485

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It's conditional probability. So Prob (2 crits, given 1 crit):

(2C2 x .5^2) / (2C1 x .5 x .5 + 2C2 x .5^2)

= 1/3

Actuarial science, motherfucker

>>

>>582914154

>guaranteed crit

This is your mistake, friend.

>>582914165

>NC or CN (same thing)

If you combine these into the same result, you must also account for the fact that the combined result is twice as likely as CC.

>>582914348

>It does guarantee one of the hits is a crit

That's not how the problem is worded.

>>

To the 33% people, explain me this. Why do you have 66% chance to crit the first time.

>>

The answer could either be 50% or 33%.

If you state which hit, 1 or 2 is the guranteed crit, the answer is 50%.

If you don't state which one is the crit, the answer is 33%.

/thread

>>

>>582914154

You are not correct.

In your calculation, there are only two possible outcomes: crit-noncrit or crit-crit.

In reality the "other" option (noncrit-crit) could also occur, regardless of which hit you decided to take for granted.

Captcha: lhouspe probability

>>

http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Gambler_s_fallacy.html

>one hit is a crit

>crit chance is 50%

>one hit left

>>

>>582914538

I hit an enemy twice.

>flip 2 coins

One hit is a crit (lands heads

>one coin is heads

What is probability both hits crit (land heads)

>other coin has 50% chance of landing heads.

Answer is 50%

>>

>>582912287

Your issue is you're not getting the events correctly. There are actually four different events we might define:

A = Critical Hit on first punch

B = Critical hit on second punch

C = Critical hit on either first or second punch

D = Critical hit on both the first and second punch

Although events A and B are independent (as the question implies though doesn't state directly), that does not indicate that events C and D are independent. The question is asking about events C and D.

In particular, it is asking Pr(D | C). Neither Pr(C) nor Pr(D) = 50% so I'm not sure what you're going on about. If A and B are independent, then

Pr(D) = Pr(A) * Pr(B) = 0.25 =/= 0.5

Pr(C) = Pr(A) + Pr(B) - Pr(A and B) = 0.5 + 0.5 - 0.25 = 0.75 =/= 0.5

We see that neither =0.5. Further C and D are not independent events since:

Pr(C and D) = Pr(D) = 0.25

and

Pr(C) * Pr(D) = 0.75 * 0.25 = 3/16

Hence Pr(C) * Pr(D) =/= Pr(C and D) which by definition of independence means C and D are not independent events. Hence neither 0.25 nor 0.75 are the correct answers either.

So to conclude, you are trying to assert two things. 1) that A and B are the relevant events are in question which is not correct. C and D are. And 2) that C and D are independent so we can disregard the condition that C is true, which is also not correct.

>>

>>582897992

250% chance this id a vidya

>>

>>582897992

One hit = 50%

Two hits = 25%

>>

OPS rule of gauranteed hit only works when you dont crit the first time. Think about that for a sec.

Therefor 25%

>>

>>582897992

25%

>>

P(2Crits|Crits=>1)=P(Crits=>1|2Crits)*P(2Crits)/P(Crits=>1Crit)

P(Crits=>1Crit)=P(2Crits)+P(1Crit)

P(1Crit)=C(1|2)*(0.5^1)*(0.5^1)=0.5

P(2Crits)=0.5*0.5=0.25

=1*(0.5*0.5)/(0.75)=0.25/0.75=0.33

I also like the 1 out of 3 possible events is 0.33

>>

Anonymous (ID: iaQIyZWk)
2014-12-03 22:40:15
Post No.582915172

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Have you seriously never been to school?

>>

>>582914885

FYI:

I correct tests with this exact curriculum for a living. Anyone who answered anything other than 1/3 to this question, would loose points.

>>

>>582914485

actuarial science is lame and gay. i know because i do it too

>>

I'm gonna put it as simple as possible for you all.

Crit change 100% & 50%

Non crit chance 50%

Crit to none crit is 150%/50% or 3/1

That makes your crit change a massive 75%

The magic of math.

>>

>>582915059

Doesn't matter. There is no if-then variable. Only that there are 2 hits and you know one is a crit. There is no chronological order driving the equation.

>>

50% you dumb fucking idiots. The first trial doesn't matter since you can assume one hit is a crit. The next hit has a 1/2 chance of being a crit.

>>

I saw the answer in the question.. Did you?

>>

>>582914741

No. There is no reason to overthink this so much. Take the guaranteed crit out. That leaves you with a 50% chance on the remaining swing. Simple. It either is or isn't. Stop complicating it.

>>

its okay retards im here.

There are 2 branches to start with, Crit and not crit.

This leaves three possible outcomes

Crit - Crit

Crit - No Crit

No crit - Crit

If the first is a crit, the second shot can either be a crit or not crit.

If the first is a not crit the second has to be a crit because atleast one has to be a crit.

Therefore if the first is a crit the second cannot be not crit.

The answer is 33%

Because some of you are retards ill show it again.

