What are some video games where you have to keep making calculations in order to win/get ahead?

Lol

I think it's a 75% chance they are both crits. I believe this question is equivalent to:
>you have a 50% probability to crit. If you hit an enemy twice, what is the probability at least one of them is a hit?
Correct me if I'm wrong, I'm not a statistician.

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Posting the one true answer. If you got anything else, you're a big, fat stupid-head.
OP is a faggot.

If you collected a real sample of data, you'd have hit hit, crit hit, hit crit, and crit crit. Crit crit would occur 25% of the time, and any crit would occur 75% of the time. Of the any crit set, only 1/3 would be double crit.

It's monty hall but with different words

Meh, at least it's not another anime gamer thread. Have a bump, OP

i have no idea, goodbye

The text implies one of the two hits is GUARANTEED to be a Crit, that only leaves another single hit uncertain. If the Crit rate is 50%, then the answer is 50%. You are hoping that the one hit that's not guaranteed to crit procs in this situation.

>wanting bait threads over titties
Queerbag

It doesn't imply shit. You made 2 hits and at least one was crit. It doesn't say "one of your hits was guaranteed to be crit"

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>At least one of the hits is a crit.
This line right here. This feels like one of those High School math problems where they give you unnecessary information to confuse you and make you overcomplicate it.
This line doesn't matter at all.
The answer is 25%, both hits has 1/2 chance to crit, and the odds of both critting at the same time would be 1/4.

is this an indian onahole

>It doesn't say "one of your hits was guaranteed to be crit"
It literally does. Lay off the hentai, it seems to be rotting your brain

>one of your hits is a crit
>THAT DOESN'T GUARANTEE ITS A CRIT
ok retard

>le feelsguy image makes my argument for me
You post bait and want bait threads to be posted, and then you complain about off-topic porn threads that are objectively less cancerous than your redditor garbage. So, yeah, kill yourself and stuff.

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>it's my only other controller, bro

I gave you two coins, at least one of them is a quarter. I'm not implying that every time I give you 2 coins that there will always be a quarter, just that's what it happened to be that time.

>You made 2 hits and at least one was cri
Read that again but slowly. That's exactly what's being implied, out of the hits one was a crit, what's the probability of the other one being a crit too? 50%. Fucking retarded niggers in this board I swear to fucking god I hate all of you.

Because when one is guaranteed to be a critical, It doesn't matter what order it's swung in, just the result.

All that matters is the possibility of which of the hits isn't automatically a crit is.

If the first swing is a hit, it's a 0% chance but if it's a crit, then it's a 50% chance

Oblivion

1/3, since I have an IQ above room temperature.

>guaranteed critfag
not this shit again

also 1/3 is bigger then 1/4
math aint hard nerds

Failed logic class huh, look at your chart again, you have one extra

The problem isn't talking about more than one round of two hits illiterate fuckhead. It's just one time, one scenario, one round where out of your two hits one already procced the crit.

The one true answer is that everyone in this thread, except me, is retarded and nothing they say is of any worth.

must be living in the arctic then

How can someone have negative IQ I don't get it

This is not the Monty Haul problem.

There are 2 possible scenarios where one hit crit. You can't discount them as the same event. This is a classic gold coin riddle. If you played this out with real data, and discounted the outcome with no crits, you'd have double crits 33% of the time.

absolute kumbrain

HC
CH
CC

1/3

If you don't get one-third, you should not be in a STEM major

You forgot /hm/

it's 50%...the hit that automatically crits doesn't affect the other hit

the answer is clearly 100% because you're not accounting for the depth of the second blow, which has an infinite amount of chances to crit along the way through its target.

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>If you played this out with real data,
>forgoes said real data
one is already criting, meaning you only need to account the other.

It depends: are you only looking at cases where there is at least one crit (and therefore discarding results without one) or are checking if the first hit was a crit and forcing the second one to be a crit if the first isn't? Both are reasonable enough interpretations and without know the intended way to read the question it has 2 different possible answers

We have two separate events that roll on their own. We know that one event out if these two has confirmed outcome so in reality it there are two scenarios

First is H/C and C/H( they share the same outcome and are no different in any way from each other. Order does not matter at all.

Second is H/H.

So it's 50/50 easy.

this

>one is already criting, meaning you only need to account the other.
Yes, but that scenario unfolds twice, whereas double crit only once.

