probability question

[entropi]

If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

[Amelia]

It depends.

[Bill Spight]

[EdLee]

Question X:

???

If you choose an answer to Question X at random, what is the chance you will be correct?

Please first define Question X.

[schawipp]

To make things more desperate you should extend this to ...

If you choose an answer to this question at random, what is the chance you will be correct?

(A) 20% (B) 50% (C) 60% (D) 20% (E) 0%

[entropi]

Question X:

???

If you choose an answer to Question X at random, what is the chance you will be correct?

Please first define Question X.

EdLee, the question refers to itself. So, "this question" is the question you read.

[DrStraw]

If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

Zero. Without further information we can only assume that all answers are possible. Therefore the choices form a finite subset of an infinite number of possible answers. The density is zero.

If, on the other hand, all possible correct answers are listed and each is considered equally likely then the probability is one third.

[RBerenguel]

If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

Zero. Without further information we can only assume that all answers are possible. Therefore the choices form a finite subset of an infinite number of possible answers. The density is zero.

If, on the other hand, all possible correct answers are listed and each is considered equally likely then the probability is one third.

Measure or (in some cases, depending on the use) capacity is zero. A set does not have density, but is either dense or not (depending on a metric.)

[DrStraw]

If you choose an answer to this question at random, what is the chance you will be correct?
(A) 25% (B) 50% (C) 60% (D) 25%

Zero. Without further information we can only assume that all answers are possible. Therefore the choices form a finite subset of an infinite number of possible answers. The density is zero.

If, on the other hand, all possible correct answers are listed and each is considered equally likely then the probability is one third.

Measure or (in some cases, depending on the use) capacity is zero. A set does not have density, but is either dense or not (depending on a metric.)

Agreed. But most people here are not mathematicians. As I was writing in English I used a term which I would expect to be clearer to everyone.

[RBerenguel]

Got it, I just read it in mathematician eyes but now I think of it I can't find a clearer English word aside from density. Weird, I bet there should be one...

[RobertJasiek]

Since it may be any random distribution with any probability, there are an infinite number of such probabilities but only a finite number of available answer probabilities. Hence the answer is: "almost 0". With a model of an increasing number of distributions and finally infinitely many, the answer is: "converges to 0 from its positive side".

[Fedya]

I'm reminded of the mathematician joke that ends with the punchline, "The mathematician picks up the flamethrower and sets the house on fire, thereby reducing the problem to a case previously solved."

[Bill Spight]

Since it may be any random distribution with any probability, there are an infinite number of such probabilities but only a finite number of available answer probabilities. Hence the answer is: "almost 0". With a model of an increasing number of distributions and finally infinitely many, the answer is: "converges to 0 from its positive side".

In English "to choose at random" means to choose so that each option has the same probability of being chosen.

[Joelnelsonb]

%0.0 Chance given that the correct answer is not actually listed.

[RobertJasiek]

In English "to choose at random" means to choose so that each option has the same probability of being chosen.

I'd say that "to choose at random uniformly / with a uniform distribution" would mean this. At least in (precise) German, one needs to say "gleichverteilt zufällig wählen".