/sci/ - Science & Math » Thread #11306405

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Why hasn't /sci/ still been able to solve this?

Anonymous Thu 16 Jan 17:08:21 2020 No.11306405 View Reply Original Report

Quoted By: >>11306407 >>11306483 >>11306484 >>11306619 >>11306705 >>11307311 >>11307316 >>11307637 >>11309015 >>11309041 >>11309045

Anonymous

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Anonymous Thu 16 Jan 2020 17:08:54 No.11306407 Report

Quoted By:

>>11306405
>I love Kurisu!

Anonymous

Anonymous Thu 16 Jan 2020 17:45:17 No.11306483 Report

Quoted By:

>>11306405
1/4

Anonymous

Anonymous Thu 16 Jan 2020 17:46:27 No.11306484 Report

Quoted By:

>>11306405
50-50 either you're right or you're wrong

Anonymous

Anonymous Thu 16 Jan 2020 18:48:50 No.11306598 Report

Quoted By: >>11306603

Choice = 25%, Chance = 50%
Choice = 50%, Chance = 25%
Choice = 0%, Chance = 25%

None.

Anonymous

Anonymous Thu 16 Jan 2020 18:54:50 No.11306603 Report

Quoted By: >>11306617

>>11306598
Then... It's 0%? Wouldn't that make the chance 25% though?

Anonymous

Anonymous Thu 16 Jan 2020 19:02:23 No.11306617 Report

Quoted By:

>>11306603
It's a paradox. If you choose C (0%), you are correct. However, that then makes your chances of being right 25%, not C.
No answer is valid.

Anonymous

Anonymous Thu 16 Jan 2020 19:03:20 No.11306619 Report

Quoted By:

>>11306405
Clamped and vaccinated.

Anonymous

Anonymous Thu 16 Jan 2020 19:30:06 No.11306666 Report

Quoted By: >>11306667

They're all correct as long as you can reasonably explain why you'd choose that answer and there's always a reasonable explanation for each. Just choose one at random and justify it.

Anonymous

Anonymous Thu 16 Jan 2020 19:30:55 No.11306667 Report

Quoted By:

>>11306666
Based satan

Anonymous

Anonymous Thu 16 Jan 2020 19:56:33 No.11306699 Report

Quoted By: >>11306717 >>11307638

Let Co be the even you picked the correct answer
And A, B, C, and D be you randomly selected A, B, C, and D respectively.

P(Co) = P(A).P(Co|A) + P(B).P(Co|B) + P(C).P(Co|C) + P(D).P(Co|D)

P(A) = P(B) = P(C) = P(D) =1/4.

P(Co|A) = 1/2 because there are two equally probable states. You probably think it's 1/4, but that's not the PROBABILITY of you being right, that is the answer that corresponds to A.

P(Co) = 4*(0.25*0.5)
The probability of randomly selecting the correct answer is 50%.

Anonymous

Anonymous Thu 16 Jan 2020 19:59:43 No.11306705 Report

Quoted By: >>11306710

>>11306405
is this a statistic problem or probability problem

Anonymous

Anonymous Thu 16 Jan 2020 20:03:05 No.11306710 Report

Quoted By:

>>11306705
It's a paradox, it's a degeneracy problem.
>I'm lying right now
You can say stuff that makes no sense, but it's just a wrong predicate, no right answer.

Anonymous

Anonymous Thu 16 Jan 2020 20:09:33 No.11306717 Report

Quoted By: >>11307638

>>11306699
But if we assume the answers that corresponds to the letters as the chance of you being correct.

I.e. P(Co|A) = 1/4.

That would be (0.25*0.25)*2 + (0.25*0.5)

Then, the answer would be 0.25

So, it depends on how you want to look at it.

Anonymous

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Anonymous Thu 16 Jan 2020 20:47:55 No.11306782 Report

Quoted By: >>11306822 >>11306825

Anonymous

Anonymous Thu 16 Jan 2020 21:13:27 No.11306822 Report

Quoted By:

>>11306782
No such process exists.
Without (3.) You can only ever have a power of 2 many l's.
With (3.) you can only ever have a power of 2 many l's minus a multiple of 3.

Anonymous

Anonymous Thu 16 Jan 2020 21:14:18 No.11306825 Report

Quoted By: >>11306832 >>11306836

>>11306782
do rule 2 like 3 times, M with 8 i, do rule 3 and get MiiUiii do rule 2, MiiUiiiiiUiii, do rule 3, MiiUiiUUiii, do rule 4, MiiUiiii, do rule 3, MiiUUi, rule 4, Miii, now you got 3 is, this was the goal the entire time because now you can use rule 3 to make it MU, I WILL NEVER FAIL YOU SHOULD BE ABLE TO SOLVE THIS PROBLEMS BECAUSE I INDEED SHOULD BE ABLE TO SOLVE THEN

Anonymous

Anonymous Thu 16 Jan 2020 21:19:26 No.11306832 Report

Quoted By:

>>11306825
>do rule 3, MiiUiiUUiii, do rule 4, MiiUiiii
you dropped an i there, friend

Anonymous

Anonymous Thu 16 Jan 2020 21:20:55 No.11306836 Report

Quoted By:

>>11306825
>MiiUiiUUiii, do rule 4, MiiUiiii
Does 7 = 6?

Anonymous

Anonymous Fri 17 Jan 2020 01:37:19 No.11307311 Report

Quoted By:

>>11306405
It is 0% because it is literally impossible to get a correct answer. Even if you answer 0% you are wrong because that changes it to 25%

Anonymous

Anonymous Fri 17 Jan 2020 01:39:03 No.11307316 Report

Quoted By:

>>11306405
I need to know how many of the answers are correct first

Anonymous

Anonymous Fri 17 Jan 2020 03:50:30 No.11307572 Report

Quoted By:

>smol brain
>guessed B randomly
>found out it was B
>i got the 50%

Anonymous

Anonymous Fri 17 Jan 2020 04:24:00 No.11307637 Report

Quoted By:

>>11306405
What the fuck is even the question? It's a non-question. There is no "correct" "answer".

Anonymous

Anonymous Fri 17 Jan 2020 04:25:16 No.11307638 Report

Quoted By:

>>11306699 and >>11306717
>this is how your brain degenerates on math

Anonymous

Anonymous Fri 17 Jan 2020 18:37:27 No.11309015 Report

Quoted By:

>>11306405
None of the answers are correct. Sci is retarded not being able to figure this out.

Anonymous

Anonymous Fri 17 Jan 2020 18:51:08 No.11309041 Report

Quoted By:

>>11306405
That’s funny
1/4 is the first answer you would chose, but since 1/4 is two given answers it’s really 50% chance of guessing 1/4 but the 50% chance occurs only 1/4 but the chances of picking 1/4 is 50% but 50% only occurs 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 1/4 but 1/4 occurs 50% but 50% occurs 1/4 but 1/4 occurs 50% but 50% occurs 1/4

Anonymous

Anonymous Fri 17 Jan 2020 18:51:32 No.11309045 Report

Quoted By:

>>11306405
Suppose we ask this question infinitely many times. There is a 0% chance that any single person is correct, but 100% chance that at least one is correct.
This could also be extrapolated from the common law of poopabilities: the answer is either right or wrong, 50/50. An interesting result emerges when we correlate across multidimensional psychosocietal performance indicators such as luck and wealth where the person is more believed - and thus more likely - to be right when we increase either parameter. Given this, it is likely there is only one person capable of answering this question, but the chances of them being in any one group the surveyor finds is nil. Ipso facto, 50% is good enough. C final answer.

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