I'd make a quick computer program to demonstrate it but you'd just say I made it up

>order does not matter
If I flip two coins and one is heads, what is the probability that the other is heads? While the question says nothing about order, the answer is determined by it: 1/3, because HT and TH are independent events.

>enemy has 3 hp
>hit does 1 damage
>crit does two
>hit has 50% chance to be a crit
>enemy killed in two hits
>what is probability both hits were a crit
I always thought of it like that, so while having two hits happen without a crit is possible, its just not the scenario we are looking at

I think it's the difference between people calculating how often one would crit if repeated(33%), versus a one time scenario where one of the two hits must be a crit, thus the only probability is in one of the two attacks.

doesn't matter, one is guaranteed to flip to heads, anything else is you complicating the problem.

one of the hits is a guaranteed crit, so shouldn't it be 50%? order of hits doesn't matter because they both need to be crits and one is definitely a crit

100% because I'm a chad warrior and you're a pussy ass bitch hunter hiding in the back. Fuck you faggot.

If youre guaranteed one and the second has a 50% thats 25% for both. Come on kids.

Based and mathpilled.
The rest should be permabanned.

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>If I flip two coins and one is heads, what is the probability that the other is heads?
50% because you already flipped once therefore next event is separate from the previous one. If you think otherwise it's because you are a victim of gambler fallacy.

physics grad at a top uni here

it's very obviously 1/3. are you guys just baiting?

Its a 50% chance.
If at least one of the hits is a crit, then that makes the chance of critting with both hits only based off of the other hit, which would be the 50% chance, making the chance of both hits being crits 50%.

>order of hits doesn't matter
Wrong. Separate events are statistically significant. In a sequence of two hits, there are four possibilities:
>1st no crit, 2nd no crit
>1st crit, 2nd no crit
>1st no crit, 2nd crit
>1st crit, 2nd crit
OP's problem stipulates that at least one hit is a crit, so the possibilities narrow:
>1st crit, 2nd no crit
>1st no crit, 2nd crit
>1st crit, 2nd crit
Of the three possibilities, only one can occur in which both hits are critical. Thus the probability is 33%.

All of you are fucking retarded.

>atleast one is a crit
>atleast one is heads
>one of the two coins is always heads

Can we can rephrase this as one coin has heads on BOTH sides, one has heads and tails?

Genuinely wondering if that would be a sufficient parrallel.

1/3

>one is guaranteed to flip to heads
It absolulely is not, it's just an observation of an outcome that already happened. That's the whole problem with this scenario, people are assuming that the critical hit is guaranteed and thus irrelevant, and thus discarding the entire hit as immaterial.

If you assume these 2 hits could have been non-critical, and this set just happened to have at least one set, then it's in a set of outcomes whereby the number of double crits is mathematically 25%.

Probability is all about information and what you know. You're making more assumptions than I am which is why I would err to 33% over 50%. You read this question that no matter what, one hit is always critical, but that is counter to the question itself: 50% crit rate. It does not state that every time you hit twice you will always critical at least once. That is just how you're reading it, even if for this particular question that's what you can take it to mean if you're an idiot.

Why is our perception of probabilities so strange? Why didn't nature hardwired Ckn into our bains?

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So what's the answer, genius?

>50% because you already flipped once therefore next event is separate from the previous one.
No, this is 2 coin flips that already happened. An observation is made: one or more was heads. You can't stop after the first flip and ask the question before the second. That's not what's happening here and that's where your problem lies.

this. holy fuck it doesn't even require any math.

>3 possible scenarios
>crit, hit ; hit, crit ; crit, crit
>1/3 of the scenarios are desired

Figure it out yourself, fuck nugget. I'll let you know.

>not a single person has posted the right answer yet
lol

So ya xcom games are great for this. Also you guys are fucking retarded.

>gambler fallacy.
Not him but thanks, just looked that up. Maybe I'm retarded but I think this would apply to the Monty Hall door problem too. Everyone and their mother on the internet still claims you need to switch doors. I just can't shake the feeling it's still 50%.

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3TXDI-RKCTC-JGD09
3T9ZZ-I23WG-8RGGZ
3PK9R-0I56T-0FC0A

The important thing to note here is that it says "one of the hits is a crit" instead of "the first hit is a crit."
I.E. 1/3 instead of 1/2.

