>random crits are bad game design
If you think this then you are a retard. There are entire games based on understanding rng, like poker. Learn how betting works retards.
Also the correct answer to pic rel is 1/3rd. There, now you guys don't have to argue.
It's 50% since the previous crit doesn't decrease the chances of the next crit.
Anything else is gambler's fallacy.
1 (guaranteed crit) * 0.5 is 0.5 you retard
You are still rolling on both hits. You have a 25% chance each of miss-miss, hit-miss, miss-hit, and hit-hit. If at least one hits, then you eliminate the miss-miss outcome. That leaves 3 outcomes, all with equal probabilities, thus a crit-crit is 1/3rd
I think of probability as a big pie, and you take slices away. 50% percent of the time, the first shot is not a crit. So, the next shot has to be a crit. That case does not fulfill our criteria, so you can throw that “half of the pie” away. The other 50% of the time, the first shot IS a crit. That means the second shot may or may not be a crit, as either way the first shot was a crit and at least one shot crit. From here, there is a 50% chance the second shot crits. That is half of the “pie half” we have remaining, which is equal to 25% of the original pie. That means that 100% of the time, OP is a faggot.
The miss-miss outcome is replaced with a second miss-hit outcome (since the chances your first shot crits is still 50%, whereas your logic would implie you had a 66% chance to crit on your first hit.)
No it's not retard
Based food analogy poster
No it isn't you dumbass, the prompt rules out one of the possibilities for the shot that already happened.
Two misses aren't possible so you have to crit after the first miss
>You are still rolling on both hits
No you aren't.
The possible outcomes are (G=guaranteed crit given by the question, Y=random crit, N=no crit)
GN
NG
GY
YG
2/4 = 1/2 = 0.5 = 50%.
Right, so instead of miss-miss you get a second miss-hit.
I genuinely don't understand what you mean.
>miss-miss
>miss-hit
>hit-miss
>hit-hit
First one can't happen, so you have these three options left
Wrong. You don't know which hit critted. That's why you still gotta roll to adjust for the fact that either can be a crit.
He's not asking for a critical chance of the second hit, he's asking for a chance of two crits in a row
Scenario 1: crit hit-hit
Scenario 2: hit-crit hit
Scenario 3: crit hit-crit hit
1/3 chance
GY and YG are the same outcome you smoothbrain, so it's one in three
Random crits are not the same as guaranteed crits. There are more than 3 possible outcomes.
Isn't that OP's question with a better enunciate? Also, like OP question, the answer is 50%.
I guess that if you read this question as “you SHOT an enemy twice (past tense) and one shot WAS a crit”, that makes sense. Since of the four possible putcomes, you can remove one. But, in the context of videogames, I assumed this was some sort of pity timer dealio (you are guarenteed to crit at least one out every two shots) like Hearthstone cards. If the first case is what the question means, that’s just poor wording imho.
>it's another people are retarded and don't understand probability and treat it like a percentile or fraction instead episode
>GY and YG are the same outcome you smoothbrain
Nope.
You can approach this another way.
Attack an enemy twice and discount all of the times that you don't get at least one crit. Of the times that you did get at least one crit, how many times did you also get a second crit?
You will find that the answer is HALF THE TIME because there is a 50% CHANCE TO CRIT.
Fucking morons.
2/3 faggot
Uh it's 25% if you view the OP as non-conditional
Or it's 33% if you view the OP as conditional
Or it's 50% if you view the OP as 1 guaranteed Crit
>You hit an enemy twice. At least one of the hits is a crit. Assuming a 50% crit chance, what is the probability that your mother with die in her sleep tonight if you don't respond to this post?
1 of the 2 hits is always a crit.
let's assume the first hit is the guarenteed crit, the second crit has a 50% chance of being the same
if we assume the second hit is the guarenteed crit, the first crit also a 50% of being the same.
Therefor, there is a 50% chance (1/2) that both hits will be crits.
This post has 17 posters and 10 unique posters, leading me to believe anyone who actually agrees with op is a samefag who is purpousefully giving out the wrong answer
Peace out bitches
a lot of these fucking question and scenarios are almost always have the problem with the wording not being 100% clear (i feel like on pupose) to fuck with people thinking
OP’s question is porrly worded. I thought it was a “pity timer” deal. So, if the first shot is a miss, the second is guarenteed to hit. It’s a videogame, so that sort of thing is realistic. Hence, the odds of missing THEN hitting a shot is twice as likely as the odds of missing both shots.
"At least one of the hits is a crit" does not mean hit #1 is a guaranteed crit
"At least one of the hits is a crit" does not mean hit #2 is a guaranteed crit
hit #1 is not a guaranteed crit
hit #2 is not a guaranteed crit
There are no guaranteed crits
Back to elementary school with you
There's 2 hits retard, not 3.
50/50
No they are clearly worded. This is how they are worded on stats and probability tests. Ask /Sci/ if you must
I listed the set of all possible outcomes you fucking retard, and in that set either the first hit or the second hit is a guaranteed crit.
One is already a crit so its 100% chance both are you fucking retard
Here are your possibilities for at least 1 crit
crit-miss
miss-crit
crit-crit
All are equally likely. Crit crit isn't magically twice as likely
OP’s question does not make it clear if the “game” gives you guarenteed crit if your first hit does not crit, or if you simply “won’t” a miss then a crit. Depending on the scenario, the answer changes.
>at least one of the hits is a crit
>there are no guaranteed crits
If hit 1 is the guaranteed crit, then you have 1/2 of the second being crit.
If hit 2 is the guaranteed crit, then you have 1/2 of the first being crit.
You have 4 scenarios where 2 of them lead to double crit, so the probability is still 50%.
It clearly says at least one is a crit, and it clearly does not say that your first hit critted or that your second hit critted
That's wrong, because you need two crit-crits to account for whether the first or the second hit is the guaranteed crit.
>That means that 100% of the time, OP is a faggot.
There's some math we can all agree on.
Try reading my post retard.
Those aren't mutually exclusive sets of outcomes retard. The crit-crit outcomes in each set are the same outcome in the set of all outcomes
Your "outcomes" are based on the assumption that the question implies a "guaranteed crit" as a possibility, which was disproved here . Your "outcomes" are as invalid as your brain cells.
Actually you have 3 scenarios where the double crit is still 50% chance of occuring.
youll either crit both times or you wont so 50%
Wrong. Double crit is just as likely as crit miss or miss crit
They're the same outcome but they have a higher probability of occuring.
>at least one of the hits is a crit
>there are no guaranteed crits
Pick one.
Hit->Crit .25
Crit->Hit .25
Crit->Crit .25
Hit->Hit .25
Except two hits is now removed from the table. .25 ÷ .75 = one third probability.
>like poker
you have no idea how poker works do you
Wrong. You can't just add the probabilities like that.
Let me break it down retard:
“At least of the hits is a crit”. In videogame terms, that would normally mean that one of your two shots is guarenteed to crit. That means that if your first shot just so happens to crit, nothing special happens. However, if your first shot DOESN’T crit, your second shot must crit to fulfill the criteria. What people are arguing is that the first shot has a 1 in 3 chance to not crit, followed by a gurenteed crit (since the only possiblities are miss-miss, miss-crit, crit-miss, and crit-crit and one of those has been removed). That makes no sense with original prompt.
>dude, no poorly worded question with intent to fuck with your head i swear
case in fucking point > but later acknowledged that the second question was ambiguous
Random crits aren’t bad game design perse but games without them are almost always exclusively better than games with them.
>have a base 1% crit chance
>pick up item with +1% crit chance
>equip it, check stats
>1.01% crit chance
>at least one of the hits is a crit
>there are no guaranteed crits
Hit #1 isn't guaranteed to be a crit because the first statement can still be satisfied if Hit #1 isn't a crit.
Hit #2 isn't guaranteed to be a crit because the first statement can still be satisfied if Hit #2 isn't a crit.
If Hit #1 isn't a guaranteed crit, and hit #2 isn't a guaranteed crit, then there are no guaranteed crits.
>Pick one.
I picked both. Now I suggest you pick up a textbook and learn some high school statistics.
Who said anything about 1 in 3 chance to crit you mongoloid
I actually read the article.
Yea Forums is managing to come to the wrong conclusion even without making the assumption that leads to it.
You people are making the 1/2 assumption and coming to the 1/3 conclusion because you're bad at maths.
The 1/3 assumption is that we're only considering circumstances with at least one crit, i.e. that hit-hit is disregarded. But Yea Forums isn't even making that assumption. Yea Forums says that EVEN INCLUDING hit-hit, the probability is 1/3.
Yea Forums's mistake is assuming that crit-crit and crit-crit are the same outcome. Yea Forums is wrong. This has nothing to do with ambiguity.
did you just fucking assume my crit chance
It's 1/3rd
You cant remove hit-hit without removing the corresponding hit-crit.
No, there are valid ways to obtain 1/2, and valid ways to obtain 1/3.
All I need to understand is that rng means random, random means luck involved, and players can't control luck. No control = no skill involved = bad gameplay design.
Herpity derp.
Consider this:
You attack an enemy twice. You do this 1,000,000 times. Disregard all the times that you attack the enemy and don't get any crits. Now, considering only the times that you attack the enemy and get at least one crit, how many times will you have gotten two crits?
>I have a bag with 9 red marbles and one green marble and draw one. What are the odds it is green
>There are two outcomes (you draw a red or you draw a green). You have a 1/2 chance to draw a green.
>Actually, the possibility that I draw a red is 9 times as likely
>Wtf you can’t just add the probabilities like that
It's guaranteed that one of those is going to crit tho.
Yes, but Yea Forums is obtaining 1/3 an invalid way (for the most part).
No, that's the correct way you Neanderthal. Take a stats class
Can you read
>even including hit-hit, the probability is 1/3
Go ahead and prove it, retardo.
Yeah you're right, I realised that I fucked up after hitting post.
