This got me seriously disappointed in you guys...

He got the answer wrong as well since it's 50% chance.

The question is poorly worded, but that user is being retarded because it's pretty obvious what it's asking "given that you already have one critical, what is the probability of the next hit also being a critical hit".

Anyway, when you're asking questions of "if X and Y occur", then you multiply probabilities.

Prob., critical hit 1 * Prob, critical hit 2 = Prob., hit 1 and 2 being critical

We already know that there is one critical hit, so the probability of that is 100,%, or 1/1.

1/1 * Prob., critical hit 2 = Prob., critical hit 1 and 2 being critical

We know that, under normal, non-guaranteed circumstances, the critical hit probability is a half.

1/1 * 1/2 = 1/2

Which makes perfect sense, because unless we have evidence to the contrary, the events of the previous hit (the critical) will have zero influence on the subsequent event. If you flip a coin and get heads, it has nothing to do with what the previous coin flip was.

Bayes only works on conditional probabilities. There is absolutely nothing in the question that indicates this to be true. There is no statement that says "if you have one critical hit you enter rage mode and you're more/less likely to get more criticals". It was just "what's the probability of two criticals, oh btw your first hit was critical".

It's the issue of the poor wording again, but acting like a smug dick about it because you read the question wrong is honestly a bit silly.

It makes double heads more likely, but the inherent probably of that specific coin flip being heads or tails is still 1/2.

The double heads being more likely is akin to the double crits being more likely (i.e. 1/3 instead of 1/4).

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The original question doesn't say "oh btw your first hit was critical", it says that two hits happened and at least one was critical. There is a difference. Bayes does only work with conditional probability and the original question asks "What is the probability of getting two critical hits GIVEN THE CONDITION that at least one of those hits was a crit?"

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What 1/3fags don't understand is that you can't just drop an entire probabilistic outcome and say it doesn't count. If these were real crits and you apply their model that means sometimes your hits won't even register.

As it stands if you follow the initial wording it's basically suggesting that every hit has 50% chance to crit but one of them will be forced to become a crit if neither of them are.

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>given the condition
....Actually...shit, you're right. It is contingent. Thanks, dude.

1/3rders never respond to posts that accurately refute their stupid answer.
1/3rders are actually shitposters.

There's two 1/3rders and both have a 50% chance of being retarded.
Given that one 1/3rder is confirmed retarded, what are the chances that they are both retarded?

Answer: 50%.

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>we don't know the results of the next hit
You were born on one day of the year. If it wasn't between January first and December 30th, I can tell you it's on December 31st.

Actually, wait...shit... I'm , I change my mind again.

So this dude made the argument that the probability of double heads / hits, double misses, etc.

As it stands, there are these many potential options, given that one head is guaranteed and there are two strikes with the other having 1/2 probability.

Both hits are critical
One hit is critical.

Fuck it, I'm making a tree diagram.

Given that one is confirmed retarded, but you don't know which one, we have the following possibilities: RN, NR, and RR.
There's a 1/3 chance that both are retarded given that at least one is retarded.

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There's a 100% chance that you're a retard because the answer is 1/3.

But you forgot RR (different order to RR :^)).
That means the answer is 1/4.

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It's a paradox for a reason.
1/2 and 1/3 are both "correct"

If you write that shit down mathematically the probability you will solve to is 1/3

If you read it out and apply simple logic the answer is 1/2

You'll have intelligent people arguing for both answers because it all come down to perspective.

That's stupid, you either are retarded or you aren't (and you clearly are). There aren't two RRs. If there were, why wouldn't you have two RNs and two NRs?

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Ok fellas, let's think of an IRL examople similar to this.

The gunslinger ensures every third hit is a crit. Valve uses a pseudo random distribution that changes according to how much damage you do in game but let's ignore that and say it's all random

Let's say that instead of the third hit having ensured crits is one random out of every three hits, then it becomes basically the same problem we have here but with one more hit.

Will that help you 1/3 brainlets understand that every roll has to have a consequence.

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>49681680
There's not two RRs, there's RR and RR but they are in different orders and are thusly different.

I know who you are though, not giving you a (you).

>At least one hit is a crit
One hit has 100% chance of being crit.
The second hit has a probility of 50%

The answer is 1/2

Samefag.

