Only the geniuses get this one

If there are x questions on a multiple-choice exam with 4 possible answers (A,B,C,D), what is the chance that 3 C's will occur in a row at least once in the exam? Write a generalized formula with x questions.

Ok i have the answer for x < 3 gimme a while to type out the equation

Should be 0%...

0.25 × 0.25 × 0.25
> should be 0%

mongoloid. Cant tell me this isnt homework.

(x-2)/64 for x>3

Ez

Have A B C D the same probability?

2 (3/4 * 1/4 * 1/4 * 1/4)

t. 213 IQ regular Rick and Morty watcher

Ok so obviously if x = 3 your chance is just 1/(4^x) = 1/(4^3) = 1/64

Makes sense, your only possible permutation is

CCC

If x = 4, questions 2 and 3 have to be C, and at least one of the other two have to be C as well. To calculate the chance of at least one, that's 1 minus the chance of 0.
So that would be [ 1/(4^2) ] * [ 1 - (3/4)^2 ] = [ 1/16 ] * [ 5/16 ] = 7/256.

Im pretty confident of that, but just to be sure, i'll list the permutations.

ACCC
BCCC
CCCA
CCCB
CCCC
CCCD
DCCC

Ok, that's 7 so i'm on the right track.

How can we generalize this for x though. Hm.
The chance that any preselected 3 in a row questions are all C is 1/64, as determined by when x = 3.
There's x - 2 ways you can pick exactly which 3 questions, say for example x = 5, there's 3 ways; the first few questions, the middle few questions, and the last few questions. This much seems pretty firm.
But for each question that isn't part of the 3 we selected, you can have 4 possibilities; A through D.
It's gotta be more complicated than [ 4 ] * [ x - 2 ] * [ 1 / 64 ], otherwise x = 4 would have yielded 8/64 rather than 7.

I think it's gotta be that CCCC is only 1 permutation, but with the formula above is counted twice; the first time being if we select the first 3 questions to be our 3-in-a-row C streak with the 4th irrelevant question that just happens to also be C, and the second time being the last 3 questions being our 3-in-a-row streak and the first queston being the irrelevant question being C.

Ok, so generalized we have to subtract redundant permutations.
Let's investigate x = 5 for more clues.

>Write a generalized formula with x questions.
Most people with IQ above 100 can read. Interesting.

Incoherent question - it doesn't account for the agency of the person setting up the test. Not a pure chance scenario.

Most people with IQ above 100 can understand that I'm memeing.

Also It's actually 2/(4*4*4) = 1/32

-1/12

I count C C C C twice so ACTUALLY

3/4 * 1/4 * 1/4 * 1/4 +
1/4 * 1/4 * 1/4 * 3/4 +
1/4 * 1/4 * 1/4 * 1/4

which is 7/256

If you want to find the generalised version just google the abracadabra problem

So for x = 5 I'm gonna take a different route to calculate the probability.

I'll list the possible permutations below, but instead of writing A, B, or D, I'll write N for not C.
This means that the chance of each permutation will instead of 1 in 4^x , be equivalent to [ 3^n ] / [ 4^x ], where n is the number of N's that appear for that given permutation.

CCCNN n = 2, p = 9 / 1024
NCCCN "
NNCCC "

CCCCN n = 1, p = 3 / 1024
NCCCC "

CCCCC n = 0, p = 1 / 1024

9 + 9 + 9 + 3 + 3 + 1 = 34

So when x = 5, our chance of success is 34 / 1024.

I'm noticing there's 1 way we could have 0 N's, 2 ways for 1 N, and 3 ways for 2 N's. It would appear that the formula for number of j permutations of k N's is simply j = k - 1. I'm wondering if this changes when there's more questions. I can't think of why it would.

I'll look at x = 6 's permutation list to see if that's true or not.

CCCNNN
NCCCNN
NNCCCN
NNNCCC

CCCCNN
CCCNCN
CCCNNC
NCCCCN
NCCCNC
NNCCCC

CCCCCN
NCCCCC

CCCCCC

Hmmm......

So where j is the number of permutations with k N's given x questions, it definitely is a function of x and k, and when x > 5 , it's a lot more complicated than k - 1. It looks like there's another layer of combinatorics to it.
I need to think more about this.

Looking back it looks like I made a mistake calculating x = 5's probability, I forgot
CCCNC and
CNCCC

it looked like j = k - 1 because i was only considering permutations where all C's were in a row.

>x < 3
>less than three questions
user, how do you get three answers of C in two or even one question?

basically look for questions that you know the answer 100%, then check and see what is the most common correct answer and just mark those that you dont know with the most common answer (a, b, c or d) its a guaranteed pass

Kek

Assuming the answers are uniformly distributed:

Define the predicate Q_x to mean "there is no sequence of 3 Cs anywhere in the questionset".

Then:
Q_x = 0, x < 3
Q_3 = 7/8
Q_x = Q_{x - 1} * 7/8 since the probability a question set doesn't have a run of 3 Cs is that there is no run in the questions excluding the last, and the last three answers including the last question are not all C, which is 7/8.

So the probability there is a run of 3 cs is 1 - (7/8) ^ (x - 2)

This assumes that given there is no run of length three, the probability the last two answers are both C is 1/4. This... feels right, but I'm too sleepy to prove it, and could be a brainlet assumption.

Correction, that's assuming P(C) = 1/2.

The corrected version should read Q_3 = 63/64

etc, etc,

with the answer being 1 - (63/64) ^ (x - 2).

That seems a little low, so is probably a brainlet answer.

50/50, 3 C's either occurs at least once, or doesn't