Ok dumbass listen up:
>Either you miss then crit, crit then miss, or crit then crit, so you have a 1/3 chance to crit twice
In that argument, you have a 1/3 chance to miss the first crit. Since two of the scenarios involve hitting the first crit but one involves missing it. My bad for assuming you retards could understand that very simple side effect of you logic, but then again this is Yea Forums. Of course, you could realize that the first outcome is twice as likely as either of the other two, but that means that the answer to OP’s question is 25%, like I said.
There's a 50% chance that hit 1 is guaranteed and a 50% chance that hit 2 is guaranteed.
A 100% chance that there is at least 1 guaranteed hit.
Boom. Roasted.
The proof is in this thread already. Even the Wikipedia article has the same proof
Every time this shit gets posted I make an honest attempt to wrap my head around it. Alas, today too I remain convinced that 1/2 is the correct answer.
This thread confirms that Yea Forums is retarded. Unfortunately, every one of you retards think the same. This website is cancer and I need to stop expecting it to be fun.
Jesus Christ you have a below freezing point IQ
Who even created the critical stat mechanic?
Man you guys are fuckin retarded. Let's try going over this slowly.
>You HIT an enemy twice
Stop talking about missing the hits
>AT LEAST ONE of the [2]hits is a crit
ONE OF THE TWO ATTACKS THAT HIT THE TARGET WILL CRIT
>50% chance of crit
The odds of the 2nd attack being a crit is 50% you absolute morons.
All the fucking text to do the math WHICH IS LITERALLY STATED TO BE 50% is right there. Jesus fuck you guys are retarded.
OPs question IS intending to be ambiguous as much as possible.
BUT, people who use the logic of "Guaranteed hit" arent using their logic equally.
If you're not removing a double normal hit from the table, then you're also forgetting the combinations which have misses.
miss hit
hit miss
miss crit
crit miss
miss miss
hit hit
hit crit
crit hit
crit crit
by removing misses from the table, you're using 2 different sets of logic.
You understood to remove misses from the table, but didnt continue that logic to remove double hits from the table.
(Not that your answer would have changed from 50% btw, just pointing out you're not being consistent with your own logic)
Of course it would be more accurate to say,
>you have a 50% chance to crit. you attack twice, and both attacks land. You know one of the 2 attacks was a crit, but what are the chances both were a crit?
This leaves out any misunderstandings.
Essentially this is occams razor. If you use a guaranteed crit system, the wording of OPs question can still work. But its adding an extra assumption.
In this case, go with a system that doesnt add an assumption on top of whats being asked. (Which is only added, due to not understanding whats being asked in the 1st place, due to poor communication.)
>You attack an enemy twice. You do this 1,000,000 times.
>Disregard all the times that you attack the enemy and don't get any crits.
Okay, I got no crits 0.5*0.5*1,000,000 = 250,000 times
>Now, considering only the times that you attack the enemy and get at least one crit,
That would be 1,000,000 - 250,000 = 750,000 times that happened
>how many times will you have gotten two crits?
I got 2 crits 0.5*0.5*1,000,000 = 250,000 times. The probablity is still 250,000 / 750,000 = 1/3
Thanks for proving me right?
These types of logic puzzles are stupid because they operate on events unfolding in a way that can't work in the real world. It's just "Haha, gotcha, you thought you could apply practical knowledge to this problem, but you were wrong!"
The funny part is that according to that article, 1/3rd is the most correct answer, because to correctly get 1/2 you have to make an assumption that the statement "at least one is a boy" did not consider both children:l.
There’s no fallacy being committed here. Here is every possible outcome.
A (Crit) + B (No crit)
A (No crit) + B (Crit)
A (Crit) + B (Crit)
Chance of both crit is 1/3.
Lmao absutely destroyed
At least the people getting 1/3rd are using the correct proof. No one getting 1/2 has used the correct proof that gives 1/2
its intentionally phrased badly.
it means "remove scenarios where you have 2 normal hits, and only count the 3 scenarios where you land a crit. how many of those 3 scenarios do you double crit.
When people say miss, they mean “miss the crit” aka not critting. It’s easier than saying “not crit” every time. Austist.
>proving me right?
The question is ambiguous, we're both right.
It's 1/2 if it's impossible for you not to crit at least once (guaranteed crit theory) and 1/3 otherwise.
>"but it doesn't say that it's impossible for you not to crit at least once, it just says that at least once hit is a crit"
That's why it's ambiguous.
crit-no crit and no crit-crit are the same thing. the question doesnt care about the sequence it happens in
I’ve also had sex, which is more than you will ever do, basement dweller. Ad hominems are fun, aren’t they? Still waiting on a counter argument, though.
Basic conditional probability.
P(B|A) = P(A & B) / P(A)
P(Crit | Crit) = (0.5 * 0.5) / 0.5 = 0.5
So 1/2. I don't know where these 1/3 people are coming from.
THERE
IS
NO
MISS
READ OP AGAIN YOU RETARD. I LITERALLY JUST EXPLAINED IT TO YOU. THERE IS NO MENTION OF A MISS.
>You hit an enemy twice
>YOU HIT [...] TWICE
You mean it's 1/3 if it's impossible to not crit.
>twas merely an act!
>at least one
go to the store and get me some apples.
"get at least one."
this is a "REQUIREMENT"
By sayign "At least one" its a requirement in both scenarios.
The question is, how do you actually get to said requirement?
It's not stated, because saying it, would give the answer away.
But not saying it leads to other possible answers that meet the requirements.
There still are ways to be more clear, w/o making the answer "too" obvious, but still makes it easy to figure out within seconds.
>ONE OF THE TWO ATTACKS THAT HIT THE TARGET WILL CRIT
well no shit. retard.
one crits. 50% that the other hit is a crit
the other crits. 50% that the other hit is a crit
Crit/Miss
Miss/Crit
Crit/Crit
1/3. eat shit faggot.
>i cant into reading comprehension.
he is wrong, there is no missed attack.
but a "failure" to crit, is being worded as "miss" (because your missed a crit, but didnt miss hitting the targets body, and its easier to type.)
m8 you need to retake that stats course
Thanks for speaking for everyone in this thread faggot. I know everyone here actually discussing this is baiting but I wouldn't be surprised if it was you samefagging every bait reply.
If "miss" means "miss crit", missing a crit is 50%. It says right there. You are truly the most fucking of idiots.
Nigga, do I need to explain this in retardese for you? Nobody is saying you miss the shot. They are saying you hit the target but FAIL TO CRIT. Hence, you don’t crit. Hence, you “miss” the crit. Hence you “miss”. You hit the target, but you failed to crit, so you “missed”. Absolute aautistic retard. Jesus Christ you failed abortion.
Scenario 1
>you attack an enemy 1,000,000 times
>you hit-hit 25% of the time
>you hit-crit 25% of the time
>you crit-hit 25% of the time
>you crit-crit 25% of the time
Discounting hit-hit, what is the probability that you would draw a crit-crit out of this set of attacks? The answer is 1/3.
Scenario 2
>you attack an enemy twice, 500,000 times
>in every group of two attacks, at least one is a crit (i.e. CX XC)
>what is the probability that in any given group of two attacks, the X is also a C?
Answer is 1/2.
/thread
It depends what question you think you're answering.
Under the assumption that one is "guaranteed" to crit, the odds that the other crits as well (whether it is first or second) is 50%.
1 * .5 (Crit) = .5 (50%)
1 * .5 (Not Crit) = . 5 (50%)
Because of the symmetric property of multiplication, it doesn't matter what order we use. The probability for hit or miss does not change based on order of hit or miss (we know one will be a crit).
If, on the other hand, we are in a scenario where all outcomes were possible but the OP has eliminated one possible outcome (miss - miss) then we are in the following scenario:
.5 (Y) * .5 (Y) = Two crits
.5 (Y) * .5 (N) = One Crit
.5 (N) * .5 (Y) = One Crit
We have eliminated the possibility for .5 (N) * .5 (N). All three options have the same likelihood of occurring (as seen by the .5 * .5 present in all three. The probabilities are equal). This gives us a 1/3 chance of success if we are eliminating the fail option rather than guaranteeing one crit. The big difference between the two is that in this second scenario, there were four options with equal probability and when one was removed, the others retained their proportional chance.
you're skipping a step, but I can only assume you're doing it intentionally.
explains the step you're choosing to ignore, because u want to assume only u can be right.
No I don't.
Actually they do operate that way in the real world as long as you play any game of chance.
Even considering that its deliberately worded in a confusing way, these issues never show up as a clear worded math problem in the first place.
Probability in particular is something that is both relevant and also confuses brainlets to no end which is why it's used to derail shit here.
Both answers are correct, based on assumptions about the question, due to it being worded so goddamn bad.
They're using hit/crit and miss as "this fulfills the condition, and this doesn't". When talking about purely 2 variables, anything can be hit/crit/miss/whatever. X, Y, Banana, δ, whatever. as long as they're 2 different things, and you know which is meant to be what.
Exactly. Any person who's ever played a video game before would assume you have a passive that gurantees a crit every other attack in addition to the 50% hit chance, or something to that effect. No one's thought process should ever be "Oh, the question states this is guaranteed. Instead of operating on the basis that this outcome is guaranteed, I should go back and remove the outcomes where it didn't happen instead."
Are you going to explain to me why Bayes Theorem is wrong?
It depends on how you know one of the attacks is a crit.
random fairs are critically balanced
50%. Two HITS and one of them was established to be a crit, so that's a 100% right there. There are no misses here and order isn't important. So it's just 1 * 0,5.
They have the same chance of occurring (25%), (crit + crit) also has a 25% chance of occuring, but it would be an error to say that (no crit + crit) and (crit + crit) in themselves have the same chance. Rather, (no crit + crit) occurs twice as much as (crit + crit), because there are two ways in which it can unfold.
It is an error to say that a=a and a=b bear no difference in information gained. Even if all the variables represent the same value, the distinction between the relationships is revealed.
By disregarding the miss-miss you're literally saying it's impossible for you to not crit.