So here's my tree. There are three options available. If the first hit is critical, then the second hit has a 50/50 chance. If the first hit is not critical, the second must be critical.

The probability of double criticals is....1/4?

I do think wording is an issue (I already talked to another user of whether the problem implies contingency) but there are some dumb motherfuckers here.

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Thats an
>ugleh
weapon

The only perspective shift that matters is whether you're calculating all this before or after the hits. If I flip two coins, each one has a 50% of being heads and I have a 25% chance of getting two heads in a row. However, if I flipped two coins at some point in the past and then tell you that at least one of those flips landed heads, it changes the probability of there being two heads. The ambiguity comes from the fact that "hit" is the past tense of "hit".
This argument also applies to , who would be right if OPs question didn't clearly imply that both hits have already happened.

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If one is already 100% crit, and the other has 50% crit chance, then they both add up and it will be 75% that both are crits.

The explanation is meant to confuse you. You can get 10 crits one ofter the other with a 50% chance. It's not important what happened before, the next hit will have a 50% chance of crit.

If I get two coins and flip them over and over at the same time, I garuntee you that the ratio between times where they both land heads and times where at least one was heads is going to be 1/3.
Refute this.

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Anybody recognising the pattern with some of these posts?
The contain certain numbers and letters combinations, and were previously posted as well.

They also act to provacate in an efficient, calculated matter.

>The only perspective shift that matters is whether you're calculating all this before or after the hits.
This, basically.

And this user's explanation is the one that makes more sense to me:

The real pattern is how the ratio of double crits to at least one crit is 1/3

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>then they both add up and it will be 75% that both are crits
>Imagine being this retarded
If you flip a coin once and it tails, there's still a 50% probability that the next flip will be tails, therefore the chance of getting a second tails in a row is 1/2 not 3/4

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Yes but the problem isn't asking you what are the chances that the 2nd shot is going to be a crit or a hit. In that case, 50% is right.
It just asks you "there were 2 shots, at least one is a crit, what are the chances that both are crit?" which is a different question.

>tfw smartfag
Originallingvey

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It's still 50% chance of crit, like I said, it's not important what happened before, the chance remains the same.

The way it's worded is why makes it a paradox. That's why it's an interesting problem

Almost.
It's more on how a person reads the problem
>of the 4 possible outcomes flipping 2 coins how many will have at least 1 coin land heads = 3
>of those 3 possible outcomes in which at least one coin landed heads would both land heads = 1
>1/3

Flip 2 coins
At least 1 coin will land heads
What are the odds both land heads?
>since at least 1 coil is to land heads up you just take 1 coin and set it down heads up.
>now you're only flipping 1 coin which means 1/2 chance of getting heads

>at least one is a crit
>thinks you can get two non crits

Even your break down relies on past v. future. Your 1/2 argument assumes that "one coin lands heads" = "one coin will land heads".
It doesn't, that's why it's 1/3.

50%. At least one hit means one hit is 100% going to be a crit every time. You have a 50% chance to crit and we have 1 hit left that isn't known.

That latter ((perspective))) is flat out worse than wrong.

>Your 1/2 argument assumes that "one coin lands heads" = "one coin will land heads".
>It doesn't
It literally does. Like 'literally' in the literary sense, literally.

You're getting confused, let me ask you like this. You swing 100 hits for 50% chance of crit. 99 hits were crits, what's the chance of all 100 being crits?
>it all depends on the last one, which is still 50%

Of course the chance when looking at every individual hit remains the same.

Let's make it with coins because it's clearer.
>You flip two coins. The first one is head. What are the chances that they both are head?
Your outcomes can be:
>C1 Head, C2 head
>C1 Head, C2 tails
So the chance is 50%, and you're right.

>You flipped two coins. At least one was head. What are the chances that they're both head?
(which is our problem)
Your outcomes can be
>C1 head, C2 head
>C1 tails, C2 head
>C1 head, C2 tails
[C1 tails, C2 tails can't happen]

The chance of it both being head are 1/3.

Except since we know one will always be heads means the odds of both being heads changes to 50%.