You are just as wrong as the people who are assuming the first hit is a critical hit. The entire problem with this, as with most of the things of this nature, is that the OP is intentially vague so that both interpretations could be right.

bad luck and xcom has taught me that if it's not 100% then it's 0%
and then there are some games where if it's not more than 100% then it's 0%

The question is asking about a specific set of 2 hits, in which the first is a critical. It's asking for the chance that both hits from this set are critical. This set that has already been defined as .

The question is also vague enough that your interpretation is correct. Welcome to internet bullshit.

t. too scared to even bother answering the question

And Darkest Dungeon has taught me that if it works, the Devs will break it.

en.wikipedia.org/wiki/Boy_or_Girl_paradox

I don't think it's vague at all. I think it describes an observation of an occurrence that already happened. How else could it state that at least one is crit? If the observation happens after the first hit and before the second, then that statement is impossible: it would have to be the first hit that was crit. For it to be "one of" means that it already happened.

It doesn't have the ambiguity necessary to be a paradox

except no because the order of such thing isn't stipulated nor required and its a your factor of it being a sequential roll intead of an isolated roll like the problem states.
its 50% because one has already been decided as crit and its entirely dependent of the other roll

Larger than zero, aka the odds humanity takes on a regular basis and succeeds. What else do you need to know?

I cast chaosbolt and crit both attacks. 100% chance.

OH NO NO NO 50% BROS

math.stackexchange.com/questions/991060/flip-two-coins-if-at-least-one-is-heads-what-is-the-probability-of-both-being

this question is based on a paradox.
the questions creator intended you to answer with 33% but phrased it poorly, and allowed for 50%.
this version just changes it into game terms, to make it even more vague.

you do need to switch doors still though, its been tested many times, the difference between the monty hall and the gamblers fallacy is that in the monty hall problem the door that was removed could have never had the prize behind it because the host knew where the prize was because of that the core odds that you picked wrong in the first place is the real odds of the situation still; however the gamblers fallacy isn't actually relevant for the problem presented in this thread because the question didn't state that "the first one is a crit" it stated that "one of them is a crit" which results in the H H result being removed but still leaving H C, C H, and C C, so the answer for this thread is 33% still

this

now for an actual problem which has a proper solution

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the fuck did red hook do?

>this question is based on a paradox.
if you're retarded.

You forgot HH

t. 50% fag

>if you're not retarded enough to argue with genuine mathematicians
fixed
en.wikipedia.org/wiki/Boy_or_Girl_paradox#Analysis_of_the_ambiguity

Let's do some 4channel law instead.
OP made a thread to ask a question, but posted a picture more interresting than his thread. People commented but the janitor deleted the thread and warned OP for off topic thread. Who was in the right, who was in the wrong?

Hasn't this been discussed enough? I guess OP got tired of making portal threads.

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Janny is always in the wrong.

question said at least one was a crit, so we get to ignore that outcome

>from all sets of 2 hits in which at least one is critical, a set of 2 is selected. The probability is 1/3
>from all sets of 2 hits, a critical hit is selected. The probability that the other is critical is 1/2
It depends entirely on the selection procedure, there is no correct answer given the current wording

Everyone go home, I'm better than you

You go into an item shop to buy some potions. The potion seller has 3 health potions and 3 mana potions in stock, and all 6 of them are resting in boxes you can't see into, 2 potions for each box. The potion seller will give you 1 potion for free if you buy a potion and reach in to one of the boxes at random and grab 1 potion. You buy one and reach in to pull out a health potion.

What is the probability that you reach in and pull out a second health potion?

1/5? honestly curious about this one.

I'm only interested in cases where at least one attack will crit (for my own personal reasons, I decide to ignore times where I don't crit at all).

On those cases where at least one out of my double attack crits, what's the chance that I actually get two crits?

This isn't ambigiuous OR paradoxical. There's only 3 scenarios, each of which is as likely as the other two.

The answer is a third of time, the probability is 1/3.

>1/3 as the better answer
50%fags on eternal suicide watch

Janny for doing it for free.

now delete the erroneous 2nd procedure, and you've got the right answer

so close buddy

75% statistics are fucked and I hate that I spend part of my life studying them

Also I'll type reddit, fuck yall.