Crit - Crit

Crit - Not crit

Not crit - Not crit

There are 3 possible outcomes, and only one is both hits being crits. So 1/3 = .33

>>

Anonymous (ID: /UUi5dA4)
2014-12-03 22:45:44
Post No.582915953

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50%ers, consider this:

By the OP's question, the first hit has a 50% chance.

If the first hit is h, the next hit is forced to 100% H.

If the first hit is H, the next hit has a 50% chance.

...which makes it 25%, since hH is 50% and Hh is 25%. Not 33%.

But the point is that Hh and hH need to be treated separately. We also need to know the reason for the certainty of this 100% chance of at least 1 being critical in order to get an answer.

>>

>>582915172

The W at the bottom should not be 100%.

True, we know LL did not happen, but the probability left over from LL is spread out evenly across WW, WL and LW.

You are solving a problem based on the following premise:

"We have scored at least one crit. We don't know where it happened, but we know that you always get a crit on your second shot if you didn't get one on the first."

This is not the question.

See:

>>582906656

>>

>>582911195

Son of a guy born in 1701

Could have added medium to your list of qualifications

>>

>>582915789

No tree, 2 hits are separate events. One event is a crit. Other event has 50% chance of being a crit. like I said, 2 coins flipping, one lands heads, other has 50% chance of landing heads as well.

>>

>>582915191

then your test question is probably more specific, or you're not recognizing the ambiguity of OPs question

you can interpret the question to give an answer of 1/3 or 1/2

>>

>>582897992

1/3, duh

>>

>>582915789

Yeah but the question is not how many possible outcomes are there, so you look at your outcomes

Crit - Crit

Crit - No Crit

No Crit - Crit

With thess outcomes you can assume only one of two things will happen, either both will be crits, or only one will be a crit. Doesnt matter which order they come in.

Odds of only one Crit 50%

Odds of two Crits 50%

>>

>>582897992

A lot of people are probably going to say it's a monty hall problem where you take the permutations and delete the one that is no crits and get 33%, but that's not right.

You have to set up two cases, case 1 where your first hit is a crit and your second is unknown and case 2 where your first hit is unknown and your second is a guaranteed crit. In both cases you observe the you have a 50% chance of having both hits being crits. Because these two cases represent all of the possibilities and both show that there is a 50% chance of 2 crits, your answer is 50%.

>>

Anonymous (ID: qKokdyvE)
2014-12-03 22:51:01
Post No.582916765

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if it's fire emblem, less than 80% = 0%

>>

>>582897992

One cannot die twice, therefore one cannot receive two crit hits. UNLESS, however, your enemy is Jesus Christ. Out of the 108 billion humans that have ever existed, the probability of randomly selecting Jesus Christ to fight and then crit hitting him twice consecutively is about 0.0000000000023148148, or 1 in 432 billion.

>>

Guys its 50% because there are only two possible outcomes. Outcome 1: a crit and a non-crit (so either CN or NC it doesn't matter which way round). Outcome 2: is both crit (so CC).

>>

>>582916791

No one ever said a crit = death.

Your argument is invalid.

>>

>>582916321

Well, yeah, if he by "crit chance" means "chance of both being a crit". If the chance of both being a crit is 50%, then it's 50%. But that's not what it says.

>>

>>582915414

hmm you are right, i somehow imagined they were after eachother. but that would make more sense. how can you now one is a gauranteed hit if they are fired at the same time. then it must be 100%.

do realise im losing grasp on reality tho.

>>

50%

The first and/or second crit doesnt affect the other.

It's 1/2 chance for the first, 1/2 chance for the second. If the first one is a crit, then the second one is a 1/2 chance, they're separate.

The combined chance is 1/4

Think of it using the "boy" example.

I know the first one is a boy, the chance the other one is a boy is 1/2..

If the first one is a girl theres already no chance that both are boys.

Im confused

>>

start with a possibilities table

c n

n c

c c

c c

it is still 50%. The chance would be different if the FIRST was a crit, but it simply states that one is a crit.

>>

>>582916286

That it doesn't state whether the guaranteed crit is on the first shot so therefore you would use a tree but exclude outcomes without crits. If it said the first shot had to be crit then yes there is only a 50% chance the second is a crit as well. However the first shot can be a non crit making the second shot the crit. If the first shot is a crit the second can be either. This makes 3 possible outcomes of which only 1 is double crit

1) Crit - Crit

2) Crit - Not Crit

3) Not Crit - Crit

4) Not Crit - Not crit

Since atleast one hit needs to be a crit outcome 4 is not possible. Therefore leaving 3 possible outcomes of which 1 is double crit.

>>

>>582916321

How is that?

With a 50% chance and two draws, there are four equally weighted outcomes:

CC HC CH HH

And yet, we are told "At least one is a crit." Thus, we have removed one of the outcomes from consideration. Remaining outcomes are

HC CH CC, and thus one in three is the answer.

>>

>>582913685

>you could

he wasnt providing an answer he was providing tools to help one come to a conclusion themselves

>>

Lets make it a bit simpler for you all.