Is English even your first language?
The problem is written to imply a guaranteed outcome for 1 of 2 events, which is what makes the entire problem a paradox.
If you don't actually see that then you aren't quite as smart as you think you are.

You're getting confused on the guaranteed part. It's not 100% chance on the guaranteed one, it's still 50% just like the rest, it just already happened.

How? By a wish granted by our lord and saviour?

I showed you the logic behind it in super plain words, there's mathematical proof here: .

No. it's (1/2)^100. That's the probability of 100 hits being crits. You are pretending that a given conditional is the same as a single outcome. If all hits are critical, out of 100 swings, the odds are (probability of all critical among the set) / (the set of all outcomes with at least one critical). In the case of 100 swings, there's only a single outcome without crits - 100 non-crits in a row. This set would be total - 1. The outcome of all crits is (1/2)^100 of the total set. The probability of all crits given at least one crit is exactly (1/2)^100 / (1 - (1/2)^100)

This is basic math.

We know because the problem says it.

>By a wish granted by our lord and saviour?
Yes, he said one would be a crit. It's a constant and not a variable.

It doesn't though. Not in the way you're using it.
Wrong.
The problem implies that two hits happened and at least one of those hits was a crit. It doesn't guarantee that a specific hit critted.

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We don't know that the first shot was crit, we know at least one of them was.
It could be the first, and the second hit.
It could be the second, and the first hit.
This scenario (C|H) is twice as likely to happen than C|C.

How to prove yourself wrong:

>1.) Flip two coins 1000 times and record the results.
>2.) remove all instances of Tails + Tails
>3.) Sum all instances of Heads + Heads
>4.) divide the number by 10

You will have a number between 28-38 (percent) of the total, proving that the actual probability is, indeed, 1/3.

I don't know why maths is so difficult for you Americans.

The 99 others are a known constant, not variables.

Yes, which is why we're not considering a outcome where H|H happened.
We're considering only outcomes where H|C or C|C happened.

We know one is and we know the other is a 50% to be, thus giving us 50%
It says in the problem we know that

>One is already 100%
>Other has 50%
>if both were 100% than it would be 100% that both are crits
>if both were 50% crits than thatbe 50% crit for both
>100+50=150
>since there are 2 hits divide it by 2
>150/2=75

Fucking brainlet

In your question it is true that the last hit has 50% chance of being crit. That is not equal to at least 99 being crits. At least does not mean exactly the first or exactly the second. You are worse than retarded.

Except they're not. You don't know which of the 100 are crits or not. You know only that at least 99 (any 99) are crits.
The problem says "At least one the hits is a crit".

.) Flip two coins 1000 times and record the results.
.) remove all instances of Tails + Tails
That's wrong though, tails + tails NEVER HAPPEN to begin with.
They were never a possibility.

They were a possibility. That's how probability works man.

Look, you can either crit or not. If you do, all 100 are crits, if you don't not all 100 are crits. The chance is 50%. Math is frying your brain, think logically.

Again, really, the thing with the outcomes is the easiest way to see it.

You're in this scenario:
>H1 crit, H2 hit
>H1 crit, H2 crit

When the scenario is this one:
>H1 crit, H2 crit
>H1 crit, H2 hit
>H1 crit, H2 hit

You're not considering the possibility that the "at least one is crit" is the 2nd hit.

But that's not the question and that's not how programming crits works

P(A|B) = P(A) because A and B are independent events. Likewise, P(B|A) = P(B) because B and A are independent events. P(A) = P(B) = .5 by the definition of the problem.

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

>Mfw a thread about this picture is always gonna be on r9k now

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Now you're dismissing the fact that at least one of them has to be crit.

No, one critical is guarenteed.

That's not what the problem is.

You replied to the wrong person.

They're not independent events.

"At least one is a critical" is a conditional on the outcome of the probability. That's a definition. If you do not follow the definition you are doing the problem wrong.

My point still stands if at least 99 will be crits you dense retard or if the first 99 will be
See above and the problem revolves around a single flips probability due to everything else being known, so its 50%

Easy.
Denote (x,y) in {0,1}x{0,1} for the possible outcomes.
Let C={(0,1),(1,0),(1,1)} denote the set "at least one hit was crit" and A={(1,1)} "both hits were crit".
Then our question is what is P(A|C), the probability of A given C. This is the same as P(A intersect C)/P(C) and since A is inside C it's just P(A)/P(C).
P(A)= 0.5^2=0.25 and likewise P(C) = 1 - P(({0,0)}) = 1 - (0.5^2) = 1-0.25=0.75 and 0.25/0.75 is one third.

