/sci/ - Science & Math

>>11438180
50%

1/3

25%

>>11438183
>>11438196
If either of you two are reading this, elucidate your responses or your mother dies in her sleep tonight

>>11438197
You too

>>11438199
>>11438202
Jk immunity shiba cancels the mom dying but still I'd like to hear your thoughts.

>>11438180

(π*luck+Illuminati)/27

>>11438180
Bayes Theorem y'all

>>11438180
1. 1st hit normal, 2nd hit critical
2. 1st hit critical, 2nd hit normal
3. 1st hit critical, 2nd hit critical
The probability both hits are critical is 1/3

>>11438221
how does the 50% figure calculates in?
what if the chance of a crit is something else?
what if the first crit chance is different than the second crit chance?

can you elaborate on this?

holy fuck /sci/ is full of literal retards. since you already know one result is a crit, all you need to know is the odds that the unknown attack will be a crit. it is 50%. it's the same thing as asking what the odds of coming up heads x times in a row is after youve already flipped heads x-1 times.

all these monty hall faggots can fucking kill yourselves.

>>11438221
If both hits happen at the same time, the odds go up to 1/2.

>>11438318
why then get hit critical today 4 times if the chance was 10%?!

If you are that smart explain this to me?

>>11438318
>undergraduate education

>>11438318
There's two different scenarios where only one result is crit, dumbass

>>11438318
Learn to read before you call other people retard kid

>>11438348
those two scenarios are equivalent for the purposes of this task, dumb statnigger
the only thing that matters is the crit chance on the one hit that wasn't a guaranteed crit, which is stated to be 50%

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

The absolute state of /sci/

>>11438180
It's 1/3.
Here are the possibilities overall:

MM
MH
HM
HH

if we know that at least 1 H occurred, the possibilities are now:

HM
MH
HH

So, there's a 1/3 chance we get HH, or two hits.

>>11438180
100%. Since they are both just as likely to receive the same outcome then the second one will receive the same outcome.

>>11438196
fuck off retard the order doesn't matter bayes does not apply

>>11438348
H/m is the same as m/h, Monty hall btfo.
Think of it this way. If you are throwing 4 dice at the casino the odds are different than if you are rolling one dice several times. Because 1,2,2,5, is an order, but if you throw all the dice at the same time, there is no order.

50%
either it happens or it doesn't

>>11438221
NO

this question is absolute brainlet filter, even a youtube channel who specialises in math riddles fucked it up

The probability space is

1, hit, miss
2, hit, hit
3, miss, hit
4, hit, hit

1/2

>>11438180
It's 1/4 ma dude

HOW TO INCORRECTLY APPLY BAYES THEOREM THE THREAD

>>11438980
I bayes your mom right in her theorem

>>11438984
Bayesd

>>11438976
Good heavens, diseregard my response sci for I have misinterpreted the qeuestion

>>11438964
DING DING DING!!!!
WE HAVE A WINNER!!!!

>>11438964
>If I win the lottery this week I will win the lottery next week

>>11438964
Finally. I can't believe it took this long.

Looks like 50% chance to me.

We want to know if both of them are critical. One of them is definitely critical. Therefore, the chance that both of them are critical totally depends on the odds that the other one is also critical. The odds of that are 50%.

>>11438180
(1/2)^2=25% chance

>>11438964
This isn't even slightly convincing, not sure why a couple others are saying this is the answer, unless trolling.

>>11438971
>H/m is the same as m/h, Monty hall btfo.
I thought I'd find you here after seeing this thread on /v/ last night. You're wrong. Sure, it's the same if you're checking the outcome and determining whether they both hit. One did, one didn't. But the point is, you can get "one did, one didn't" in two ways, whereas you can only get "both did" in one way. Nothing to do with Monty Hall, my friend. The "order" is completely arbitrary. The point is both attacks are separate events. If you throw four dice all at once, each die still gives its own result. If each die has a different colour, you can definitely say blue was 1, green was 2, etc. Or maybe green was 1 and blue was 2? There are a lot of options for 1, 2, 2, 5 there. But there's only one way to get four sixes.

>You flip a coin twice. At least one of the flips is heads. Assuming a 50% chance of heads, what is the probability both flips are heads?

If anyone can explain to me how this problem is substantially different from the one in the OP, only then can they begin to justify why the answer is anything other than 1/3.

>>11439039
"atleast one of the flips is heads" is a precise mathematical statement that doesn't really happen in real life scenarios naturally very often. A lot of the real life ways people will interpret "one of the flips is heads" are not actually "at least one of the flips heads" and so the probability is not 1/3. It's entirely how the question is phrased.