Stop me.

Ahh fuck you mine's at least Vidya related

>g/g
>g/s
>s/s
>already pulled a gold, so double silver doesn't even matter

Gotta be 50% senpai. Doesn't logically make sense to include double silver.

i think it's a 25% or i may be a mathtard

>need to switch doors still
>the core odds that you picked wrong in the first place is the real odds of the situation still
Thanks, I appreciate this small gesture of you, trying to explain math to a retard like me. Maybe I'll hit up Khan Academy some other time.

Room temperature in Celsius? Sounds about right

I don't understand why the order matters. You know that one of the hits is 100% guaranteed to be a crit. So all that matters is the chance of the other one, which is 50%, and thus the answer is 50%.

What am I missing here?

2/3.

sadly no, this one is significantly less intuitive than the other one; since the first ball you pull out is gold, the silver only chest is instantly removed as it could not possibly be the one you picked; from this point the correct answer "feels" like 50% but its not, had you grabbed the G S box the odds of getting the gold on your first grab would have been 50% but had you grabbed the G G box the odds where 100% that the ball you grabbed was gold, this leaves the chance that you grabbed the G G box at 75% based on the ball that came out

50%
Monty Hall relies on the ability to change your choice after you select and then have a wrong option removed from the remaining pool.

is your answer including using the same box?

>g1 g2
>g2 g3
>g3 s1
2/3

It literally says you're using the same box my guy.

retards trying to act smart.
if at least one is a crit, that means one of the hits has a 100% chance of being a crit. the odds of both being crits is only dependant on the one that doesn't have a 100% chance.
the answer is 50%

50% obviously

>g1 g2
>g2 g1
>g3 s1
I fucked that second scenario up.
still 2/3

None, because this is clearly a disguised gambling activity and therefore there wasn't a potion purchase.

its pretty impressive that you tried to call other people retard's while getting it wrong yourself, I mean you're both wrong in different ways I suppose.

oh. didn't read that bit

2/3

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imagine putting this much thought into such a simple problem and still being wrong. embarrassing

You have a significantly higher chance of picking the left box instead of the middle box.

Separating by cases:

>G1G2
>G2G1
>G3S1

2/3

How is it not 50%? One hit is already guaranteed a crit

>retard trys to pretend to understand math

why are you counting the order of crits when isn't relevant?

it's literally the more correct answer you retard
learn math

See: The ultra-condensed version is that it matters how you interpret the question. If you read it to mean "in any set of 2 hits where 1 was crit, what is the chance the unknown hit was crit?" The answer is 1/2. But this is making an assumption: the computer only presents this question after a crit has occurred

If we read it to mean "in only sets of 2 hits where 1 was crit, what is the chance of 2 crits?" the answer is 1/3 because we must count all possibilities, and in fact, the real data would support this. If we have say, 10000 sets of 2 hits, filter out all of them with no-crits, we'd have 3333 double crits, 3333 crit-hit and 3333 hit-crit. It doesn't matter that the question doesn't ask about order. The fact that there were two hits doesn't become inconsequential to probability, it's still extremely important

ok retard

the order is relevant in both questions

>significantly higher chance of picking the left box

Nani the fuck? Says who? You can't just insert bullshit like that.

One of the hits is a guaranteed crit. So the only outcomes are
>hit crit
>crit crit

yes but you don't know which of the two, either one could be the crit, its not the first one crits whats the chance the second one crits, its if one crits whats the chance the other one crits
imagine all possible outcomes, theres the outcome where the first attack crits and the second one doesn't, the outcome where the first attack doesnt crit and the second one does, and the outcome where both crit. now how many possible outcomes were there? 3. from that point you can figure it out really easily if the outcome you're looking for is one of three possible outcomes

why? all it says its that one is a crit, doesn't say which is first, which is the entire foundation of all the 33%ers

>not building 100% crit rate

if it said the first or second hit is guaranteed to crit then it would be 50% but since we don't know which will crit, it's 25%

Why does this "whiich of the two" bullshit matter? You're given the crit. That's like arguing with someone trying to give you a blowjob; you're still getting head, retard.

have fun at school tomorrow, I hear you're learning about odds soon

>One of the hits is guaranteed
In the sense that a flipped coin that landed on heads is guaranteed to be heads? Are you a moron? That's what you're assuming. That just because it was heads means it was guaranteed.