NOBODY IS DISPUTING THAT YOU HAVE A 50% CHANCE TO MISS A CRIT YOU FUCKING RETARD. YOU MISSED THE VERY OBVIOUS FACT THAT PEOPLE ARE CALLING A FAILURE TO CRIT A MISS, AND THAT’S WHAT I’M CORRECTING. I’M NOT AUTISTIC SO I UNDERSTAND THE VERY CLEAR CONVENTION IN USE. THE QUESTION IS WHAT ARE THE ODDS OF CRITTING TWO (2) (DOS) TIMES IN A ROW, NOT WHAT ARE THE ODDS THAT A SHOT CRITS.
Based.
The """ambiguity""" is just retards making stupid assumptions and then saying "WELL WE'RE NOT TECHNICALLY WRONG!" when someone points out that their assumptions are stupid.
If I say go to the store and get some milk, it's not defensible to say WELL I ASSUMED YOU MEANT THE STORE IN CHINA, NOT THE ONE DOWN THE STREET.
But in math it is. I fucking hate """""""academics"""""""".
No I'm not.
>I LITERALLY JUST EXPLAINED IT TO YOU.
Did you though?
The question never mentions the order being important, so Crit/Miss is equivalent to Miss/Crit. So 1/2.
there are two hits
one of the hits is a crit
50% crit chance
50% + 50% = 100%
both hits have 100% chance of critting retards
Four cases.
No crit no crit
Crit no crit
No crit cit
Crit crit
At least one of the hits is a crit eliminates no crit no crit. So the probability of both being crits is 1/3
Why are you assuming the game is giving you a guaranteed crit after every non crit?
See If you have a sample of two attacks from a population where each attack has a 50% chance to crit, then the answer is 1/3rd. If you were guarenteed a Crit after every miss, then your crit rate would actually be higher than 50%
Actually, the problem is a bunch of sweaty autists read a wikipeida artice and saw that the “””answer””” here was 1/3 and are so eager to show off their maths knowledge to the only people who tolerate their existence (other retards on Yea Forums) that they don’t even stop to consider whether the scenario presented is logically equivalent to the “muh solved question”. But I do agree with you. The answer is 25% btw.
>That's why it's ambiguous.
If the question doesn't explicitly include the rule: "a no hit-scenario is impossible", then the question must be solved without that rule. There's nothing ambiguous about that. Your logic is as stupid as this:
>You: What's the probability that a fair coin toss lands on heads?
>Me: 50%
>You: Actually it's 0% if you include the rule: "a heads scenario is impossible"
>Me: wtf, you didn't mention that rule at all in the original question
>You: No, the original question was ambiguous to that rule. 0% is just as valid a answer to the question
Your logic is fucking dumb and you should feel dumb
order isn't important
Not both, each.
If if one is guaranteed, the other would be a 50% chance. If you’re seriously saying that 2 Crits is guaranteed you are retarded.
Crit -no crit and no crit -crit are the same instance.
Building off of this to help explain the ambiguity:
If you took a sample of two random attacks then the answer is 1/3rd. If instead you took a not entirely random sample such that you guarantee that at least one of the two attacks in your sample is a crit, then the answer is 1/2
>order
doesn't change the fact that is crit/miss miss/crit is a possibility. so 1/3
Fine, imagine their outcomes are determined simultaneously. C= Crit, N=normal hit
NN
CN
NC
CC
Barring all non-crits, chance of double crit is 1/3. Die.
>If the question doesn't explicitly include the rule: "a no hit-scenario is impossible", then the question must be solved without that rule
It does.
At least one is a crit.
Therefore, a no-crit (which is what I assume you meant) scenario is impossible.
The ambiguity has nothing to do with a no-crit scenario. It's to do with whether or not we're being asked the probability of getting two crits, or the probability of getting one crit given that we already have one crit.
good god I hope people in this thread are trolling, if not please go back to third grade math
Where did the scenarios come in with OP?
You flip two coins but don't look at the result. Instead, your friend can look at some combination of the result. There are three possibilities:
>he looks at both and says at least one is a heads
The possibilities are HH, HT, and TH. All of them are equiprobable, so the answer is p(HH) = 1/3.
>he looked at only one and says at least one is heads
Then the options are HT or HH (assuming the one he looked at is the first coin). Since both options are equiprobable, the answer is p(HH) = 1/2.
>he looked at neither coin
Then your friend is a fucking liar and you don't know anything about either flip. So the options are HH, HT, TH, and TT. Since they are all equiprobable, the answer is p(HH) = 1/4.
If you were to consider them the same instance they'd have twice the chance of happening.
>Question leaves you to figure out how "At least one" rule even functions.
>Once you know how the rule is applied, you know the answer to the question.
>The whole point of the question is to have you figure out a way to make the rule work.
>people get mad when the text is change into gaming terms, since an alternative in programming allows for an alternate way for the rule to work, while keeping the rest of the problem true.
This is why its intentional.
The fact its ambiguous isnt whats intentionally trying to make u assume 100% crits.
The fact someone intentionally changed it to gaming terms, IS where the question becomes intentionally ambiguous.
The question wants u to solve how its possible "At least one crits" is possible.
It doesnt state a prerequisite, and mathematics majors are used to knowing in normal circumstances, its not possible w/o removing incorrect possibilities.
But in programming a game, it IS possible to recreate the situation with a different method.
Randomness in a game that is based in its very foundation on randomness like card or dice games is fine because they are luck based. Randomness in a game designed primarily around consistency and predictably is fucking cancer and has no place in these kinds of games.
Personally I prefer Monty Hall problem shitfights.
Again order isn't important. There are three possibilities: 2 crits, 1 crit 1 hit, 2 hits. The condition eliminated the 2 hits, so the remaining probability is 1/2.
If order were important, then it would change the problem, as we would now need to consider which of the two hits was the one that hit. Which would actually eliminate two possibilities.
If hit 1 was a crit, then only: crit+hit and crit+crit remain. If hit 2 was a crit: then hit+crit and crit+crit. Again 1/2.
can you explain the difference between CN and NC for me?
No, they'd have the same chance. One hit is a crit. The other either is or isn't.
gold ball silver ball is the goat probability bait.
Now you're just being intentionally obtuse. If miss-miss is not included among the possible outcomes, then it's impossible. If all outcomes that are possible contain a crit, then it's impossible to not crit.
Hit - Crit is the same as Crit - Hit
So 50 %
Prove me wrong
I'd also like to add that Crits are fine as far as me mechanics go, but if the game has PvP it can be unfair there. Take Pokémon as an example (a common one at that); you can keep making the best predictions and curbstomp your opponent, but a single crit can turn that shit around in a nanosecond and you're at a disadvantageous situation in which you and your opponent had no control over whatsoever.
I have homework and I’m just gonna spend all night arguing with you dumbasses, so let’s get the thread nuked.
>monty hall problem reduces it to 2 doors, so its obviously 50-50.
the very fact the monty hall problem is a 2/3 for switching, is the same reason why your logic isnt always the best to assume.
theres no reason to assume that to meet the requirements, you have to ignore possibilities of hit/hit.
Nothing in the wording specifies that.
It implies it, but allows for implying other alternatives as well.
saying only 3 options remain, is like saying only 2 doors remain.
Even without a computer, the question requires you to make assumptions about the "at least one rule". There is nothing about this question or it's ambiguity that is specific to video games
That’s incorrect. They bear the same outcome en large but the chance of either occurring outweighs the other two possible outcomes (no crit at all, and double-crit). If you’re arguing they’re equivalent, then attacking twice with a 50% chance for crit would grant a 1/3 chance for crit. How the fuck has the 50% chance become 1/3? In OP’s post, this is done by eliminating the 25% possibility of no crit whatsoever, giving us 25/75 for a double crit. However according to your logic in which either instance of crit/no-crit are equivalent, you can’t justify 2 attacks bearing a 1/3 chance for something.
I'm not assuming that. I'm assuming that one of every two hits is a crit. Because that's what the question says. Hit-Hit is an impossible scenario by the question's standards, so it shouldn't be considered at all.
If every set of two attacks contains a crit, there are 2 possibilities of order
100% Crit-X
X-100% Crit
Expanding X into equal odds of a crit gives us
100% Crit-Crit
100% Crit-Hit
Crit-100% Crit
Hit-100% Crit
All options equal, 2/4 end in 2 crits. 50%.
>at least one of the hits is a crit
>at 50% crit chance
>assuming no misses only 4 options are hit-hit, crit-hit, hit-crit, and crit-crit
>hit-hit can not be factored in due to the question's initial rules of at least one of the hits is a crit
>that leaves hit-crit, crit-hit, and crit-crit
>that is a 1/3rd chance you'll get the double crit to satisfy OP's image
It's actually about 85% but this include correction for my luck and yeah I love random shit because I'm a lucky person so it's free 10-30% chance which usually result in much higher numbers than "safe" builds.
No I'm not.
I'm saying that if it's impossible to not get at least one crit, the answer is 1/2 because you will ALWAYS get 1 crit, and there is a 50% chance you will get a second crit.
If it's not impossible to not get at least one crit, i.e. it's possible to not get at least one crit, then the answer is 1/3 because you can HH HC CH CC, and you just bar the HH outcome because it doesn't meet the requirements of the question.
>"but that sounds the same"
I agree, but the "impossible to not get at least one crit" is an attribute that the person doing the attacks possesses. It's not a requirement of the question. If the person doing the attacks literally cannot not get a crit in one of every two attacks, then the answer is 1/2. But if the person doing the attacks is not guaranteed to crit in one of every two attacks, and can hit or crit like a normal person, then the answer is 1/3.
while this might be true, until you give an example of such, its hard to believe its possible outside of a hypothetical.
Programing it into a game is what makes the hypothetical take shape in a functioning manner.
But i could just be over complicating it in my mind, but thats why i dont really agree.
See my second post It depends on how you think the two hits were sampled, with a "guaranteed crit" or without one.