>>11439050
The question is phrased precisely. People coming up with a different interpretation have only themselves to blame.

>>11439039
>At least one flips heads
You have to rule out TT results.

>>11439086
So its either HT or HH. Therefore, 50% chance that both are heads.

>>11439086
That is correct.
>>11439090
TH is also a possibility. Therefore, not 50% chance.

>>11439106
Ah yeah, I get you. Was lazily just assuming that HT = TH... whoops.

>>11438196
Correct.

Interpretation of probability: Generate infinitely many sequences {0,1}{0,1} where each of 00, 01, 10, 11 are equally likely. Discard all the 00 ones. Take the fraction of those sequences 11 to other remaining sequences.
Mathematically calculated and computationally verified, the result is indeed 1/3.

I strongly believe in 50% because given X, Y (X=1 ^ Y=2 or X=2 ^ Y=1) say hit X is given to be the critical one, no matter what Y still has an independently 50% chance of a critical. Hence 50% overall.
However, my professor told me to just apply conditional probability's definition from which I got the following.

> A = both crit
> B = at least one crit
> P (A ^ B) = P (B) x P (A | B)
> P (A | B) = P (A ^ B) / P (B)
> P (A | B) = 0.25/0.75 = 1/3

What is the fallacy I made in my argument in the first line arguing for 50%, 1/3fags?
Or rather did I make a mistake in the mathematical solution for 1/3? Did I get P (B) wrong? 50%ers?

Because, if I redo the calculation with A and B as the following (more similar to my argument above for 50%):

> A = first hit crit
> B = second hit crit
> P (A ^ B) = P (B) x P (A | B)
> P (A | B) = P (A ^ B) / P (B)
> P (A | B) = 0.25/0.5 = 0.5
Which works because we have two cases: A is given to be crit, or B is given to be crit. These cases are symmetrical. So in any case the chances are 50%.

What is the fallacy in my second mathematical argument, 1/3'ers?

Please. I'm dying to know.

>>11439196
>I strongly believe in 50% because given X, Y (X=1 ^ Y=2 or X=2 ^ Y=1) say hit X is given to be the critical one, no matter what Y still has an independently 50% chance of a critical.
That would be true if you knew that X was the critical one. But you don't. It can be either X or Y, with equal probability (or both).
Essentially the problem is that the "the critical hit" is not well-defined, because there could be true of them, so you can't reason about "the critical hit and the other hit" independently.

>>11439203
Oh fuck so 1/3fags were actually right all along...
FUCKKKK...

>>11438180
>50% crit chance
what are you playing?

>>11438973
Based probabilitychad

>>11439333
Rutger in fire emblem

>>11438199
I got 50% as well. Here's my reasoning:

It is given that one of the hits was a crit. For the other hit, you have two options:
If the other hit is also a crit, then "both the hits are crits" is true. Otherwise, if the other hit is not a crit then "both the hits are crits" is false.

Therefore the probability that they are both crits is entirely dependent on the probability that the "other hit" is a crit or not. That is, 50%

it's one of these:

hit hit
hit miss
miss hit

and all of them have the same probability of happening since the strikes are independent I guess. therefore it's 1/3.

>>11439203
Brainlet detected.
You can just call the one that was critical "X". You could call it anything.
> You don't know if X was the critical one or not
Yes we do, because we're calling the critical one "X". And we know it exists too, since it is given that one of them is definitely critical

>>11439406
>the critical one
My point was that "the critical one" is not well-defined, because there could be two of them. If both hits are critical, which one is "the critical one"?

>still arguing over this
>have to take part in it

The problem is so damn simple, it is not even funny. The question boils down to: Do you know that one is a critical hit because...
...you fire one shot and if it is not critical, you try the next one: 50%
...you fire both shots at the same time or after another, and somehow just know, in a quantum universe like way (not really), that at least one is a critical and else you discard? 1/3

Classical interpretation is the second one, because the wording of the problem indicates that there is no specification for order of shots and their timing.

1/2 crit chance means 4 possibilities
crit,not
not,crit
crit,crit
not,not
4th is ruled out by the specificiations of the scenario
1/3

>>11439418
>...you fire one shot and if it is not critical, you try the next one: 50%
You mean the first shot is critical?
>...you fire both shots at the same time or after another, and somehow just know, in a quantum universe like way (not really), that at least one is a critical and else you discard? 1/3
Yes. That is precisely the problem. You don't have any information about the first shot, just the fact that one of the two shots was critical.
That's why the answer is 1/3 and NOT 1/2.