Most likely.
Bunch of retards who think they are more intelligent than they actually are arguing with other retards who think they are more intelligent than they actually are.

Have to admit, have kind of missed a good old fashioned flame war

I like how this is the epic new troll problem taking over 4channel. Saw it on Jow Forums, /sci/, /v/ and here.

Only 2 outcomes exist. 1 is a crit and the other is a hit or both a crits. Its a 50% chance

One side isn't retarded though.
And that side is the 1/3 side.

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Please get help user, your parents worry about how retarded you are.

lol, it's already said to be a paradox and both are correct answers.
I guess 1/3rders want to post the super-smart maths equations to stroke their ego.

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no, the scenario stipulates that the END RESULT of two hits has at least one crit. The crit chance is explicitly mentioned to be 50%, which makes it identical to flipping a coin twice.

We then discount one of the four possible results because we know it did not happen, leaving us with THREE outcomes. ONE of those outcomes is two crits.

The probability is 1/3.

Which part of this is difficult?

The probability is NOT conditional therefore the odds are 50%. Anyone who says 1/3 is a fucking meme tier brainlet

Nobody but 1/3ders think they are smart. The answer is obvious to anyone past the age of 10. These people are thinking too hard about the problem and have become contrived

>the only right answer is getting overly pedantic about phrasing and refusing to answer a simple question
i love this board
the 'you should be able to solve this' steinsgateposting died down eventually and this probably will too

Of course that's the problem, you just get confused because the chance of crit is 50%, yet one hit was already a crit and then it tells you at least 1 is a crit. Yes the first that already happened, the question if both are crits solely depends on the last one.

It's 1/2. If the first hit is crit, there's a 0.5 probability of that. The next hit is guaranteed to be crit as stated in the problem. If the first hit is the guaranteed crit, then the following hit has a 0.5 probability of being crit. In either case it's 0.5

There are only two results though. Either the coin landed heads or it landed tails

If Touhou is a video game, and at least some of the graphics are in the style of anime, what is the likelihood I will get shitposted at for posting anime girls?

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No, it doesn't.
Let's make another example with coins because I get confused with hits and crit and fuck me.

>I flipped 3 coins. At least 2 were heads. What are the chances of all 3 being heads?

Your idea:
>C1 head, C2 head, C3 head
>C1 head, C2 head, C3 tails

Reality:
>C1 head, C2 head, C3 head
>C1 tails, C2 head, C3 head
>C1 head, C2 tails, C3 head
>C1 head, C2 head, C3 tails
All these outcomes have at least 2 heads, but only 1/4 is the one we're looking for.

You'd be right if the problem was:
>I flipped 3 coins. The first 2 were heads. What are the chances of all 3 being heads?

If you mean you posting touhou counts as part of this, it's 100%.

>tfw I already prooved the correct answer but the thread keeps going without a (you)

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The first hasn't already happened. It's an exceptionally retarded statement: "given the outcome of the probability, I know what the first result is therefore I will discard it". I know you're being retarded on purpose, but it's not funny to anyone else but you.

But Touhou isn't an anime. It can't be 100%.

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There are three outcomes.

>C1 head, C2 head
>C1 head, C2 tails
>C1 tail, C2 head

Tell me how this doesn't fit the "at least one is head" bit.

Time for honkposting

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No you fucking retard. Flip a coin one is going to be heads. What is the probability that you will get heads the next time you flip a coin. There is no cosmic god ensuring a 50/50 spread. Each individual flip has a 50/50 chance. You could get heads 5000000 times in a row out of 500001 and the last flip will still be 50/50 regardless of the fact that we had a psychic tell us one of the flips results. The result already happened though so it no longer counts

>hasn't already happened
Then it's impossible to guarantee it will be a crit, because it still has a 50% chance.
This is the turning point, you can't guarantee chance.
The problem is shit man, let's just give it a rest.