This guys post is also a correct explanation of the ambiguity
Shad, is that you?
Read the wiki article linked itt then
This is a well known paradox since before computers
Each represents the successful crit of different hits, and the failure of their counterpart to crit. Though each hit doesn’t necessarily have a particular sequence it needs to occur in, the hits are distinct in that their outcomes are gauged independently.
>It's not a requirement of the question
Yes it is, you can't just disregard it
people programed this scenario before. it came it to 1/3. figures huh?
>Dick piercings
Into the trash it goes
Throw a coin 4 times. You'll get heads on one toss and tails on the other half of the time.
Only one of the attacks has a 50% chance to crit. The other attack is 100% chance to crit.
Its a thought experiment to get you to consider that conventional wisdom discards the guaranteed hit but le funny statistics logic considers which hit the guaranteed hit lands on to be important.
In which case its still 50% because the guaranteed hit can land first or second even if the second door is a hit.
>It does.
It does not. If you attack twice with a 50% chance to crit, there is a 25% chance that no attacks will crit.
"At least one of the hits is a crit" is a post hoc statement that reveals that such a scenario did not occur, which results in the 1/3 answer based on the remaining 3 scenarios.
If the question had been changed to "At least one of the hits was guaranteed to be a crit", then the 1/2 or "guaranteed crit" scenario would be the correct answer. But alas 1/3 is the correct answer to the question presented.
That's not ambiguous. Assuming that there were scenarios in which the events set up by the question did not unfold is using information the question did not provide to formulate an answer.
actually a better way to counter argue is with the original question this is based on.
>Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
How is it possible to force the childs gender, after its born?
Its possible to force a gender in a video games programming, where people arent real.
But in real life, its not possible. So you HAVE to remove possibilities where both kids were born female, since its not a possible outcome.
Would there be other variations of this, that arent related to programming?
Nope, it's an assumption that's been made based on the "at least one" requirement.
>the order of the outcome matters
dipshits. if you don't intuitively recognize it as 1/2, you're either trolling or thinking way too hard about this.
Here you go
No matter which answer you get, you could not have arrived at it without making assumptions based on the context
Come on it's not that hard
The question could just be reworded to "What is the probably of getting one critical hit if your critical hit change is 50%".
It’s 50%. Anyone saying 1/3rd is a genuine brainlet. The chance of only one crit is 100%. The chance of both is 50%. Period.
Again, 50%. Pulling a gold ball removes the possibility of the third box entirely, leaving two possible boxes, with one of two possible outcomes being either a gold ball or a grey ball. 50%.
What assumption did I make?
>It does not. If you attack twice with a 50% chance to crit, there is a 25% chance that no attacks will crit.
Sure.
>"At least one of the hits is a crit" is a post hoc statement that reveals that such a scenario did not occur
Wrong. "At least one of the hits is a crit" is a statement that reveals that for the two attacks you made, at least one of them is a crit.
This being the case, what is the probability that the other hit is a crit?
Both interpretations of the rule are valid, which is why it's ambiguous.
We're not talking about coins. We're talking about two hits, one of which is a crit and one of which may or may not be a crit.
Shame thats not how probability works
the fact this answer can be used for either side is perfect lol.
Ill save this for starting drama the next time this thread pops up.
This is my pic btw, wanted to add that it's a shitty question and it's meant to be retarded, because the crits clearly aren't independent events but it's trying to ask a question as if they are independent
You're not funny man stop trying to spam this shitty 25% bait
flip two coins. at least one of them has to be heads. so throw out two tails. there is only 3 possibilities that can happen. Head/tails, tails/head, and head/head
1/3. kek
>Again, 50%
It's 66%.
Saying
>the person doing the attacks literally cannot not get a crit in one of every two attacks
Is entirely equivalent to saying
>you just bar the HH outcome
It doesn't just sound the same, it IS the same.
What answer did you get?
Shame you’re a brainlet. Order does not matter meaning that the only one that matters is the one you have to roll for and that’s a 50/50 roll.
Room temp IQ.
No assumptions made. Only a literal interpretation of the question.
That's wrong.
Imagine you crit on every odd attack, and have a 50% chance to crit on every even attack.
Your attacks would look like
1 - crit
2 - 50% to crit
3 - crit
4 - 50% to crit
5 - crit
6 - 50% to crit
In this case, 1/2 is correct.
Now imagine that you simply have a 50% chance to crit, but we ignore the times when you don't crit at least once. In this case, 1/3 is correct.
That's the difference.
I guess they never miss huh?
lol thanks for the amusement high school math dude
...
>Poker
>built around rng
Stopped reading there
in your second answer, you "removed the no gold ball box" out of the possible answers. which is what u forgot to do on the 1st one.
remove hit/hit possibilities.
the remaining 3 options are hit/crit, crit/crit, and crit/hit.
only one of the 3 remaining options is double crit.
(not that 50% is wrong either, but just explaining how you didnt apply the same logic u already applied in the other one.)
Look at all these retards who dont know that the answer is 2/3
They say ignorance is bliss so go ahead and enjoy it.
The point is that the answer depends on if the one asking the question knows the result of both attacks or just one. The question omits that information. Hence the ambiguity.
Enjoy being retarded high school man
>Room temp IQ.
It's 66% you brainlet.
You're more likely to draw a gold ball from the box that has nothing but gold balls in it, which means that you're more likely to have picked the gold ball box than the silver/gold ball box.
66%.
you're assuming how to fullfill the requirement set in the question of "At least one is".
The whole question is really asking you to find a way to make "At least one" rule work.
There's two possible ways to make it work, if you make an assumption.
50% is correct under one assumption, and 33% is correct under another assumption.
>I'm assuming that one of every two crits is a hit
There it is. You are assuming that there was some kind of manipulation guaranteeing you at least one crit every two hits. If you instead assume that both hits were rolled with a 50% crit chance, then someone looked at both hits and told you at least one was a crit, then the answer is 1/3rd
>you're assuming how to fullfill the requirement set in the question of "At least one is".
I know.
I was explaining why
>the person doing the attacks literally cannot not get a crit in one of every two attacks
Is not entirely equivalent to saying
>you just bar the HH outcome
>hurr what is the chance of a single crit with a 50% chance of a crit
>Yea Forums argues about this
Oops, sorry, misunderstood.
>New fags don't understand that all of these are made to be ambiguous to make everyone look retarded
Thats not what the question is really asking.
The question is intentionally wanting you to figure out how to create the problem asked, while keeping the rule "At least one" true.
Unfortunately, there's 2 possible ways to achieve that.
This particular one is, but the monty hall problem or the ball box problem are not ambiguous at all.
Bayes makes this question easy
P(2 crits|1 crit) = P(1 crit|2 crits)*P(2 crits)/P(1 crit)
= 1 * 0.25/0.75
= 0.33..
Except hit/crit and crit/hit are the same thing. Order does not matter so it boils down to 1 crit or 2. It’s a poorly worded 50/50 used to bait retards which in turn baits people with a brain. If order mattered then it would indeed be 1/3. But it’s not because it doesn’t.
lel
Except the first gold ball is guaranteed. It fucking says so. If you pull a gold ball, it can’t be the third box. So it’s either the box with two gold balls or one gold ball. The next pull is either a gold ball or a grey ball. 50%
People here are illiterate AND bad a basic mathematics.
LOL
>Again, 50%
No. Assuming you got the first box, did you pull the first gold ball or the second one?
There's two scenarios where you land in the first box and will draw a gold ball, since you could've picked up either one, and will thus draw another gold. There's one scenario where you land in the second box, and that second ball will be silver. Three scenarios, two of which you get another gold. It's 2/3.
I'd write out the math but I do it every damn thread and I get really tired of doing it, just take a probability and statistics course sometime. Or just read this:
>How do I conditional probability?
The answer is 1/3rd.
But you don't have to make an assumption. The condition "at least one is" is entirely unambiguous. It's directly equivalent to "one or more" by definition.
One of the two hits is a crit. Full stop.
You're a retard, scenario 2 requires this to be a prediction, but the wording already explains that the event has already occured, thus only scenario 1 can apply
It's amazing when faggots that are utterly wrong act like they're talking to idiots.
Nigger get this, you're not "rolling" anything, the outcome was decided, then you got one piece of information about it, that "one of the hits was a crit".
From this information, the only outcome you can disconsider is having neither of them crit.
Which leaves you 3 possibilities, the first one crits and the second doesnt, first one doesnt and the second does, and finally both of them crit, which is now a 1/3 chance.
If this was NOT post fact information and instead you said "given one crit, what's the chance the next blow will crit" THEN it would be 1/2
Guys, make this equation work!
What is 3 _ 1 = X?
(The underscore requires you to fill in with a sign of multiplication, divide, add, or subtract!)
>everyone argues what X is, rather than which sign to use.
This thread.
If I attack twice, then these scenarios are equally likely:
HH
HC
CH
CC
>"At least one of the hits is a crit" is a statement that reveals that for the two attacks you made, at least one of them is a crit.
So I have made two attacks and HH was not the result. Therefore, these scenarios are all possible and equally as likely:
HC
CH
CC
>This being the case, what is the probability that the other hit is a crit?
2/6 aka 1/3
>which is why it's ambiguous.
There's no ambiguity in the original problem. You're confusing 'changing the question to add an "it's impossible to not crit after two attacks" clause' with 'ambiguity'. It clearly is possible that two 50% crit-chance attacks can both not crit, it just didn't happen based on the clause in the original question.
P(1 crit) = 0.5 not 0.75
>A diagram clearly showing 33% can be used for either side
based ESL retard
High school man tell me more math
That's not an assumption. The question states that is true. We're given a set of two attacks, and one is a crit. The idea of any other attacks beyond these two even being made is the assumption.
>first thread dies in a few posts
>2nd thread goes up for 200+ posts
I don't get it.
order does matter, which is why you got the 2nd one wrong.
yet again, you're not explaining how the 50% answer doesnt work.