You say the second one is "classical interpretation", yet to interpret the problem in the first way is to completely change the problem statement because you provide much more information, which then renders the problem trivial.

>>11439412
Why does it matter?

Notice: any of the following may be the case
1) first shot is critical and the second shot is not
2) the first shot isnt critical and the second one is
3) first shot is critical and the second one is also critical

The probability of it being critical is fixed at 50%. Notice: this means it is 50% regardless of wether it came first or second. I.e: order doesn't matter.

So, let's say we order them like so:
1) first shot is critical and the second one is not
2) second shot is critical and the first one is not
3) first shot is critical and the second one is also critical

Let's call the shot referred to in the first part of each sentence above as "X" (aka: "the critical one") and the second part as "Y" ("the other one")

Given that X is critical, what is the probability that both X and Y are critical?

= 1 (because definitely happens) * 0.5

P(2 of 2hits were crits |at least 1 of 2 hits was crit) = P(2 of 2hits were crits | 1 of 2 hits was crit or 2 of 2 hits were crits) =(P(2 of 2 hits were crits and(1 of 2 hits was crit or 2 of 2 hits were crits))) /(P(1 of 2 hits was crit or 2 of 2hits were crits)) =(P(2 of 2hits were crits) / P(1 of 2 hits was crit or 2 of 2hits were crits)) =(1/4)/(2*1/4 + 1/4)=1/3

>>11439427
Okay, now look at the last part of each sentence. Twice you've got "is not" and only once you've got "is also critical."

>>11439427
To clarify the last step:
P(both X and Y are critical) = P(X is critical) * P(Y is critical)
But it is given X is critical (actually we essentially defined X to be critical), so P(X is critical)=1

So P(both X and Y are critical) = 1 * P(Y is critical)
= 1 * 0.5

>>11439427
Alright, if X refers to the first one in the list
>1) first shot is critical and the second one is not
>2) second shot is critical and the first one is not
>3) first shot is critical and the second one is also critical
and Y the second one then
P(X and Y critical | X is critical) = P(Y critical| X critical) = P(Y critical| X critical and X is the first shot ) * P(X is the first shot ) + P(Y critical| X critical and X is the second shot) * P(X is the second shot )
Now
P(X is the first shot) = 2/3 (only 1) and 3) fit)
P(X is the second shot) = 1/3 (only 2) fits)
P(Y critical| X critical and X is the first shot ) = 1/2 (1) and 3) fit)
P(Y critical| X critical and X is the second shot ) = 0 (because none of the 1) 2) 3) satisfy this)
Adding up, we get
P(X and Y critical | X is critical) = 1/3

>>11439441
>P(both X and Y are critical) = P(X is critical) * P(Y is critical)
This is wrong. X and Y are not independent. You have to justify this step.

>>11439446
Ths probability of any shot being critical is fixed at 50%

>>11439423
>which then renders the problem trivial.
The problem is trivial, in both cases. What is not is the exact interpretation of how it actually happens in the real world. Think about it: If this was a scientific study, how would the study actually happen? If one group of scientists decided to shoot one shot first, check if critical, and else discard that run, it would be within the specification. It would be more specific then, you are right. What does this mean? This is a linguistic problem, because the wording does not absolutely make sure that it gets interpreted in the same way, *even* if the interpretation that both shots occur simultaneously and are being observed so that one can throw away only the cases where both are not critical, is somewhat more standard.

IMO this is just a linguistic/philosophical problem. The math is easy, trivial and solved, but the way the problem is defined is too unspecific to conduct a real world test, rendering the whole problem debateable in the first place.

>tl;dr
Rephrase the wording so that different groups of scientists could (in theory) conduct some kind of test leading to the same results - the problem will be gone.

>>11439450
Yes, but that's unrelated to my question. Why is P(both X and Y are critical) = P(X is critical) * P(Y is critical)? Remember that we're only given that 1st shot and 2nd shot are independent. X and Y are functions of both shots. It may very well be that they're independent, but you have to demonstrate this.

>>11439451
In my opinion all probability questions should be required to phrase in terms of an algorithm and asking what a given ratio tends to as N tends to infinity. It's the only way to avoid ambiguity.

>>11439457
Exactly what I mean. Well said!

>>11439451
I still maintain that there is no problem in the wording and that people are instead looking for problems that aren't there. In fact an absolute genius in the /v/ thread last night came up with a perfectly natural way of putting the problem in context.

You're playing a game. You attack an enemy with 5 hp. Your base attack does 2 damage, but you have a 50% chance of critting for 4 damage. You attack the enemy twice before it dies. Now you know at least one of your attacks had to be a crit. What are the odds that both were?