It might have been no crits. This set, which already happened, just so happened to have 1+ crit. You have to count the result of each swing independently, that's how you determine that the overall chance of a double crit is 25%. Jumping from 33% to 50% is making a guess that the problem as worded does not provide.

It says at least one is a crit, and the crit on each hit is calculated independent of each other

>implying game devs won't hard cap you
lel

because you pulled a gold orb, for you to have pulled a gold orb from the middle one was only a 50% chance while the left one had 100%; since you pulled gold first its significantly more likely that you had the left box

Because you already know that you picked a gold ball, that's a given, and since the left box has more gold balls, it's more likely that you pulled from that one instead if you ended up grabbing a gold ball. If it's still not obvious then imagine if the left box had a million gold balls instead of two.

>it's 50%, it either happened or it didn't
you're worse than the monty hall shitters

that makes no sense, if one of the hits has a 100% chance to crit and the other has a 50% chance, they will always have a 50% chance to both hit no matter which one is first
wow great reasoning

The fact that the left box has two gold balls in it makes it more likely for it to be chosen.
See
en.wikipedia.org/wiki/Bertrand's_box_paradox

no, you're counting in the possibility that neither crit however the question clearly states that it isn't a possible option so you can safely discount that outcome and remove it from the equation

so? that only proves me correct, it doesn't matter what order it goes, one crit is a constant, the other is a 50% variable, the result is entirely dependent of the variable.

33% chance

25% chance to land both crits in a normal situation
25% chance to land no crits in a normal situation
50% chance to land 1 crit in normal situation

You can't land no crits, so that removes 25% from your total probability. Now it's 75%.

Normalize the remaining:
25%/75% = 33.33% to land 2 crits
50%/75% = 66.66% to land 1 crit

you don't actually understand how the question works, you're making assumptions about the outcomes

the question isn't asking for the percentage of the chance both hits are crits out of all the potential roll outcomes, just the chance of both the hits critting as opposed to the only the guaranteed single crit. the other hit remains at 50% in either order so it's 50% regardless

its not a constant

says you, not the problem

50 + 50 = 100
Almost got me bro, give me another one.

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it says to assume a 50% crit chance, so there is no garunteed crit

the answer has been explained like 10 times in the thread and you still refuse to learn anything, I don't know what more can be done for you; there are 3 possible outcomes not 2, you're treating the question as what is the chance of getting a crit in a single attack with a 50% crit chance and ignoring the part where there is 2 attacks and the one that has to crit could be either one

>have 1000 sets of hits
>pick a crit at random
>probability other in set is crit is 50%

>have 1000 sets of hits
>limit pool to those that have at least 1 crit
>pick a set at random
>probability is 33% for double crit

Given the wording of the problem, the latter scenario seems more likely, and real data will conform to it too

>It might have been no crits.
No. There are two hits. One is a crit. The only has a 50% chance to be a crit.

it literally states "at least one of the hits is a crit", there is only one hit with a 50% chance to crit, you literally cannot rebut this fact.

of course this is what the creator wanted.
but the fact you limited the choices possible, was the real "answer" he was wanting you to figure out.
if he said to do it this way, then you knew the answer before he finishes talking.

so by leaving this step out, it causes the problem we now have.

>retards still arguing over the problem when the proper solution has been linked 3 fucking times already

that doesn’t mean one hit jumps to 100% chance of crit, it means any scenario where neither hits were a crit are ignored and we do the experiment again.

just like in the ball question, its still possible to grab a silver ball, we just reset the experiment in that case because it wasn’t what the question was asking for

Four options of equal probability: HH, CH, HC, CC.
>At least one of the hits is a crit
Only crit ones are left, so three options, CH, HC, CC. So 33%. If it said the 1st hit is crit, then yea it'd be 50%. It doesn't. It just says one of them is. American education, jeez.

"at least one" not "the first one" so both the first and second could crit while the other doesn't leaving two possible options in which it doesn't have 2 crits and only 1 that does have both crits, this is the kind of shit they use to try and teach kids that statistics don't always conform to what they feel like they should be

people who don't have time to read 40000 words btfo

It's 1/3rd.