I understand how the 33% answer works (and it does work, and is the intended answer)
But you need to actually disprove the answer 50%, not by listing a completely different system.
Use the math presented, and explain its failure.
you fail to understand the thread then.
There's a 33% chance of people falling for this bait twice.
If the first hit crits, there is a 0.5 of the second critting.
If the second hit crits, there is a 0.5 chance of the first critting.
aka 0.5 of a crit.
How can a chart of 3 possible options with equal chances possibly show something other than 1/3?
P(1 crit) = P(crit + no crit) + P(crit + crit) + P(no crit + crit)
The fact that it has a much clearer enunciate means i can call you a wrong retard with no way for you to weasel out of it by saying it was ambiguous.
The fact that you pull a golden ball first tells you two things.
First, it tells you that its impossible for the box to be the one with two silver balls.
What you fail to realize, is that it also tells you that it is more likely that you chose the box with two golden balls.
The answer is 2/3
>assuming we get the first box
It says we get the first box in the fucking image. Read, nigga, read.
Two of those three outcomes are the same. You either get two crits or one. Order does not matter. Get pissy when you’re right, faggot.
Except it doesn’t. The image itself implies such. It just says at least one of them is a crit and the other has a 50% chance of being one as well.
>yet again, you're not explaining how the 50% answer doesnt work.
I didn't say it doesn't work, retard. I said there was no ambiguity. The problem presented by the original question is quite clear.
let me try to explain why the order matters.
Lets say there's 3 boxes.
ignoring the double silver ball box,
the remaining boxes are;
A box with a large gold ball, and a small gold ball.
A box with a large gold ball, and a small silver ball.
You pick up a ball, which is colored gold. It could either be a big gold ball, or a small gold ball.
Now, what are your chances the other ball is a gold ball? (big or small)
If this is real life, you either grab the big gold ball, or the little gold ball.
2 different possibilities.
these are 2 different actions.
in the other box, you grabbed a gold ball, and didnt grab the silver ball.
so there are 3 possible balls you could have grabbed, which changes which other balls were possible to grab.
>P(crit + no crit) and P(no crit + crit) are different results
L O L
retards in this thread have no idea how probability and chance works
probably the same retards who have bought 99 csgo keys and think because it's a 1/100 chance for a rare knife that they are guaranteed one on their next unboxing.
you could have a 50% chance to crit and literally never hit a crit out of hundreds of shots. Unlikely, but that's how it works you fucking niggers.
>It says we get the first box in the fucking image
It says we drew a gold ball you illiterate nigger. Stay in school.
Order matters in this case you shithead.
Common sense is not a substitute for knowing how math works
The probability that the next hit is a crit is 50%, regardless of the previous hits success. Since the previous one hit, you now have a 50% chance to make a double crit.
The probability of both hits being crits in succession is 25%.
The probability of 3 hits in a row is about 12.5%. Each hit is still a 50% chance though.
bait thread, destroyed.
oh, youre just pretending to be retarded. carry on
>Dexterity increases crit chance
>Base rate 1%
>Weapon base 10%
>Each point of agility adds .01%
>Suddenly realize stats are useless and this is just going to be a soulless gear grinding shitfest
B-but muh guaranteed crits...
1/3 people make no sense. Why the fuck would you list miss/miss and then just remove it? That is fucking contrived as hell. In actuality, miss/miss should not even be considered in the first place because we already know it’s an impossibility. We know one of the hits is guaranteed, miss/miss should not even be something that is possible to remove because it has no chance of happening whatsoever. It’s not in the list of outcomes...
I am not reading this thread, you brainlets will just raise my blood pressure. Those of you with an actual attention span just please read my fucking post.
The OP is a trick question and it relies on "unintuitive probability". I've got a shitty MSpaint chart attached displaying the situation described. You have two hits. They're either crits, or they're not. Simple, right? Well, this gives you six scenarios:
>No crit/No crit
>Crit/Crit
>Crit/No crit (or vice versa)
You already know, for a fact, one hit is a crit. Cross out the first scenario. Only two possible scenarios remain. You either will get double crit, or the crit/no crit situation. If you ignore the face value context look at the board, look at what you have to draw from once you take away the crit you're guaranteed to already have:
>Crit
>Crit
>No crit
If one of your hits is guaranteed to crit, then in a two-hit combat scenario, the likelihood of the next hit to also be a crit is 66%. Not 50%, 50% only comes from averaging out this bizarre statistic across the entire game where combat is also influenced by a first hit being a no crit. Understand?
i understand thats what the image shows.
you're forgetting the peopel who say "50%" are arguing you apply the crit after this step.
any by showing the hit/hit in the picture, they would argue its 4 possible outcomes, where you forced a crit.
I do concede the picture doesnt show that step being done, so it leans FAR more towards the 33% answer.
It’s not asking the probability of pulling a specific gold ball, numbnuts. It just wants the probability of the second pull being a second gold ball, which is 50%.
And where the fuck are you getting this size shit from?
My bad. It’s hard to stay on track wrangling so many retards at once. What I meant to say is that we always get a gold ball first.
Go ahead and show where it matters. Use paint and circle it. I’ll wait.
>only one crit/crit
no
This is beyond wrong
you made an error
Fucking idiots. If my post doesn't help you google Bertrands Box or look at this user who programmed a simulation
it's amazing to me how many different explanations people can come up with to try and prove the chance is 50%
I see a new one every time this thread is posted
How can you count crit + no_crit and no_cirt + crit together? Only one or the other can happen, so you need to consider their outcomes individually.
You realize that user got 1/3 and you somehow got 2/3? At least 1/2 is actually an arguable answer, although most of the 1/2 retards in his thread are using the wrong aruments
>If the first hit crits, there is a 0.5 of the second critting.
Continue the statement:
>and if the first hit doesn't crit, there is a 0.5 chance of the second critting and a 0.5 chance of the the second not critting
>which means the chance that there will be two crits is 0.5 / (0.5 + 0.5 + 0.5) = 1/3
>If the second hit crits, there is a 0.5 chance of the first critting.
Continue the statement:
>and if the second hit doesn't crit, there is a 1 chance of the first critting and a 0.5 chance of the the first not critting
>which means the chance that there will be two crits is 0.5 / (0.5 + 0.5 + 0.5) = 1/3
aka a 1/3 chance of two crits.
i never said it asked the probability of pulling which of the 3 gold balls.
im saying, which of the 3 balls u grab, now limits which of the other balls u grab.
if you grabbed the large gold ball in the double gold ball box, its impossible to pull out a large gold ball.
if you pulled out a little gold ball, its impossible to pull out a 2nd small gold ball.
if u pulled the gold ball from the gold/silver box, then its impossible to pull out the other 2 gold balls.
if i said the gold/silver box was medium size, you know u cant pull out anything else.
problem is, if the problem i presented to you, doesnt specify which size the ball is, but that its gold, due to poor communication on my part, that doesnt change the fact u pulled ONE ball, out of 3, and its impossible to pull that same ball out of the 3 again.
Regardless if i worded its description poorly.
in a real life scenario, if u repeated this, u'd get a different result than 50%, and you'd see why the order matters.
Imagine you have three boxes with a BILLION BALLS EACH
One has only golden balls
One has only silver balls
One has nine hundred ninety-nine million nine hundred ninety-nine thousand nine hundred ninety-nine silver balls and ONE golden ball.
You take a ball, its golden. What's the chance the next one from the box is golden?
If you say 50% you're retarded.
THE CHANCE OF THE GOLDEN BALL COMING FROM EACH BOX CONTAINING ONE IS NOT EQUAL
you're a fucking riot dude
>only rolled once
1/3 is the inverse of 2/3, something in his code is backwards but I can't be assed to parse it right now. I've programmed this myself before and got a consistent 2/3
>and if the first hit doesn't crit, there is a 0.5 chance of the second critting and a 0.5 chance of the the second not critting
No, if the first hit doesn't crit than the second hit must crit. Its right in the problem description.
So you have two choices: does the first or second hit crit.
If the first crits, then there is a 0.5 of the second critting.
If the second crits, then there is a 0.5 of the first critting.
Of course, you can also just ignore the order altogether as it doesn't matter. And you get the same result of 0.5
>tfw the thread is still alive
It just works every time it seems.
My dick leans. It is irrefutably 50%. You either crit twice or you don’t. Simple as that. It’s a trick and you fools are falling for it.
Again, where are you getting this size bullshit? It does not mention the size of the balls at all. Fuck off.
It doesn’t matter what the chances of pulling a gold ball the first try are for each box. You pull a gold ball no matter what. It’s guaranteed. Since you know one box had no gold balls, it doesn’t matter. You’ll either pull another gold ball or you won’t. It’s a 50/50, dumbass.
Your program is flawed in that it only counts double crits that occur when the first hit is a crit. Sometimes the first hit is not a crit and the second one is, and you have to count those because they fall under the category of "at least one crit".
I have provided a comment for your edification
>What I meant to say is that we always get a gold ball first
Right, but you're ignoring the fact that you could've picked up 3 different gold balls (2 in box One, 1 in box Two). If you pick EITHER of the 2 gold balls in box One, you get another gold. If you picked from box Two, no second gold. That's what this is about - you know you picked up Box1_Ball1, Box1_Ball2, or Box2_Ball1. Two of those balls are in box One, meaning you will draw another gold 2/3 times.
I am so damn sorry for you
Do you think it really matters which order you add one and crit()?
Depends, what's the crit rate?
>So you have two choices: does the first or second hit crit.
No, you have 3 choices: the first crits only, the second crits only, or both crit
>If the first crits,
This happens in 2 out of 3 scenarios
>then there is a 0.5 of the second critting.
that means that the answer is 2/3*0.5 = 1/3
>If the second crits,
This happens in 2 out of 3 scenarios
>then there is a 0.5 of the first critting.
that means that the answer is 2/3*0.5 = 1/3
Either way, the answer is 1/3
It doesn’t matter which gold ball you pick up. You pick up a gold ball. What is left is either a gold ball or a grey ball. One of two choices. Aka a 50/50. You’re thinking of how many permutations of the scenario there are, not the probability. Once the first ball is pulled, box 1 and box 2 exist in the same hypothetical space.