WHY is this so debated? The question is well defined and the answer is as absolute as any other mathematical expression.
The 4 possibilities are:
(hit, hit)
(hit, miss)
(miss, hit)
(miss, miss).
The question eliminates the fourth tuple, so only the other 3 are valid within the domain of the question, and the answer is 1/3.

>>11439462
In that case there is an added condition to the problem: You do know *how* you know that one was critical. The problem of OP does not specify this, which is why you can interpret that part.

>>11439472
You know because the problem tells you.

Here are the possible outcomes, with 'C' denoting a crit, 'c' denoting no crit and 'G' denoting the given crit that we know exists:
>Gc
>cG
>GC
>CG
It's 50/50

>>11439475
>At least one of the hits is a crit.
Do you just know or is there some kind of real world connection? Like how do you know? You can shill for your opinion that you just know "because the problem tells you" all day long. That is a linguistic problem, we can start a new thread about semantics and interpretation of words, if you wish. Still, >>11439457 totally nails it. If you just want to one interpretation to be correct, then you should use a well defined mathematical one.

>>11439463
>The question eliminates the fourth tuple
look everyone I found the retard

>>11439390
Yes but what if the "other hit" is the crit and the first hit is not?

>>11439499
Except it unarguably does you shit for brains midwit fucking moron and denying this doesn't change that.
The question is well defined and no one who isn't either genuinely stupid or purposefully denying it to save face would argue against this.

>>11438221
>Odds of crit = 50%
>Odds of no crit = 50%
>1. 1st hit normal, 2nd hit critical = 0.5*0.5 = 25%
>2. 1st hit critical, 2nd hit normal = 0.5*0.5 = 25%
>3. 3. 1st hit critical, 2nd hit critical = 0.5*0.5 = 25%
25% + 25% + 25% = 75%
You can't have a probability tree that doesn't add up to 100%

>autistic people who don't speak english reply to a troll 5000 thread anniversary edition

>>11439497
There is no linguistic problem as far as I'm concerned. And given that semantics appears to be at the core of the issue here, there's no need to make a new thread about it when this one will do. The problem tells you that you have the knowledge that at least one hit is a crit. Without making any further assumptions about how you obtained that information, the most straightforward interpretation will yield the result 1/3. Anything else requires you to speculate about what if actually, it's not merely that you have this knowledge, but maybe it was guaranteed all along, or whatever, and that's going beyond what the problem states.

>>11439511
>implying troll threads aren't the best content /sci/ has to offer

>>11439503
Wow I was not expecting you to get baited this hard. Thank you.

>>11439513
Exactly, which is why it is tempting to think about the problem in the first place. Make it another thought experiment: How could you translate the problem to the real world? If you wanted to conduct a study, you would need to define the procedure.
If that should not happen at all, then a mathematical expression is needed. The problem of OP is not in any way mathematical expressed.

>Without making any further assumptions about how you obtained that information, the most straightforward interpretation will yield the result 1/3
That is just like your opinion. I beg to differ. I know that my interpretation that interpretation is allowed is probably the minority interpretation, but the problem is not expressed specific (in mathematical terms) enough, that I could not think about it that way.

>>11438180
Let X be the guaranteed hit out of the two hits that will crit, and Y the unknown variable that is likely to crit. Regardless of the order of hits, there will always be one X and one Y that is not bound by the statement and thus has only 50% chance of crit. So the answer is 50%.

>>11438969

Bayes' does not require order to work. It certainly is used when an order is imposed and you know about a later event, and are asking about a prior event. However, the statement of Bayes' Theorem does not have an order as a precondition. This is a common mistake I see students make.

It's very obviously 1/4.
P crit = 0.5
P nocrit = 0.5
Scenario 1: First hit is a crit (0.5), second hit is a crit (0.5). Total probability = 0.25
Scenario 2: First hit is a crit (0.5), second hit is not a crit (0.5). Total probability = 0.25.
Scenario 3: First hit isn't a crit (0.5), and therefore the second hit MUST be a crit (1). Total probability = 0.5.
The scenario with two crits has a probability of 0.25

>>11438180
P(both crits|1 or more crit) is
P(1 or more crit |both crits)P(2 crits)/P(1 or 2 crits)
is p(2crits)/p(1 or 2 crits)
is .5*.5/(1-.5*.5)
is .25/.75
is 1/3. If that seems low, recognize that the probability of getting two crits without a guarantee is .5*.5 is .25. So you do get a nice little boost from your passive.
>>11439150
>fraction of infinity
'no'

You always get to roll twice. You can prove that the second roll is a guaranteed hit given that the first one didn't but you can't prove, if your first one hit, it wasn't the guarantee. There's no reason to only enforce "at least one hit" on the second roll. Since the chronological order is irrelevant here, both rolls occur at the same time and you only ever get to play with one variable.

