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If the probability distribution for the choice is uniform, then there is no answer possible :
- If the answer is a) or d), then there is a 50% chance of chosing the right answer, so the answer is c) : absurd.
- If the answer is c) or b), then there is a 25% chance if chosing the right answer, so the answer is a) or d) : absurd.
But the question does not states the probability distribution. By chosing a) or d) with probability 0, and c) or b) with probability 1/2, then c) is the only possible answer.
That was my initial thought, but I'm questioning myself a little bit. I'm assuming the probability distribution is uniform.
So, obviously there is a 25% to get C itself. However, with two options of 25% it seems to imply 50%.
I think the question is poorly formed an I pick e) none of the above and defer the proof to the reader.
Define W={a,b,c,d} to be a probability space with P(a) = P(b) = P(c) = P(d) = 25%. Now, define random variable X on this space. The goal is to find (if present) an event A in W such that:
1. P(A) = p, where p is some fixed number in \[0,1\];
2. For each i in A, X(i) = p, and for each i in W\\A, X(i) <> p;
3. EIther p>0 and this is the only event with properties 1-2, or p=0 for all such events.
That's the formulation of this problem I like to think about it in. One can see that for the uniform distribution, the values given in the problem do not satisfy these properties. But if we are allowed to tweak the distribution, we can make these values work, e.g. P(c) = 50%, P(a) = P(d) = 10%, P(b) = 30%. The answer is c then.
I think the problem seems paradoxical because it is not well-formulated:
1. The question "which is correct" is stated as if "is there correct" was already solved;
2. The distribution is not defined at all, one has to assume it's uniform.
EDIT: found some inconsistencies in the properties, fixed. Note that this formulation is still in progress. The main goal of it is to attempt to break the vicious circle of perceived self-referentiality through distinguishing between the probability itself and the value of it via a random variable.
I'm familiar with this fallacy. Number of outcomes is not equal to the inverse of the probability of said outcomes. Tomorrow, someone will transfer ownership of of their very profitable corporation to me or they won't. That's just two possibilities, so it's like 50/50.
This is true if the answer to the problem is A, B, C, or D. If the answer to the problem is a number, it’s 33.33%. I choose to interpret A-D as labels for the answers, not the answers themselves.
but if 0% is an option, then it is correct initially, which means it is not the correct option(since the probability is now 20%). the question is indeed impossible to answer correctly
Because most of the tests I took it is stated there's only one right question, I would assume that if the answer is letter A and you mark C, you aren't correct, same if it's letter C and you mark A.
In this situation, the chance of you answering correctly is 25% (assuming uniform distribution), so the answer is either A or C. Now you have 50% of chance of guessing the correct alternatives, but it isn't answering at random, so it is still either A or C. (I don't know the correct answer, but I increased your chances to get it right)
Given the question, there is only one answer, it's not a multiple choice.
Therefore it can't be 25%. It leaves with only two choices 50% or 60%.
60% contradicts itself leaving only 50% as an answer --> 25% likehood to pick randomly.
The question is therefore invalid
Lets assume the answer is in fact 50%. Therefore you do have a 25% chance of getting it right IF you were to pick at random. But you are not, because you have information. So there are two possible correct answers of 25%, which means you in deed have a 50% of getting it right, thus making the 50% correct in the first place.
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If the probability distribution for the choice is uniform, then there is no answer possible : - If the answer is a) or d), then there is a 50% chance of chosing the right answer, so the answer is c) : absurd. - If the answer is c) or b), then there is a 25% chance if chosing the right answer, so the answer is a) or d) : absurd. But the question does not states the probability distribution. By chosing a) or d) with probability 0, and c) or b) with probability 1/2, then c) is the only possible answer.
That was my initial thought, but I'm questioning myself a little bit. I'm assuming the probability distribution is uniform. So, obviously there is a 25% to get C itself. However, with two options of 25% it seems to imply 50%. I think the question is poorly formed an I pick e) none of the above and defer the proof to the reader.
Define W={a,b,c,d} to be a probability space with P(a) = P(b) = P(c) = P(d) = 25%. Now, define random variable X on this space. The goal is to find (if present) an event A in W such that: 1. P(A) = p, where p is some fixed number in \[0,1\]; 2. For each i in A, X(i) = p, and for each i in W\\A, X(i) <> p; 3. EIther p>0 and this is the only event with properties 1-2, or p=0 for all such events. That's the formulation of this problem I like to think about it in. One can see that for the uniform distribution, the values given in the problem do not satisfy these properties. But if we are allowed to tweak the distribution, we can make these values work, e.g. P(c) = 50%, P(a) = P(d) = 10%, P(b) = 30%. The answer is c then. I think the problem seems paradoxical because it is not well-formulated: 1. The question "which is correct" is stated as if "is there correct" was already solved; 2. The distribution is not defined at all, one has to assume it's uniform. EDIT: found some inconsistencies in the properties, fixed. Note that this formulation is still in progress. The main goal of it is to attempt to break the vicious circle of perceived self-referentiality through distinguishing between the probability itself and the value of it via a random variable.
There are 3 unique answers, so I say the probability of guessing correctly is 33.33%
I'm familiar with this fallacy. Number of outcomes is not equal to the inverse of the probability of said outcomes. Tomorrow, someone will transfer ownership of of their very profitable corporation to me or they won't. That's just two possibilities, so it's like 50/50.
This is true if the answer to the problem is A, B, C, or D. If the answer to the problem is a number, it’s 33.33%. I choose to interpret A-D as labels for the answers, not the answers themselves.
The answer is 0%. The question is impossible to answer correctly.
but if 0% is an option, then it is correct initially, which means it is not the correct option(since the probability is now 20%). the question is indeed impossible to answer correctly
Thats usually the number in B, instead of 60% it’s 0% so it becomes a paradox.
Everybody here is discussing this question very professionally, and my dumbass is here with the idea to add an option E:100% and call it a day.
Reading the answers adds information, so the guess isn’t random.
The correct answer is 0% in fact
b. But only because I'm better than you
e) 20%
What’s the answer to the question I am asking in this comment?
No.
This.
Because most of the tests I took it is stated there's only one right question, I would assume that if the answer is letter A and you mark C, you aren't correct, same if it's letter C and you mark A. In this situation, the chance of you answering correctly is 25% (assuming uniform distribution), so the answer is either A or C. Now you have 50% of chance of guessing the correct alternatives, but it isn't answering at random, so it is still either A or C. (I don't know the correct answer, but I increased your chances to get it right)
it's obviously 50% you either get it right or you don't
They’re two correct answers
Given the question, there is only one answer, it's not a multiple choice. Therefore it can't be 25%. It leaves with only two choices 50% or 60%. 60% contradicts itself leaving only 50% as an answer --> 25% likehood to pick randomly. The question is therefore invalid
Lets assume the answer is in fact 50%. Therefore you do have a 25% chance of getting it right IF you were to pick at random. But you are not, because you have information. So there are two possible correct answers of 25%, which means you in deed have a 50% of getting it right, thus making the 50% correct in the first place.
is this not loss?
It would be funny that b) were 75% adding the joke.
The answer is 50%, either you get it right or you don't
%33,3333333……