I'm gonna split each hit for you, crit and none crit. So you have 4 at 50%

2 of those are definitely hits and the other two are 50/50 (1/1)

So now we have 2/0 and 1/1

Add that together and you get 3/1

That's 75% crit chance.

>>

Anonymous (ID: GmYt19yk)
2014-12-03 22:56:18
Post No.582917537

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>>582907235

>>582908990

>>582911864

>>582912017

>>582912649

say that to my face faggots not online see what happens. yall jelly of my math prodigy status.

>>

50% - if the first one isnt a crit, theres no chance for both of them to crit.

>>

50

>since one hit is a crit, we don't care which one, we are only interested in the probability of the other hit being a crit, which is 50%

>>

>>582916609

You are right in a way but wrong in the fact that Not crit-Crit and Crit-not Crit are two separate possible events.

>>

>>582916696

>>582916609

>>582917125

over complicating it

see >>582917661

>>

>>

Half the people here are autistic.

1/1 * 1/2 = 1/2 (50%)

>>

I don't see why some of you see the need to consider NC and CN to be different outcomes

>>

chance of first hit being crit: 50%

chance of second hit being crit: 50%

possibility of two events happening a certain way: probability1 times probability2

.5 x .5= .25

25%

>>

>>582917326

The question does not ask for a percentage of outcomes for a double crit, it asks for probability of both hits being a crit knowing that one hit has crit. We don't care whether hit 1 or hit 2 is the crit, just that both either are or are not crits, which means the probability of both hits being a crit is 50%.

>>

>>582917735

You are right, they are but when it comes to grouping them in similar outcomes, there are only two possible out comes. One where both are crit one one where only one is a crit.

This makes the chance of getting crit-crit a 50 percent chance.

Where as your crit chance in total would be 75% if every other one is assumed a definite crit.

>>

>>582897992

>At least one of the hits is a crit.

Does the intelligence responsible for providing this knowledge know the nature of both hits, or only of one of them? Until OP clarifies, this thread is cancer and you can all move along, because necessary information is missing.

>>

>>582918108

1/2 * 1/2

Am retarded and don't remember how to multiply infractions of the penis code, but isn't it 1/4 times both will crit (1 of every 4 sets of 2 attacks.)?

>>

>>582918352

Finally, someone with a brain that can see it's 75%

>>

>>582918173

Because it means there are twice as many ways for there to be one crit as there are to be two crits. However as is the question is ambiguous

>>

gun is incapable of firing 2 not crits: 33%

one is crit, other is 50% is obv 50%

if the first one isnt crit, the 2nd is its 25 %( but doesnt evenfit ops vague description so wrong)

lack of info and context. troll thread/

>>

>>582918360

Lol, the flew by me, didn't read it correctly.

Fucking reading is for fucking fags. Jk, I'm a fag for missing it, now i need to sharpie the pooper

>>

Think of it this way people.

If this was to occur and hit the enemy here what can happen

I hit and crit. I hit again and crit

I hit and crit. I hit again but dont crit.

I hit and dont crit. I hit again and crit.

In only 1 instance do I double crit.

In the other 2 I dont.

So 1 out 3 times I double crit.

>>

>>582917350

the phrase "at least one of the hits is a crit" doesn't specify whether you're looking at both hits before considering whether it's a valid outcome, or whether you're only looking at one hit before considering whether it's a valid outcome

you could interpret the problem like you've said:

CC

HC

CH

HH

if I look at both outcomes before saying "one of these is a hit", then my probability of considering all the CH and HC outcomes is 100%

but if I only look at one hit before saying "one of these is a hit", then I have a 50% chance of missing the HC and CH outcomes

so with your interpretation it's

CC = 33% * 100%

CH = 33% * 100%

HC = 33% * 100%

in the alternate interpretation it's

HH = 25% * 0%

CH = 25% * 50%

HC = 25% * 50%

CC = 25% * 100%

>>

>>582897992

75%

>>

>>582918576

If you say its 50% that means the only thing that will happen when I shoot is either

I hit and crit. I hit again and crit

I hit and crit. I hit again but dont crit.

Or

I hit and crit. I hit again and crit

I hit and dont crit. I hit again and crit.

This doesn't make sense

>>

>>582897992

Come on, use your /b/rains. It is called conditional probability. Given that C=Crit and N=Noncrit, and given that at least one of the hits is a crit, there are three possible outcomes: CN, NC and CC. Because P(C)=.50, each of these outcomes is equally likely. Therefore P(CC)=1/3.

/thread

>>

>>582900478

/thread

>>

>>582918330

that would be correct if neither was guaranteed to be a crit

but one is guaranteed to be a crit

>>

>>582918409

>At least one will crit

That makes it 1/1, and with a 50% crit chance it'd be 1/2. 1*1=1, 1*2=2, so it's 1/2

>>

The answer is 1/3. Here is a simulation (run in your favorite JavaScript repl)

var count2crits = 0;

for (i = 0; i < 10000000;){

var firstCrit = Math.random() < .5;

var secondCrit = Math.random() <.5;

if (!firstCrit && !secondCrit){

continue; //can't happen

}

if (firstCrit & secondCrit){

count2crits++;

}

i++;

}

count2crits/10000000;

>>

Logically it would be 50%, since you can disregard one of the two hits as it is stated to be a crit. The answer to the question now lies solely on the results of the second hit, which is 50%.