Dilate and study more math you fucking faggot

except it doesn't matter, you already know that one is going to crit, the order doesn't matter to the problem, the only thing that counts is the 50% unknown variable of the other hit

>study more math
Take your own advice.
en.wikipedia.org/wiki/Boy_or_Girl_paradox#Analysis_of_the_ambiguity

where does the order of the crits matter to the problem? that's not what's being asked, it holds no weight in the problem

both hits are calculated independently, and neither magically jump to 100% so it does matter

What's the hit rate? Chance to miss or get blocked?

thread should have ended here

100%
0%

No you can't that isn't how probability works.

again, how does that disprove me? we already know one (first or second hit) crits while the other (second or first) has a 50% chance of criting, if its individual how does it the 1/2 2/1 2/2 argument come from?

how many times must something like this be explained to your directly before you understand it? is it simply a refusal to learn or are you so stubborn that the idea that you might be wrong a concept you refuse to even consider. when approached with the monty hall problem do you just shout "THERES ONLY TWO OPTIONS IT MUST BE 50%!"

Replace crits with heads in a coin flip

The coin has no memory of previous flips

Just because the first flip is tails, doesn't make the next automatically be heads.

Tails-Tails is a possible result, but it is not taken into account

TT - doesn't matter
TH
HT
HH - all three of these have an equal chance

Therefore it's 33%

One crit is guaranteed so is irrelevant to the probability and obviously the sequence in which it occurs does not matter. If you don't answer with 50% you are retarded.

Just curious, what's your answer for
?

>Thus, if it is assumed that both children were considered while looking for a boy, the answer to question 2 is 1/3

Hurrrrr

because its the correct awnser and 33%fags try to unessesarly complicate the problem when constants and variables have already been explicitly stated.

probability questions have parameters, and if a scenario happens outside of the parameters we get to reset and try again.

>if it is assumed
The very fact that you made the assumption instead of pointing out the paradox makes you just as retarded as 50%fags.

>it's a Yea Forums NEETs discuss probabilities episode
This should be outlawed

100% because im not a lucklet

This would be it if a crit wasn't a guarantee you dumb flaming homo.

see
for a full breakdown on why you don't understand math

that one that crit wasn’t magically a 100% crit, it was still 50%

But the problem already tells you that one of the hits is a crit, so you can eliminate that possibility that both attacks are normal hits. Hit+Crit is the same thing as Crit+Hit so you dont count that twice. You're left with a 50% chance to do two crits because once again the problem tells you the crit chance is 50%.

Regardless, the answer is not fucking 25% in any universe.

read the wiki page you actual troglodyte

>try to unnecessarily complicate the problem
Taking possible outcomes into account for statistics isn't "unnecessary" its quite explicitly how math works you fucking invalid

says you, person misreading the problem so it can suit your retarded mathematical solution

This, people discard HH but don't discard CH and HC hahahah, only pick the winning team.

This is the mathchad's correct answer

50/50 is correct too but less because the image never stated which hit is the critical one so first and second hit are dependent events.

Absolutely correct. I didn't read the post you were replying to originally and just assumed it was a 50%fag. Wow.
You have my sincere apologies.

H C and C H are not the same thing you fucking idiot, they are two different outcomes you fuckwit both of them having the same number of crits doesn't make it the same result

>Hit+Crit is the same thing as Crit+Hit so you dont count that twice.
that's where you're wrong kiddo, the event is clearly ordered and while they're technically the same result, there are two ways to achieve it

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you mean purposefully adding constants to a problem that doesn't have them? what's next, 2+2 = 5 because you decided to add a +1 coming form your ass?

no u

>irony

This isn't a Monty Haul problem, and it's not of crit chance in general. It's one specific outcome, and because there's two states for two actions, it means there's four possible states, so it's 25%

i admit that I usually feel very stupid whenever I'm posting on here. I'm very bad at debating/argumentation and a lot of anons, when they're not shitposting, seem to have an air of intellect around them, but then again, English isn't my first language so that might affect my perception.

then threads like this pop up and make me feel so good about myself.

I hope for the sake of your local educational system that you are trolling and not actually this incapable of understanding a fairly simple problem with such dedication to being wrong, the idea that you went through school and actually ended up like this is a dire sign for future generations assuming you are actually this dumb and not trolling; which I truly hope isn't the case

show me all 4 states you think can happen. it's only 3.