Lets test how smart Cred Forums is

Lets test how smart Cred Forums is

Attached: 089267E7-83AD-4890-9FE2-FFE7FC339EED.jpg (1218x740, 212K)

Attached: image0-39.jpg (1280x932, 104K)

2

Third

Bump for any takers

Retard

Two

Attached: 20200220_232106.jpg (1024x622, 125K)

1 (arrays start at zero)

Fuck you all its third

all wrong, the 2nd one is closest but its the lighter shade of blue.

2 retard

Second one.

3 and 2 are both correct depending on if you fold it inwards or outwards

4th

its the correct answer retard
the grey would be on the right, not the left, of it.
it's an understandable mistake.

The second people are folding down, as if to make the vertex away from us, because we only see the topside with color. Your claim of the third is folding up, and assuming the colors are the same on the other side. You may be right, but maybe not. The 2nd is definitely right.

the answers are folded outwards and we have no reason to believe the colors are the same on the back.

2

but it's only reasonable to fold it outwards
(me)

Not if you fucking fold it the other way

Wrong, it's the same shade.

It's #3 folded inside out you cuck. Don't get mad because you're a failure at life and are wrong

2.

Fine fuck you its second

Either of the middle two depending on how it's folded.

>the answers are folded outwards

You don't know that for a fact. They could be folded "inward" and rotated.


>we have no reason to believe the colors are the same on the back.

It's reasonable to assume they are the same color front and back. Changing the colors on the back would add an unknown variable to the problem, in which case any answer could potentially be correct and it would be a trick question.

lol dunning–kruger

The second one and also Big Black Cocks.

The only correct answer for certain is number 2. It could be number 3 if you fold it the other way, assuming that the colors are the same on the other side but then again you could also assume that the colors are different on the other side which gives you any number of possibilities and any of the shown answers could be correct. That’s the problem with making assumptions. Anyway, number 2 is the only correct answer based on the evidence we are given without having to make any unknown assumptions that can’t be verified.

No, assuming the color is the same on the back is just that, an assumption. It is unwarranted. No unknown variable is being added. We simply don't know the color on the back, nor do we need to. This is not a trick question.

OP here. If we get to 50 replies I will give th answer and points.

I say tell us now. Before a fight breaks out.

>No unknown variable is being added.

...

>We simply don't know

It's amazing you can get out of bed in the morning without tripping over yourself. Though, to be fair, that's an assumption too.

> Reasonable to assume
> Could be a trick question

I’m gonna assume this is a trick question and I can reasonably assume that the creator assumed that we would assume that the colors are the same on both sides and therefore changed the colors in a completely random way. Correct answer is number 4!!!

Its 2, anyone saying anything else a nigger or a retard

After some thought plus some rethought because Cred Forums is the type to pull shenanigans such as not even listing the right answer, I do believe the answer is 2nd from the left.

Bumping to get to 50.

Bumping to see what points/ IQ I get.

The Left-center diamond is the assembled version of the template.

You're a retard. It's not a variable, it has no impact on the problem. Kys nigger

Only 10 more replies

OP here. Gonna go ahead and give the answer. The correct answer is B.

If you choose B give yourself 3 points
If C give yourself 2 points
If D give yourself 1 point
If A give yourself 0 points

3 = 120+ IQ
2= 90-120 IQ
1= 60-90 IQ
0= >60 IQ

Unfair i looked at C first thinking it was correct and didnt look at the other options thinking they would prob all be scrambled fuck you fuck you

How about a different problem? Suppose I flip two coins in secret and tell you one landed heads up. What is the probability that the other one also landed heads up?

One in two. The truth of your statement wouldn't affect the outcome of the second coin flip because you only mentioned the result of the previous coin flip.

Wrong. The probability that both are heads is 1 in 3. Thank you for playing though.

It's an uncut cock diagram

Attached: _1989.jpg (1267x590, 310K)

how?

2nd one too easy

Maybe if the colors were a little less monochrome. Goddamn.

i bick b

It's intentionally a strange question. You are told one of the coins is heads but not which, so we have to look at all remaining possibilities:

Coin1 Heads & Coin2 Heads
Coin1 Heads & Coin2 Tails
Coin1 Tails & Coin2 Heads

Coin1 Tails & Coin2 Tails is not a possibility because we know one or the other is Heads. So there is one scenario in three where both are Heads.

oh, neat.

Attached: laugh.gif (400x275, 361K)

Bullshit.

None of them u dick

a solid refutation if I ever saw one

2 or B.
t. rookie artist

Yes.
the second one

Attached: B44768D4-D74F-4173-98C9-3CD073308C59.jpg (1024x622, 52K)

The lack of shadowing is fucking with your eyes

2

All right, let me show you what I mean with a simple function set to run 1,000,000 times.

Each time this function executes, 2 integers are generated, and one of them is randomly set to have a value of 1 (akin to flipping heads on a coin). Then, the other integer is randomly assigned a value of either 0 or 1. Each time the function's execution ends with both integers having a value of 1, a counter tics up by 1. After the function iterates 1,000,000 times, the total number of times where both integers possessed a value of 1 is printed in blue text to the right.

Is that solid enough to debunk this deceptive brainlet 'trick question'?

Attached: Bullshit.jpg (967x1193, 118K)

Attached: Screenshot_20200221-042005_Chrome.jpg (1439x2524, 482K)

That would be correct, if the question didn't explicitly state that at least one of the coins landed on heads.

What you're doing in your simulation is flipping the coin 1,000,000 times, but simply ignoring whenever both happen to land on tails. To accurately represent the question at-hand, you'd have to ensure each trial that at least one of the coins lands on heads, not just cease to acknowledge the iterations where neither of them do.

The answer to the original question is 50%.

What you've done in your script is immediately make one coin or the other heads reducing the other to a 50/50 flip. This would be fine if the question read "coin 1 is heads, what's the probability coin 2 is heads?" or vice versa. For the question, as it was posed, your method is incorrect. Your method removes the 1/2 probability of one of the coin flips sonce you hardcoded it based on the results of the other coin.

Second and third are both valid depending on if you fold up or down.

non of them is correct you autistic fuck. which can be seen in less than 3 sec if not an American.

here the corrected version. Which one is it?

Attached: nigger.png (1218x740, 327K)

All right, I can't tell if you're trolling, but I'll give you one last (You) before I just say fuck it for the night.

The question posits the following hypothetical:
>Two coins are flipped
>One of them lands on heads, though you aren't told which of the two it is
>Given this information, you're asked to state the probability that the other coin also landed heads-up

From this, we KNOW for certain that one of the two coins has to be heads. It doesn't matter which.

What then, at this point in the hypothetical situation, is the probability that the other coin also landed on heads?

The sub-100 IQ response Cred Forums likes to tout around is that:
-[H][T]
-[T][H]
-[H][H]
are the only three possible outcomes (this is correct), and because [H][H] comprises 1 of the 3 of these possible outcomes, there's therefore a 1/3 chance that the other coin landed heads-up (this is both misleading and blatantly wrong).

It's wrong because as soon as you're given the information that one of the coins landed on heads, you're then able to arbitrarily discard either the [H][T] or the [T][H] outcome, because it does not matter which of the two coins landed on heads--at that point, you're only focusing on the other coin.

Assuming you're not trolling, the correct question to ask that would warrant the solution you're trying to justify is more like this:

"If two coins, Coin A and Coin B, are flipped, and you're told that at least one of the two coins landed on heads, what is the probability that you could guess how both Coin A and Coin B respectively landed?"

Does that make more sense? I'm sincerely trying to explain this the best my 4 AM sleep-deprived brain will let me.