Mathematically I don't give a fuck.

>>

>>582918352

yes they are similar events but it is possible for those events to occur is separate orders and so they each have their own probability of occuring.

>>

>>582918347

>The question does not ask for a percentage of outcomes for a double crit

exactly, if that is the question, it'd be 25%, because

CN, NC, CC, NN

but then the question ask given we crit at least once, NN becomes impossible to happen, while other outcomes are still equally likely to happen because 50%chance each hit, so NC, CC and CN are exactly equally likely to happen, chance for crit-ted twice just increased from 25% to 1/3

>>

>>582897992

50% crit chance on the only unknown hit.

All the rest of the data is irrelevant.

>>

>>582919231

that's why you fail maths, m8, your answer would be both logically and mathematically incorrect

maths is logic

>>

>>582898103

First reapoinse wins.

>>

>>582919050

ah... I see. makes sense. thanks.

>>

>>582919469

But CN and NC are the sameeee

>>

>>582919231

again it depends how you interpret it

if you're saying that one of the two hits was always guaranteed to be a crit, then you're right, you only have to look at the chances of the other hit, and it would be 50%

but if you're saying that it was never guaranteed that either was going to be a crit, and that you've made both hits fair and random and now you've just gone and selected the outcomes where one was a crit

then you must see how that's different from the first scenario

because you have to consider how you got to the outcome of having one crit, since it was never guaranteed

>>

Enough for a kill

>>

>>582919120

this is a good answer, except that it is a little odd that you chose to increment i outside of the main for declaration.

>>

>>582919231

Your logic is flawed.

With only hits considered, the set of potential cases is {HC CH CC HH}

The instructions do say one hit is a crit. They do NOT say that the FIRST hit is a crit. If they had, the choices reduce to {CH CC} and you'd be correct.

But so long as it is implied that both hits have an equal chance to crit -- and it is flat out stated this is true -- you can't determine which of the two draws will be a crit. This means considering the full set of choices.

We can remove HH. But we still need to consider that the first hit could be normal or a crit because the SECOND hit could be normal. Thus, three tries, or 1/3.

>>

in an hour or 2 you will finally realise ops description is intentionally vague and everyone is right.

except 75%. u retard?

>>

>>582919930

well, you are throwing out cases where there's two plain hits. If you were to consider those for purposes of the increment, you'd wind up with the answer 1/4

>>

>>582919443

That is true, but then we look at the question.

What is the probability that both will be crits, since at least one is a guaranteed crit and the other is 50/50 then it doesnt matter what order, the probability of a specific order would matter it would change the probability.

Because the question asks only about crit-crit we have to group into two possible outcomes, regardless of their order.

One with a only one crit.

One with two crits.

Only one of those two statements will be true in this scenario.

>>

>>582919879

how are they even the same, m8?

if you say it's 50%, what you're saying is it'd be either NC/CN or CC, but then using your brain, you'd realize that NC/CN combined would be more likely to happen than CC itself, just because I might win lottery or not, doesn't mean I have 50% chance of winning

>>

>>582918352

I actually take back what I said and see my mistake.

However I noticed something else as well.

Just check out my math instead of me describing it....

.5 C â†’ .5 C =.25

.5 C â†’ .5 NC =.25

.5 NC â†’ .1 C = .5

So a 25% chance of a double crit.

>>

>>582912552

EXCEPT IN THIS CASE IT DOES...

Because, if in the past, you DIDN'T get a crit then it is impossible to get two crtis, you fucking retard.

NC C

C NC

C C

where c = critical and nc = non-critical. Easy

>>

its fucking obvious, there are initially four cases:

CC

NC

CN

NN

where N if not crit and C is crit.

There is a guarentee that atleast one is a crit so NN is eliminated. Therefore there are three possible cases:

CC

NC

CN

therefore it is a 1/3 chance that both hits are crits.

>>

>>582920027

No, it isn't. It's a straightforward statistics question, but most fuckers are basic.

>>

you could also interpret the question like this:

you hit an enemy twice

at least one of these hits must be a crit

therefore, if the first hit is normal, you automatically get a crit on your second hit

if your first hit is a crit, your second hit can be either a crit or normal

in this case, if your first hit is N, then your second hit is always C, so

NC = 50%

CN = 25%

CC = 25%

>>

>>582920246

I don't think I follow. I formatted the for declaration as:

for (i = 0; i < 10000000; i++){

which is obviously the more conventional notation, and it is equivalent.

>>

>>582920570

Yes but that's only if you include the chance that there could be 2 no crits, a 25% chance.