50%.

>>11438180
Both hits will crit because I anointed my computer in holy water.

>>11439631
There was never any guarantee. You are looking at things after the fact.

I think you can rephrase the problem to be intuitive as follows:
Your friends flips 2 coins behind your back and tells you that at least one of them is a tail. What's the probability that both are tails?
This is a precise formulation of the same problem and clearly the answer is 1/3.

>>11438180
...this is Pokemon-tier difficulty. Elementary school children can figure this one out and strategize around it.

>>11439658
Something's not right here, but I can't quite put my finger on it...

>>11438975
>>11439391
>>11439463
You got a boyfriend, I bet he doesn't kiss ya.

>>11439661
>>11439647
The statement is computationally realized by forcing one of your rolls to be 100% crit, and it doesn't matter which one if they occur at the same time or are preassigned.

The world can't check if one of the hits will crit before it decides if the other one will, since its own time frame does not allow this.

>>11439682
>The statement is computationally realized by forcing one of your rolls to be 100% crit
Why would you assume this to be the case?
>The world can't check if one of the hits will crit before it decides if the other one will
Whether or not one hits or crits is entirely independent of the other.

You're assuming a lot of things here that were never stated in the problem to change it into something else entirely.

>>11439687
>Why would you assume this to be the case?
How else would you assign a model on which the statement can be enforced?
>Whether or not one hits or crits is entirely independent of the other
If one doesn't hit, the other must hit. This is literally the key point of the question as it forces a propositional relation between the two.

>>11439708
Have you considered that it's merely describing a situation to you that happened by chance and nothing actually has to be enforced?

>>11438180
>at least one of the hits is a crit
>assuming a 50% crit chance

Two incompatible statements.

>>11439724
>Flip two coins
>At least one is heads
wtf how is this possible!

Threads like these always remind me how many literal retards hang out on /sci/ with 0 reading comprehension. I can't believe I still visit this place.

>>11439502
doesn't matter:
1.00 * 0.50 == 0.50 * 1.00

>>11439714
that's literally how pseudo-random distribution in video games works though

>>11439742
Who says it's pseudo-random and not random?

>>11439749
People who understand why it is only pseudo-random.

>>11439749
video games dont use quantum random number generators

>>11438180
50%
1*0.5

>>11439753
So idiots.
>>11439754
You're technically correct and yet entirely unhelpful. You must be an engineer.

You mathfags can get fucked, if one of them is always a crit then the outcome only depends on a single roll.
1/2

>>11438958
Correct
>>11438973
>>11438975
Incorrect

>>11439506
Those problems are in the wrong sample space dingus. The anon you replied is considering the sample space in which at least one critical hit has occurred

>>11439785
>>11439754
>>11439753
chad engineer
>>11439789
>>11439708
>>11439566
virgin mathcuck

>>11439798
>mathcuck
>someone else does your hw for you
o_o

>>11439797
yes i realized that after posting but it was too late to delete

>All hits have a 50% crit chance
>Attacks only twice
>One hit already happened and it crits
>Only have to worry about figuring out the crit chance for the other hit, of which you've already been told the answer

50%. Anyone that says anything else has no reading comprehension.

Just fucking type
x=1/2
x

>>11439890
>One hit already happened and it crits
No, you're misreading it. Both hits already happened. At least one is a crit.

>>11439897
The point fucking remains.

If you remove "At least one of your hits is a crit." Then you can bother doing anything else beyond (1/2)*100.

>>11439902
No, the difference matters, because now you're left with, is the first hit a crit, the second, or both, and all those options are equally possible.

>>11439917
You fucking troglodyte. You're figuring out the chance on a 50% crit happening. The first crit is out of the equation entirely. Second attack has a 50% chance or either being a crit or not crit. As such there's a 50% chance you get 1-0. A 50% chance you get 1-1 and a 0% chance you get 0-0.

>>11439927
Chill the fuck out, spergy.
>The first crit is out of the equation entirely
Wrong. "at least one of the hits is a crit" is NOT THE SAME as "the first hit is a crit"
Therefore, you must consider whether the first hit is a crit or not, and then whether the second is a crit or not.
With heads and tails instead: HH, HT, TH .... therefore it's 1/3 not 1/2.