Don’t worry about it. It just lets me hide it in my pants leg easier.
YOU ARE ****NOT**** GUARANTEED TO PULL A GOLDEN BALL FROM WHICHEVER BOX YOU CHOOSE.
THE PROBLEM ONLY TELLS YOU THAT YOU DID. LEARN THE DIFFRLERENCE.
HOLY FUCKING SHIT
This is why I stopped entertaining these threads. Probability brainlets can't be taught even with shit like
>YOU ARE ****NOT**** GUARANTEED TO PULL A GOLDEN BALL FROM WHICHEVER BOX YOU CHOOSE.
>THE PROBLEM ONLY TELLS YOU THAT YOU DID.
Which means the exact same thing in the context of the scenario. Are you stupid?
The question has three equally correct answers since the question is vague. Attached is an image of three simulations, each one fulfilling the rules of "at least one crit" in their own way. First one flips two coins until at least one of them is a crit, gives 33.33%. The second flips two coins and if neither of them is a crit then it overwrites the result of the first flip making it a crit, this gives 25%. Final one just always sets the first to crit then rolls the second, giving 50%
There are three outcomes of which only one is crit/crit.
>def who_crit():
> return random.randint(0,1)
But that's wrong you fucking retard. You don't assign a crit and then roll the other one. They both get rolled simultaneously, and discount the roll if you don't get at least one.
I fucking love these threads.
>tells you that it did
Are you really coming at me with semantics right now? The actual probability of pulling a gold ball from the boxes initially is inconsequential. It doesn’t matter. All it does is tell us box 3 cannot exist in this situation. Calm down.
25% is a completely incorrect answer brainlet
>The second flips two coins and if neither of them is a crit then it overwrites the result of the first flip making it a crit, this gives 25%
Which is not how it works in reality, so it's wrong.
>Final one just always sets the first to crit then rolls the second, giving 50%
Which is not how it works in reality, so it's wrong.
I think 50% and 33% can both agree that 25% is absolutely retarded
>50.01
??
No the order doesn't matter, what matters is that he's not simulating crits properly. Every round he assumes there's one crit, and rolls for another. In reality, you roll twice and only count where there's at least one.
If you don't like the code, there's a fucking PROOF at the bottom of the original image he was responding to.
Possibilities: nocrit/nocrit, crit/nocrit, nocrit/crit, crit/crit. nocrit/nocrit gets converted to crit/nocrit which is still just a single crit, so out of four posibile outcomes, only one is a double crit, so the chance is 25%
NYGGYR
It's simulation not a theoretically perfect answer.
Do you not understand probability?
If I flip two coins, do you think it always has to come up one heads and one tails because it's 50/50?
>gets converted
u wot
It doesn’t exist. It can’t happen. It doesn’t get converted you mouthbreather.
Take Bayes theorem: P( A | B ) = P( A and B ) / P( B )
Now lets use your numbers
P( A | B ) = 1/3
P( B ) = 1/2
Therefore P ( A and B ) = (1/3)*(1/2) = (1/6)
So there is a 1/6 chance of two crits.
Now lets count up all the equally likely possibilities:
C/C, C/H, H/C, H/H ... wait we are missing two.
A contradiction, therefore the answer cannot be 1/3.
Maybe cuz floating point arithmetic?
No, you always get one. There is no discarding, read the question.
Rules only apply to the final result, doesn't matter how you got there. These are functions that roll a 50% chance twice then makes sure that at least one of them hit, which is impossible without manipulating the result or re-rolling.
>Final one just always sets the first to crit then rolls the second, giving 50%
If the crit chance is 50% like the problem states, then this scenario gives the first a crit chance of 100% which is wrong
>The second flips two coins and if neither of them is a crit then it overwrites the result of the first flip making it a crit, this gives 25%.
If you chance the first hit from a crit into a regular hit some times that would mean the probability of a crit is less than 50% which is wrong
>First one flips two coins until at least one of them is a crit, gives 33.33%.
This is the correct answer
>The actual probability of pulling a gold ball from the boxes initially is inconsequential
No, it isn't. That's the whole point of the problem.
Fpbp
Only dopey pseudointellectuals will disagree
en.m.wikipedia.org
You are all fighting over something that has never had a definitive answer to begin with.
If the first roll is a non-crit, then the second one will be guaranteed to crit. 1/2 chance of NC-C happening. If the first roll is a crit, then it has 1/2 chance to crit a second time and a 1/2 chance of not crit.
NC-C=1/2
C-C=1/4
C-NC=1/4
Thus the CC probability is 1/4. There should be a mistake here.
>mobile site
yikes
2/3
someone posted this last month and I learned that the majority of Yea Forums never even studied high school math. very low IQ board.
and I spent my life time with these people.
You think we care about that? We are here to argue not to think.
>nearly 300 replies for a thread that shouldn't have gone past 30 replies.
Based and fbpb
dont save thumnails nigga
>last month
you sure?
If you flip a coin irl it's not gonna alternate between heads and tails every flip. Only if you flip it thousands of times is it going to give you an overall 50/50 ration of heads to tails. The number gets way higher when you start flipping two coins like these functions do.
The problem only states that in this video game, if you attack twice, at least once will you get a crit. If you are gonna assume that this doesn't affect the crit chance, then you introduce a fourth answer: 0%. Only by alternating between crit and nocrit will you both "crit at least once every two hits" and maintain an overall 50% crit rate, so you will never hit two crits in a row.
probably, but not sure.
Only because it was originally worded ambiguously.
The answer to the box problem though is unequivocally 2/3
> P( A | B ) = 1/3
>P( B ) = 1/2
wtf? P(B) is the chance that the first hit is a crit, which is 2/3
P (A | B) is the chance the the second hit is a crit given the first hit is a crit, which is 1/2
So P ( A and B ) = (2/3)*(1/2) = 1/3 which is the correct answer
You absolute dumbass. Do you understand what you're reading? It's similar to Bertrand's Box and the answer is still 2/3, 66%. The people who continue writing dissertations about it being 1/2 are assblasted brainlets without a leg to stand on, just look at that fucking chart.
ONE IN THREE
ITS THAT SAME FUCKIN MAGIC QUARTER SHIT
I AINT GETTIN FOOLED TWICE
>wtf? P(B) is the chance that the first hit is a crit, which is 2/3
No P(B) is the chance of the second hit being a crit. Which is independently 0.5, unless you want to argue that crits are dependent.
>P (A | B) is the chance the the second hit is a crit given the first hit is a crit, which is 1/2
No its 1/3, as this is what the entire question is about: the probability of a crit given a grit.
>66%
no one is this retarded to believe this.
Close but wrong.
There are five possibilities:
>gCrit = guaranteed crit
>sCrit = successful “roll” for a crit
>fCrit = failed “roll” for a crit
gCrit, sCrit
gCrit, fCrit
sCrit, gCrit
fCrit, gCrit
The fifth is if both are failed crits, which is an impossibility, leaving 4. After which there are two with at least one crit and two with 2 crits. Your chances are 50/50 on whether there is 1 or 2 crits in this situation. It should also be noted that the order of the guaranteed crit and the non-guaranteed crit is inconsequential so there really only two possibilities to worry about.
Yes, it is. The first pull is made for us. We only have to worry about the second in this hypothetical.
0% because a hit isn't a crit
Get help. There are two examples in this thread spelling it out for you and a chart on that very page. Again, look up Bertrand's Box and fucking READ. Don't reply to me, don't dig yourself further in this hole you down syndrome faggot, read. And then read it again. It's 66%
NOTHING IS GUARANTEED
nigger just en.m.wikipedia.org
There are three major camps for this: 1/2, 1/3 and 1/4, and the logic for each can be broken down as such.
1/2: neither hit has bearing on the other, so if one is guaranteed, the other is a 50/50 shot, so 50% chance.
1/3: There are three options: Crit hit, hit crit, and crit crit. Hit hit is impossible, therefore it's a 1/3 chance.
1/2: The two rolls are sequential, and both are a 50% chance, with one being changed to a crit if the other fails. Thus, since you're still effectively rolling two 50% chances for it, it's 25% total still.
The fun thing is all of these are correct depending on one key fact: WHEN the crit is applied.
If which hit is a guaranteed crit is chosen beforehand, then yes, it's a 50%.
If the guaranteed crit is applied only in the case of the first hit being a normal hit, it's 25%
If you just reroll until you get something that's not a hit hit, it's 1/3.
The most important thing his is how the guaranteed crit works, but it's cleverly left out.
>poker is just RNG bro
There's way more to poker than just random dice rolls. The betting aspect is what makes it entirely possible to do consistently well since the point is to win hands, not to get the best luck. You can't control RNG in vidya not can you play around it other than taking steps to minimize random chance, which is impossible to do when you're not the one rolling the dice.
Not him but you're mixing up the problems, the answer to the box one is 2/3, the boy thing is 1/3
The game would only give a guaranteed crit if you didn't roll a crit the first time in order to prevent both hits from being non crits, why would it give one on the first roll? That I don't understand.
>it's Yea Forums don't know the difference between probability and odds episode.
so much for the elite secret club
actually, looking over that now, a better way to describe the 1/3 probability is that after rolling, you pick one of the two rolls and apply a crit to it. if it's already a crit, nothing changes. if it isn't it becomes a crit. You choose that at random between the two.
That makes a neat before-during-after application of the guaranteed crit that fulfills all the common answers of 1/2 - 1/4 - 1/3 respectively.
unless i'm an idiot, which is possible.
>The problem only states that in this video game, if you attack twice, at least once will you get a crit.
No it doesn't, why do so many mongoloids itt think that? If you attack twice again, there's a 25% chance that you don't get any crits because each individual attack has a 50% chance to crit.