We are told to modify the game so that this outcome is removed. This alters the odds.

>>

>>582920018

That would be true if the question was which would be a critical hit.

The question is only will 1 or 2 be a critical hit, there is a 50% of each being critical.

let us put it this way

we have two coins

i lay one heads up

I flip the other one.

what is the chance both will be heads?

We know we have 1 true so the truth table instead of being

00

10

01

11

we instead have

10

11

or if you want it explained more cleanly one coin can be called true

TRUE 0

TRUE 1

See what happens? we don't care which coin is true but we know one will be so it's only down to the second

is this a troll ive not notice dover the years?

i mean please tell me this bothers me so fucking much as it feels like i'm with a load of people who just saw this problem and learnt it but never learnt anything more.

>>

>>582897992

You hit an enemy twice.

>Oh shit nigger that's 2 times!

At least one of the hits is a crit.

>oh shit nigger that's 1 hit out of 2! 1 crit no matter what! NO MATTER WHAT!

assuming(lol) a 50% crit chance

>wow, ok guys one's a crit no matter what, we know that. we have a 50% chance to land a critical hit on the one that isn't a crit. we know a crit happens no matter what

what is the probability both hits are crits?

>I want to say 50% but after reading the thread idunno

See the problem here is I the public schooling system failed me or I failed the public schooling system. Either way..

Things get more complex with

>at least one of the hits is a crit

>at least one

Which one? And why is that relevant? how do I solve that?

>>

>>582920841

never mind I'm stupid

>>

>>582920570

In your math you cant assume that both chances for a crit are .5

You have to assume that at least one is 1 because one is a guaranteed crit.

>>

>>582897992

You gave away the answer by saying 50% crit chance, meaning that the probability of both hits being crits is 50%

>>

>>582920841

No, you're going to want to run that. You'll get .25

>>

69% you fucking idiots

>>

>>582921071

I'M SORRY ANON I'M A BAD PERSON I WAS WRONG :(

>>

>>582920912

Actually i did take that into account

Ill put all in math again.

.5 C â†’ .5 C =.25

.5 C â†’ .5 NC =.25

.5 NC â†’ .1 C = .5

.5 NC â†’ 0 NC = 0

There is no probability of NC occurring This is what makes NCâ†’C a 50% chance. There is only thing that can happen if the first hit is NC that being the next shot is C

>>

Anonymous (ID: n1M8aZbI)
2014-12-03 23:21:46
Post No.582921375

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

>>582902272

The PHD stuff is BS, but your answer is correct.

>>

>>582920763

Flawed.

You are not twice as likely to flip NC then flip CC or CN.

>>

Anonymous (ID: CLb5M5kl)
2014-12-03 23:22:10
Post No.582921437

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

>>582919469

>>

>>582920998

if you want to know what the probability of it being the first or second hit then that is 1/3

The probability of both being a critical is only 50% as it really doesn't matter which is a critical so long as the other ones 50% chance pays off.

That's either the troll or the confusion as many people seem like they have to prove the odds for a different question altogether

>>

Anonymous (ID: n1M8aZbI)
2014-12-03 23:22:23
Post No.582921470

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

>>582905551

Correct.

>>

>>582921465

C NC

C C

NC C

1 out of 3 idiot.

>>

>>582921409

again, it depends how you interpret the question

the question says "at least one of the hits is a crit"

there is a 50% chance that your first hit is normal

if your first hit is normal, your second must be a crit, because that's what the question states, one must be a crit

so there is a 50% chance of getting NC

again, and again, depends how you interpret the question

>>

>>582920763

no NC = 50% meaning you're twisting the data given to you, it's 50% simple probability, you don't get a change in probability in second hit, you just made that into 100%.

the question is saying that "given that NN didn't happen....." it just said NN did not even happen, doesn't mean that after a non-crit, second hit becomes garenteed to crit, it's still miss, so it's still 50%, not 100%. just like what is given to you, 50%

It's is just that this very variation is ruled out, and it'd not be happening

>>

>>582920649

how is this straightforward. you could easily interpret this 5 differentways. it says 50% chance except it really isnt. it doesnt state the exact rules. u could say u ignore results where are no crits, or u could say it doesnt even happen. u could say one of the shots will crit for sure and the other one is 50%. its troll

>>

1 is going to be a crit no matter what

1 is 50%

since the first one is 100% we don't have to care about it leading us to either crit or non-crit which was 50%

this would be the most logical or am i retarded?

>>

25%

>>

>>582921437

see this

>>582921718

I'm an asian and I got A in maths for 6years of my "highschool" there

>>

>>582921034

in that case the probabilty would go over 100%

1 C â†’ .5 C =..5

1 C â†’ .5 NC =.5

.5 NC â†’ .1 C = .5

this already = 150% of total probability.

when one crit is guaranteed it does not get 100% chance of occurring during a single event unless the event before it does not meet its demands.

>>

>>582921718

the question isn't saying either of those things, that's why I keep saying it depends how you interpret the question

you could interpret it both ways

you're assuming it's possible to get NN, but we're just going to ignore those outcomes for the purpose of this question

i'm assuming that the rules of the game do not allow for NN to be a possible outcome

OPs question doesn't have enough information to determine which interpretation is correct

>>

>>582921708

Don't combine them.

NC is 50% then 100%

CC is 50% then 50%

CN is 50% then 50%.

According to your logic, that makes 150% probability then 200%. I can see where you make the mistake, so hopefully this makes it clear.

>>

>>582919469

CN=NC by reflexive property

leaving CC and CN

50%, faggot

>>

>>582920639

wrong

if first hit is N, then chance of second hit being C is 100% and chance of it being N is 0%

if first hit is C, then chance of second hit being N is 50% and chance of it being C is 50%.

(0+100+50+50)/4 = 50%

>>

>>582921465

Thank you

>>

>>582921839

No, because if the first one isn't a crit it's impossible to get double--crit.

>>

Some of you fags are idiots. One of the hits is a guaranteed crit. So really you're calculating the probability of only one hit to be a crit - 50%

>>

>2 hits

>CN, NC, CC, or NN possible results

>1 guaranteed crit so really CN or NC

>We only roll on 1 hit. Either C or N

>50%

>Profit

inb4 someone tries to bate me into responding

>>

You are all idiots, it's like flipping a coin, doesn't matter if you already got heads once, or a thousand times, it's still 50% chance

>>

>>582921354

Again you assume that each hit only has a 50% chance.

1 C â†’ .5 C = .25

.5 C â†’ 1 C = .25

1 C â†’ .5 NC = .25

.5 NC â†’ 1 C = .25

0 NC â†’ 0 NC = 0

0 NC â†’ 0 NC = 0

.5 NC â†’ 0 NC = 0

0 NC â†’ .5 NC = 0

These are all the possible outcomes. 4 of these outcomes are impossible because you must have an outcome with at least 1 crit at 100% chance.

Assuming that we see that a double crit, in both orders still takes a 50% chance.

>>

>>582922627

Except you're the idiot. If the first hit is non-crit it's impossible to get double-crit. duh

>>

>>582921704

did they ask you to tell them which would be a critical? no

Which side of the equation holds the critical is irrelevant

the question only uses the variables that give

C NC | NC C

C C | CC

It doesn't matter which side of the equation so long as the condition is satisfied/

This is where everybody is falling down by not reading the question.

This is not short circuit evaluation order is not relevant just the final result 0 or 1

>>

3 outcomes

No Crit -> Crit

Crit -> Crit

Crit -> No Crit

Therefore chance is 1/3

>>

as an asian, I finally see the stereotype is real, my teacher didn't lie to me, americunts are real bad in maths

>>

>>582922276

i'm not combining anything, and you're not seeing what i'm getting at

you and I disagree on how the outcomes are set, because it's not clear in the rules

you are saying there are 4 outcomes: NN, NC, CN, CC, and we're going to ignore one of them

i'm saying there are only 3 outcomes: NC, CN, CC

if your first hit is normal, then in order to satisfy the rules of the game, your second hit must be a crit, that's what the question says, at least one is a crit

so if the first is normal, the second must be a crit, that's in the question

>>

The probability is 50%

The probability the first hit is critical and the second as well.

P(c1^c2)=P(c1)P(c2)

We assume at least one hit is critical, arbitrarily take this to be c1, P(c1)=1

There is a 1/2 critical chance, then P(c2)=1/2

therefore the answer is (1)(1/2)=1/2

50 percent.

>>

>>582922805

Well its 50% for the next flip to be heads.

However the probability changes when you talk about the event occurring multiple times

>>

>>582922276

I think what you meant to say is

NC is 50% x 100%

CC is 50% x 50%

CN is 50% x 50%

so

NC = 50%

CC = 25%

CN = 25%

which is what I said to begin with

>>

>>582922849

Let me fix that for you:

50% C, 50% C

50% C, 50% NC

50% NC, 100% C

Chances of CC? 1 in 3

>>

>>582922869

where does it state that? nowhere.

we just know at least one crit will happen and that 2 crits are possible.

>>582922969

order is not a matter of the equation so combine your c nc resutls

>>582922986

again you answered correctly for a different question as it never stated in what order must the crit be done or which result will be a crit

>>

>>582923136

The question only asks for two hits.

>>

>>582922849

Your math makes no sense.

You said 1*.5=.25 on 4 occasions.

>>

67%.

Here is how it's done.

Only possible outcomes are

Crit

Crit/no crit.

Or

No crit/crit

Crit.

Either way it's two crits to one no crit.

2/1 or 66.6'% (67%)

>>

>>582923263

At least one has to hit so one has too be 100% C, 50% C

>>

>>582922303

Let's play a slightly different game.

Two coins, one has a 0 and 1 and the other has a 10 and an 11. We flip each once. If the result is two low numbers, we reject the toss; otherwise, we write it down.

What is the probability of getting 12?