There's nothing that indicates a "guaranteed crit" is a constant factor. All the question says is: "You did two hits, and one or two of those were crits"
>50/50 = educated virgin
>1/3rd = chad truck driver
It is in the question. We don't need to consider any other scenario (like hit/hit or missing) being possible because they aren't in the solution space.
My mistake, I assumed it was the same as the girl question (which is the same as Bertrand's Box)
The question doesn't state that you get a crit and then roll for only the other crit, therefore you must roll both crits. Since you are rolling for both, you will occasionally get a result that doesn't match the conditions (at least one crit). You must discard those results that don't match. This is how you get results that match the actual probabilities represented by the question.
>50/50
nigga that's 1 you baka
>The question doesn't state that you get a crit and then roll for only the other crit, therefore you must roll both crits.
Why would you roll something with a guaranteed outcome? The problem doesn't say the order of the crits matters, so we can assume the guaranteed one is first and just roll the second.
There are four outcomes: hit-hit, hit-crit, crit-hit, and crit-crit. So it's 1/4 (25%) to get perfect crits.
It doesn’t matter if it is or not. All that matters is the knowledge of the first pull and what it dictates can be hypothetically possible beyond that point. If you pull a gold ball, it can’t be box 3. That means it is no longer a factor. After that, if you pull a gold ball, that means it was box 1. If you pull a grey ball it was box 2. It’s a 50/50 chance assuming that the first pull was a gold ball, which the image states is the case.
Because it would have to give one on the first hit if the second one has a possibility of missing. Since order doesn’t matter, it’s possible for either hit to not crit but not both. You’re not a programmer trying to make sure a player is guaranteed at least one crit, you’re just calculating the probability in this specific situation.
But skill could come into play with how you react to rng. Like xcom and dealing with a shitty rng roll.
That's right, because if you were educated enough, you would be able manipulate the RNG in such a ways as to always get both crits every time, regardless of the actual probability assigned to that RNG.
>"At least one"
>Doesnt indicate "Guaranteed crit".
"At least one" can either be;
"And 1 or 2 of them crit"
or
"1 of the 2 has to meet the requirement of being a crit"
"At least one" is a requirement.
How you fill that requirement is what the creator of the problem wants you to solve. Not if the end result is 50% or 33%.
Once you know how to meet that requirement, then you know how to solve the problem.
The original question was with a father, and his kids. You cant change the gender of a kid, once its determined.
But it is possible to change the outcome of a crit multiple times in a games programming.
The original question didnt take this into account, and when its possible to recreate the problem, and meet the requirement for "at least one" but with a different method, both answers are true.
(Even the original creator agrees, but his agreement is actually wrong, since it was based on children/genders, and the disagreement was that the father could be any father, despite being named as a single individual.)
>and players can't control luck
You can control which side the roll is in favor on, and when it happens enough times, you're about as likely to have it average to missed rolls as you are to win the lottery. It's not complicated.
I think its the wording that's screwing you up. Think of it as "If you hit twice, how likely is it to get two crits assuming 50% to crit per hit".
hit crit
crit hit
crit crit 1/3
You have to look at it as a whole because we can't assume to which hit crits.
Correct answer is 50%. Anyone saying anything different is a tryhard that is trying to sound smart while actually being dumb.
you forgot to meet the requirements of "at least one" of the hits crits.
hit-hit doesnt meet this requirement by default.
If we're including combinations that arent included, you forgot the times you missed as well.
miss miss
miss hit
hit miss
miss crit
crit miss
You're wrong, because you didn't consider that my answer was the right one. If you had, you would have seen why your answer was wrong.
>If you hit twice, how likely is it to get two crits assuming 50% to crit per hit
That’s now what it’s asking at all.
How so?
if anything, this shows the importance of how you go about programming something in a game. If you turn this into "one of the crits is guaranteed", you can make it happen in a number of ways that all change the probability of both being crits.
a sword with this as the effect can be radically different in effect depending on how you program the game to determine this.
>Why would you roll something with a guaranteed outcome?
Because there ISN'T a guaranteed outcome.
>The problem doesn't say the order of the crits matters
Doing probabilities correctly means not adding any extra information that isn't already present in the question. It states that both rolls have a probability of 50%, which means either can fail. We must therefore roll each one.
>we can assume the guaranteed one is first and just roll the second.
THERE IS NO "GUARANTEED ONE". They can both fail, just never at the same time.
When did you realize that OP's problem is fundamentally impossible for us puny three-dimensional humans to solve?
Not to mention blocks, dodges, parries, counters, critical failures, and spontaneous changes of heart
Im well aware of the Boy and Girl paradox.
You said it doesnt indicate guaranteed crit.
which it technically does, while originally having no intention of doing so.
congrats. this is the most brainlet response of the entire thread
How are you guys so dumb? It's 25%
The first hit is 50% hit, 50% crit. If it's a hit we know the next is a crit and we're done. HC has a 50% chance of happening. If the first one is a crit (50%) then the next could still be a hit or a crit and 50% each, so .5*.5. The probability matrix is:
HC: 50%
CH: 25%
CC: 25%
Just because you have 3 outcomes doesn't mean they're all equally likely.
There is a 4th scenario, no crit twice, so the answer is 1/4
One crit is guaranteed at all times. See There is a guaranteed crit, just not a guaranteed order.
>THERE IS NO "GUARANTEED ONE". They can both fail, just never at the same time.
This is a contradiction.
No there isn't, the prompt rules it out.
Just read the wikipedia article.
That troll faggot Russel is the one who created the problem and he says it's 2/3.
So go disagree with him instead.
correct, which is why the OP phrased the question in RPG terms.
The original question was
>Mr smith has 2 kids. At least one of them is a boy, what are the chances both kids are a boy?
In programming you can set it up to remove a predetermined double hit possibility, after it rolled for this.
Or you program it to change one of the 2 results at random, which also has to be after the crit has been determined.
for kids, u cant change it once a gender is determined, so u have to assume a girl girl option never existed.
they aren't independent variables then they're linked.
No, he just means that it can be either "first crits" or "second crits." You don't know which one is the non-critting one, which is important because it affects the number of possible outcomes.
a block, parry, or counter doesnt change a crit that im aware of. (Though maybe some games do make this effect a crit, to which you'd be correct if so.)
now, does a crit failur bonus count? i think that can go either way, just like the original paradox.
changes of heart, i refuse to believe any true warrior would ever do this!
>so you need to consider their outcomes individually.
No you don't, because they're the same outcome. One hit and one crit.
This is exactly what the gamblers fallacy is, if the chance is 50% it does not mean a guaranteed hit, you could go for 100s of hits with no crit if your luck is shit. The only correct answer is 2 times 50%, 50/2=25%
Yes you do, because there's a chance you get 0-1 and there's a chance you get 1-0. You're more likely to get one or the other than 1-1.
>you could go for 100s of hits with no crit if your luck is shit
why would you ever let yourself be such a lucklet?
>OP trolled us all (or is just retarded) by turning a real world paradox into a vidya one in a lazy way and fucking it up do to screwing up the unwritten assumptions
THE ACTUAL CORRECT RESPONSES. The original paradox works because of real world physical constraints, but on a computer there are multiple ways to implement the behavior described here which result in different probabilities. There is no "right" answer because it's arbitrary and it's a game.
If anything the real interesting part of this is pointing out precisely the kind of wiggle room programmers have that can trip things up.
If you count order this way, then you need to add an additional 1-1, because there are now four outcomes.
Crit1 -> Hit, Crit1 -> Crit2
and
Hit -> Crit2, Crit1 -> Crit 2
You can't treat the two double crit events as the same if you reach them sequentially.
OP didn't mention anything about stacking luck so we got to assume its 0
>for kids, u cant change it once a gender is determined
omg that is so insensitive the entire paradox is discriminatory against trans, intersex, and asexuals!
This makes sense on the idea that each hit is calculated.
If it's 3 planned situations in rotation then it's definately 1/3
Otherwise HC has a 50% chance of happening while CH and CC have 25% because of the first roll.
Given it's videogames and combo hits, it's probably the latter.
This is just wrong thinking and over labeling in general. The only possibilities are a hit and crit
hit crit
crit hit
crit crit
This model of probability above does show a guaranteed critical in all scenarios but shows that only 1/3 can be double crits. We just don't assume which is the crit. But more importantly how is my rewording wrong?
While this is true, this isnt the only way to achieve "At least one is a crit"
You can just remove all hit/hit possibilities, and only look at 1/0, 1/1, and 0/1
which is 3 options.
Or you can do something even more complex.
When you program a simulation using real world logic, you get 1/3. Flip two coins, if both are misses, abandon the trial as indicated by the prompt. If it's two hits, add a tally to pass; if one misses, add a tally to fail. You'll get an average of 1/3.
Stop being retarded.
Miss/hit is not the same thing as hit/miss, but hit/hit is the same thing as hit/hit.
Nobody can be this stupid
Nothing is ruled out, it must add to 100%.
technically the original paradox actually addresses those complains by saying "this is only under the conditions there are only 2 genders, hermaphrodites dont exists in this example. The chance of the genders is 50-50, ignoring any other possible changes to percentages." and some more stuff.
you can tho because it's a computer program. a lazy programmer could easily go (0 = hit, 1 = crit)
hit1 = rng(01).
if (rng(01) == 0)
then hit2 = 1
else
hit2 = rng(01)
that's semi-deterministic, hit 1 is 50/50, but hit2 is dependent on the outcome of hit1.
This scenario you're creating doesnt exist.
The first one crits or doesnt.
The second blow crits or doesnt.
There is only one combination where both of them crit.
But there are two combinations where only one does.
There is only one combination where both crit.
>There is a guaranteed crit,
No there is not. Either hit is capable of failing to crit. Neither is guaranteed.
>they aren't independent variables then they're linked.
This is true. By the conditions of the question these ARE in fact dependent variables.
>"At least one" can either be;
>"And 1 or 2 of them crit"
>or
>"1 of the 2 has to meet the requirement of being a crit"
"At least one" can ONLY be "one or more." One or more of the hits was a crit.