>>ans: 1 in 3 and you can simulate it with a coin and enough tosses

You are correct that the VALUES of NC and CN are equivalent -- IN the game you are playing. Their identities and thus their probabilities are not, and that becomes obvious when you modify the game. This is an indication that your solution is incorrect.

>>

>>582897992

H1 H2

NCrit NCrit

NCrit Crit

Crit NCrit

Crit Crit

25%

>>

>>582897992

>/b/ can't do simple binomial distributions

>>

>>582923310

you're just refusing to realize that you had to make some assumptions in order to interpret the question your way, and you're refusing to accept that there's other assumptions that could be made

your answer is correct for your assumption, and your assumption is possible

my answer is correct for my assumption, and my assumption is possible

again, the question just says one must be a crit

you're assuming that it's possible to have an outcome where neither is a crit, and the question is asking you to ignore those outcomes

i'm assuming that the question is saying it's not possible to have an outcome with no crits

both assumptions fit the question

>>

>>582922968

>did they ask you to tell them which would be critical?

No, but only one of them must be critical. Stop assuming that's going to be the first one. The order is relevant, because if a non-crit is hit first then it is impossible for a double-crit. You aren't very logical.

>>582922986

Yes, but you're just as likely to get any of them. NC isn't twice as likely, because the first flip is 50% no matter what.

>>

>>582923603

Wait I'm wrong. Since probability of both Crits is conditioned on the event that one is for sure a crit. Ah it was long ago I took that course.

>>

To all of you who think it's 50% because you get a 100% garanteed crit on second hit because you didn't crit on first hit, fuck you

Did you not even read the fken question, it says 50% chance crit,niggas, not 100%

it's still 50%chance crit, you still need to luck for it to proc a crit, it's just that since the number of possible outcomes has reduced from 4 to 3 , i.e. NN CC CN NC to CC CN NC, so from 25% to 1/3

If you keep saying one is garanteed to crit, you're not using simple probability on each individual hits

>>

>>582923442

Because thats the different order that these hits can take place.

A 100% crit hit, then a 50% crit hit.

A 50% crit hit, then a 100% crit hit.

A 100% crit hit, then a 50% crit miss.

A 50% crit miss, then a 100% crit hit.

The only four options you have when at least one has to be a crit hit.

>>

THis isnt about math. No one in this thread made any calculation error. this really is my last attempt to tell you you are being trolled. i really dont think one explanation is more accurate than the other one. at least one isnt an absolute term in probability.

>300 replies

wp OP

>>

>>582897992

What raid buffs do I have?

>>

Anonymous (ID: Heaven)
2014-12-03 23:37:34
Post No.582924035

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

use a probability calculator

http://stattrek.com/online-calculator/probability-calculator.aspx

>>

>>582923241

Don't multiply, wtf are you doing?

>>

50%.... u dumbasses..

>>

>>582923831

NC is twice as likely because there is no possibility of NN

what are the chances the first hit is N? 50%

since one hit has to be a crit, then if the first hit is N the second must be C

so if the chances of the first hit being N is 50%, then the chances of NC is 50%

because there is no such thing as NN

then the chances of CN are 25%

and the chances of CC are 25%

>>

>>582923310

It doesn't state it, that's the part. It's a possibility. A possibility to hit NC or C first. You can't assume the first hit will be a crit.

>>582923558

Except the first can be NC.

>>

>>582924085

i'm trying to fix your math but I have no idea what you're trying to do because you're not making sense or listening to reason

>>

Anonymous (ID: CLb5M5kl)
2014-12-03 23:40:59
Post No.582924543

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

>>582922969

>>582923263

>>582923310

>>582923481

>>582924198

>>582924296

you're all wrong

>if first hit is N, then chance of second hit being C is 100% and chance of it being N is 0%

>if first hit is C, then chance of second hit being N is 50% and chance of it being C is 50%.

(0+100+50+50)/4 = 50%

>>

>>582924296

If the first is NC then the second has to be C with 100% certainty. Still giving you only two possible outcomes.

One where you crit twice, and one where you only crit once.

>>

>>582923831

okay this must be a troll fuck this but one more time

>hit an enemy twice

> at least one of the hits is a crit

> assuming a 50% crit chance what is the probability that both hits are crits

i never implied that as that means i didn't read the question

again

2 hits

hit a

hit b

We can assume one will be a crit wether that is a or b we do not know and have no need to know

we can assume a 50% hit chance and i'd assume for both.

so if we know one will be a crit we only need to know if the other will be a crit or not, there is no wordplay outside of that other than the other coin wether first or second is or isn't a crit.

Again the result is

T 0

T 1

0 T

1 T

Either way it's 50%

If we wanted to know the probability of coin A being a crit then definitely it would be 1/3 but that is not what was asked.

>>

>>582923972

when one crit is guaranteed it does not get 100% chance of occurring during a single event unless the event before it does not meet its demands.

Now lets say your equation did make sense what you said is there is a 200% total probability.

You made 1 mistake which resulted in multiple

Because you're using outcome numbers as your probability you are saying 1*.5 =.25 because 1 out of 4 outcomes is that option.

>>

>>582923241

it's

NC 0.5 x 0.5

CC 0.5 x 0.5

CN 0.5 x 0.5

it's still the chance of them happening individually

because we' ruled out one possible outcome in this question, they each takes up a larger percentage out of all the posible outcomes

so NC1/3 CC1/3 CN1/3

>>

>>582898139

It's only 25% if you don't know the outcome of one of the two shots. Otherwise it's 50%

>>

>>582924394

>>582924198

Not making sense?

NC

CC

CN

What's the likelihood of CC occurring? You say 25%, I say 33%. Do retarded pseudo-math if you want, but it's obviously 1/3.

>>

>>582923972

The crit never has 100% possibility unless the first shot is a non crit

>>

How are you guys so retarded? 1 guaranteed crit leaves only 1 factor undecided, this factor is a 50% chance. There is no math.

>>

>>582924543

It's decided by three not four. Only three possible outcomes.

Hit

Hit

Mis

>>

alright kiddos:

>c/nc

>c/c

>nc/c

>nc/nc

they are all 25% possibility

when you have to different 50% situations you divide the first 50 by the second and get 25%

>>

>>582924854

yes that is why one is 0%

>>

>>582924637

------------One Hit, One Guaranteed Hit-------------

A 100% crit hit, then a 50% crit hit.

A 50% crit hit, then a 100% crit hit.

----------One Guaranteed hit, one miss--------------

A 100% crit hit, then a 50% crit miss.

A 50% crit miss, then a 100% crit hit.

There are only ever two outcomes in my math, you are just failing two see it in two groups.

>>

>>582924706

I guess i'll say it again - it depends how you interpret the question

although I'm not sure anyone is capable of understanding there are multiple ways to interpret the incomplete question

look at your outcomes

you are saying that each of your individual hits are twice as likely to be a crit than they are to be normal

the question explicitly states that they have a 50% chance

>>

>>582924854

fullretard.jpg

>>

>>582924562

>>582924610

NC

CC

CN

That's 1/3 not 50%

>>

Anonymous (ID: gHpJPd/c)
2014-12-03 23:45:36
Post No.582925201

[Report] Image search: [iqdb] [SauceNao] [Google]

[Report] Image search: [iqdb] [SauceNao] [Google]

ALL YOU NIGGERS ARE WRONG!!!

The answer is Apple

>>

>>582924823

you're making the same mistake as the guy two posts above you

look at your outcomes for the first hit

you're saying that there's a 67% chance for the first hit to be a crit, and a 67% chance for the second to be a crit

that's against the rules of the question

>>

P(an event that happens 100% of times)

P(an event that happens 50% of times)

Experiment: You hit an enemy twice. At least one of the hits is a crit.

Sample space = {(C,NC),(C,C),(NC,C)}

What's the probability of them both happening, one after the other, no matter which one first?

P(C,C) = P(1) * P(0.5) = 0.5

No need for Baye's Rule because crits are independent events so there's no conditioning.

No need for PaaschĂ©'s Index because we're given enough information on success.

>>

>>582925123

why do you think all three of those have an equal probability?

>>

>>582924836

One must assume it is an outcome for the sake of math, that all four orders are possible.

When your first is a 100% crit, your second is a 50% chance. You would never know it unless you missed the second crit.

Just like youd never know which was your guaranteed crit if both were crits

>>

I don't know why I'm doing this but I'm really bored.

Basically its a conundrum. The chance isn't 50%, although it says it is, its a paradox. One of the hits has a 100% chance of being a crit, as stated, it just doesn't use a percentage. Then, whichever hit ISN'T the 100%, only has 50% crit chance. So, to satisfy all possible outcomes, if I hit the enemy once, and that hit is a crit, then my next hit has a 50% chance of being a crit. Thus, since one crit is already 100% confirmed, and we know the second is 50%, the chances of both are 50%. And then if the first hit isn't the 100% crit, then we know that the second one absolutely is, and so the first one has a 50% chance, but the second is already 100% confirmed. However, you could also argue the logicality of the situation. Its both solvable and unsolvable at the same time.

>>

>>582925369

if its a 50% crit chance, its a 50% non crit change... They are equal

>>

>>

>>582925212

No, I am not. I am listing the only three possibilities:

Critical, Non-Critical

Critical, Critical

Non-Critical, Critical

There are the only three possibilities available. As such, they are all equally one-third as likely to occur. Simple. I design software, I know how things work.

>>

>>582925441

it's an assumption

you could equally assume for the sake of the question that it is not possible to get an outcome with no crits

>>

ITT: Everyone make up their own rules of probabilities

>>

>>582925048

If you have ever taken a stats class you would know that after calculating your probablities the total must equal 100% no more no less.

In your equation its 200%

Your mistaking the guaranteed shot as in it must have a 100% in at least on situation per 2 shot event. This is not true however since if the other event is already a crit the other shot is no longer guaranteed.

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