>Nothing is ruled out
user the prompt says "at least one of the hits is a crit"
double miss is ruled out
both of your statements hold true, just so you know.
"At least one is a crit" applies to removing hit-hit possibilities, and reducing it now to only 3 possibilities.
AND a system that changes a hit into a crit.
Both equally meet the requirements the question wants you to figure out.
Once you figure out how to meet this requirement, you can then solve the equation. (The real test is meeting this requirement)
>tfw always wanna post in these threads
>not intelligent enough to contribute anything worthwhile however
im so fucking useless bros
>When you program a simulation using real world logic
But programmers constantly do not do this user.
>Stop being retarded.
No, you stop acting with no experience. Vidya in particular are PILES OF HACKS, hacks upon hacks done for speed heavily by underpaid overworked entry levels operating under death marches. All sorts of exciting bugs and surprising behavior comes from people making things work in weird ways and then LETS SHIP IT.
>Neither is guaranteed.
At least one is a crit. This is information provided to us by the question itself.
I'm sorry but its completely unreasonable to bring hit/miss into the equation unless we get to know those numbers. The question only mentions chance to crit, therefore there is still a 50% crit chance.
If you wanna bring hit/miss into it than why not also dodge, parry, block, glancing, or whatever other fucking thing you can think of?
%1006
>But programmers constantly do not do this user.
But this specific example is using real world logic.
Holy shit, kill yourself.
>I'm sorry but its completely unreasonable to bring hit/miss into the equation unless we get to know those numbers.
Nobody's doing that
>Nothing is ruled out, it must add to 100%.
50 + 25 + 25 = 100
so does 1/3 + 1/3 + 1/3
so does 50 + 50
There are only 3 possibilities, but they don't have to be the same probability in a computer program. Adding to 100 does not help you user.
(You)
Actually, pursuant to the idea of these being independent variables, the 1/3 paradigm makes the true probability of any individual hit being a crit 2/3. The only way to correct this and make the actual rate of critting on either hit 1/2 and still always have at least one crit is to take this reasoning to its illogical extreme. I present to you:
Hit Crit
Crit Hit
This is the only probability table that has the crit rate of either hit being truly 50% AND always has at least one crit. I submit to you, that there is a 0% chance that both hits are crits.
>real world crits
People who still fall for this are blatant newfags.
"one or more" as a descriptive requirement, applies to
>"1 of the 2 has to meet the requirement of being a crit"
Just adding a bit more text to clarify how, doesnt stop it from matching up.
>You program a game.
>You tell it to roll 2 landed hits.
Now
>You either tell it to remove a double hit, and go with a new roll that has 1 or 2 hits.
or
>you program it to change one of the 2 landed hits into a crit. (even if its already a crit)
both work, both meet this requirement of "at least one crits"
but gets different answers.
I suggest you look up "The boy and girl paradox" to understand this better.
>real world crits
Real world 50/50 chance you stupid retard
This has to be bait
every game out right now is undeniably awful, so we need something else to argue about
Those scenarios you gave are not how real world probability works, so of course they don't give the right answer. What a stupid post.
an interesting take on this, though the person you're quoting wont understand how thats possible.
>But this specific example is using real world logic.
Real world logic WHICH THIS VERY THREAD DEMONSTRATES HUGE NUMBERS OF PEOPLE DO NOT UNDERSTAND. Do you think this is somehow unique to this place? That every random self-taught or community college/podunk tech school online """degree""" person knows it?
>Holy shit, kill yourself.
Holy shit get out of your ivory tower and spend some time in the piles of garbage upon which most modern computing is built.
Seriously, it wouldn't be a thought experiment/"paradox" if it was instantly obvious that's the whole fucking point. Fucking idiot.
>50/50 chance
are you confusing probability with odds again?
I mean you can stab someone in the leg.
OR you can stab someone in the leg, piecing their femoral artery and they bleed out in 10 seconds.
>At least one is a crit.
That is not the same thing as there being a "guaranteed crit". "Guaranteed crit" here being used to mean "a hit that does not need to be rolled because it has already been determined to be a crit".
Considering I quoted myself on that, I think he'll get it just fine.
0/10, be more subtle next time
a 50% chance is a 50% chance
You're adding a bunch of extra shit to complicate an incredibly clear and simple question.
There are two attacks. One will crit, one may or may not crit. Everything else is irrelevant.
>One will crit, one may or may not crit.
No, that's not how it works. If you roll it all out with real world probability attempts and just discard the test if it's a miss/miss as indicated by the prompt, you get a 1/3 chance.
You have an equal chance to get miss/hit, hit/miss, or hit/hit. It's 1/3.
Even better:
>uh manager said this sword has to do at least one crit 50% of the time
>uh uh
>x = random(0,1)
>if x = 0 then hit_matrix = {hit, crit}
>if x = 1 then hit_matrix = {crit, crit}
>holy shit I'm a genius ok commit
so clearly it's 50/50!
>That is not the same thing as there being a "guaranteed crit".
Yes, it is. There WILL be a crit. The question states that very plainly. At least one of the hits is a crit.
>No, that's not how it works
That's exactly how it works. You're over complicating it on purpose.
I never said it was a real world possibility.
Nor does the original question even hint at the idea of being a real world scenario.
Its a video game.
The only place a video game exists, is in a programmed code.
The "Real world" probability can be programmed to the programmers liking, as long as it meets the requirements of the problem.
In a real world situation, such as the boy and girl, only 33% is the correct answer.
But since its a game, 50% is possible to reach.
I agree that Occams Razor should apply here, for the sake of keeping things simple, as the alternative methods dont "need" to be done.
The problem is, they CAN be done.
If they CAN be done, then the correct answer to the question is both 33% and 50%, depending on the method chosen. Regardless if its overly complicated. (The original question doesnt specify to use the least complicated method or not. And what is complicated to one person, can be simplistic to another, which is another issue in this debate.)
see
All the rolls still happen, you just discard the test if it comes up fail/fail. There is no magic guaranteed crit, that's not how the results are being determined in a real-world way. Three equally-chanced outcomes could happen, one of them passes the test. It's 1/3.
>One will crit, one may or may not crit.
That is not true.
"One may or may not crit, the other may or may not crit. We will never have a result where neither critted"
Fuck, these bait threads get me every time. And you just know that OP doesn't fucking care which it is.
>1 month after release
>HOLY SHIT SWORD 1.0 IS SO FUCKING OP NERF
>patch replaces sword 1.0 with sword 2.0 which now crits 10% of the time
>publisher makes sword 1.0 paid DLC
>You're over complicating it on purpose.
No, you're oversimplifying it at the cost of reality. Your answer isn't true because it's forcing a value instead of actually rolling the tests for real.
It all depends on wether you flip 2 coins and discard all double misses or if you set up one as a hit and only flip the other. Both are possible, real world trials and satisfy the wording of the question.
>as indicated by the prompt
There's your problem right there. It NEVER indicates that.
the OPs question never says to use "real world" ways.
It just asks you to make sure its possible to do.
And what they suggest is possible to do.
(and technically the creator of the question argues its possible for either to work for real life, due to some other BS i wont bother going into.)
You rule out the 25% from the double no-crit. You have 75% left. A 25% piece is double crit. So 1/4 is the answer.
>There are entire games based on understanding rng, like poker.
The rng in poker makes possible things like bluffing. Crits just involve a 'flat' optimization problem that reduces the chances of a meaningful strategy working out, and hence are generally bad, because there's nothing you can do about them apart from just see them occur and respond to them when they happen.
>It all depends on wether you flip 2 coins and discard all double misses or if you set up one as a hit and only flip the other.
The former is a realistic natural test, the latter is forcing a result instead of actually doing out the tests. Of course the one where you fake it is wrong.
I'm not forcing anything. The question has supplied that information. At least one hit is a crit. There is a crit. You can bitch about it all you want but that crit remains there, smiling and waving and fucking your wife.
>All the rolls still happen, you just discard the test if it comes up fail/fail
No you don't that sounds really wasteful. If you want to simulate the real world you just do a random number gen with 012 that way you get a 1/3 in a single operation. You don't discard anything. If someone does it with some branching they'll probably end up with 1/4.
What part of "at least one of the hits is a crit" do you not understand?
Some games have true to form percentages. POE has a method for calculating dodge rolls where you dodge perfectly with your percentage. Example, you have 25% dodge chance. It is guaranteed that your first time being hit will be dodged. The next three will pass your dodge roll. Then the 5th hit will be dodged and so on and so on.
To all those guarantee hit 1/2 shitters
hit crit \ crit hit \ crit crit
This way of thinking does guarantee one of them is a crit and has all the possible scenarios. Stop thinking one hit after another you dumb fucks and look at the two hits together as a whole.
Do it your fucking self then. Flip a coin twice. If you get tails tails, do it again. If you get one heads and one tails, put a tally mark on one side. If you get heads/heads, put a tally mark on the other side. Keep repeating this and you'll see the ratio get closer and closer to 1/3.
>Your answer isn't true because it's forcing a value instead of actually rolling the tests for real.
So it's a typical video game programmer implementation is what you're saying? Is rolling for real going to get us more sales?
You can't give the outcome and then say something is a 50% chance, that doesn't make sense unless you are saying there's some sort of system in place to manipulate the odds to make one a definite outcome instead of 50% chance.
If you have that sort of system yes it's a 1/3 if not it's 1/4.
You have a higher percent chance of choosing the car if you change your answer to the other door.
Because which one is guaranteed and which one is not matters. You can’t say both 2 crit scenarios are the same when the two types of crits are different in terms of probability. The outcome is the same either way sure, but then we’d have to look at the other possibilities the same way, too. Either way it results in a 50/50.
Yes, there is. A guaranteed crit is required in each of the non-impossible possibilities.
I understand it perfectly. You seem to think that means you have to run trials until you get those results when the prompt gives you that result for the outset.
see
and see for yourself that you're wrong.