Probability of guessing

[BMAD]

If you were to choose an answer to this question at random what is the probability that it would be correct?

a) 25%

b) 50%

c) 100%

d) 25%

[bonanova]

It is paradoxical because it concerns both (a) facts and (b) truth values of statements about facts.
There are standard ways to deal with self-referential paradoxes.

Tarski eliminated conflict by permitting statements that did only (a) or (b) but not both.
Thus, a statement can either address facts, OR the truth values of statements, but not both.

<edit>

• It's more restrictive than that, or course.
  Conflict arises when a statement refers to the
  truth value of its own statement about a fact.

<end edit>

For this particular question, here's my two-cents:

First, "correct" needs to be defined. I can think of four ways:

1. The chosen answer is "correct" if it follow the instructions for making a choice.
   That is, if it is chosen at random.
   Correct means correctly following the OP instructions.
   In this sense any random choice is correct.
   The number associated with that choice is irrelevant to its "correctness" as a choice.
   
2. The chosen answer is "correct" if the number I choose gives the probability that I chose it.
   The chosen answer is correct therefore if my choice is (a) or (d).
   
3. The chosen answer is "correct" by the previous analysis if I choose (a) or (d).
   The probability of choosing {(a) or (d)} is 50%.
   If 25% must be the probability that I chose {(a) OR (d)}, then {(a) OR (d)} is not the correct answer.
   
4. Alternately, if (a) or (d) is the "correct" answer, then (b) is the correct choice.

When more than one condition applies, there is the possibility of conflict.

In cases 3 and 4, this happens; thus the classical self-referential paradox.

By the way, my choice of definition for "correct" is the most literal one: the first.

OK.

So I went ahead and made a choice at random; therefore it was a "correct" choice.

What was my choice?

(b) 100%

Edited February 20, 2013 by bonanova
Clarify my opening statements

[ThunderCloud]

If the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.

If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.

Edited February 18, 2013 by ThunderCloud

[BMAD]

If the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.

If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.

I think you are assuming too much

[ThunderCloud]

If the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.

If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.

I think you are assuming too much

I did assume that if (a) is correct then (d) is also correct, and vice versa. If this is not the case, then exactly one of them must be correct, and the answer is 25%.

[BMAD]

If the answer is 25%, then because 2 out of 4 answer choices are "25%", the answer must actually be 50%. This is a contradiction. So the answer cannot be 25%.

If the answer is 50% (or 100%), then because 50% (or 100%) is 1 out of 4 answer choices, the answer must be 25%. This is again a contradiction. So the answer cannot be 50% (nor 100%).

So none of the provided answer choices are correct. Therefore the probability of choosing the correct answer is 0%.

I think you are assuming too much

I did assume that if (a) is correct then (d) is also correct, and vice versa. If this is not the case, then exactly one of them must be correct, and the answer is 25%.

I still disagree with your analysis

[Yoruichi-san]

Bon-chan brings up some good points, but thinking about them gives me a headache, so...

...and calculate the expected value

.

If every answer really corresponded to a probability of being correct when randomly chosen, then the expected value of being correct would be (1/4)*(.25+.5+1+.25)=.5, so I'm going with b)50%

Migraine averted

[BMAD]

It is paradoxical because it concerns both (a) facts and (b) truth values of statements about facts.There are standard ways to deal with self-referential paradoxes.Tarski eliminated conflict by permitting statements that did only (a) or (b) but not both.Thus, a statement can either address facts, OR the truth values of statements, but not both.For this particular question, here's my two-cents:First, "correct" needs to be defined. I can think of four ways:

• The chosen answer is "correct" if it follow the instructions for making a choice. That is, if it is chosen at random. Correct means correctly following the OP instructions. In this sense any random choice is correct. The number associated with that choice is irrelevant to its "correctness" as a choice.
• The chosen answer is "correct" if the number I choose gives the probability that I chose it. The chosen answer is correct therefore if my choice is (a) or (d).
• The chosen answer is "correct" by the previous analysis if I choose (a) or (d). The probability of choosing {(a) or (d)} is 50%. If 25% must be the probability that I chose {(a) OR (d)}, then {(a) OR (d)} is not the correct answer.
• Alternately, if (a) or (d) is the "correct" answer, then (b) is the correct choice.

When more than one condition applies, there is the possibility of conflict.In cases 3 and 4, this happens; thus the classical self-referential paradox.By the way, my choice of definition for "correct" is the most literal one: the first. OK. So I went ahead and made a choice at random; therefore it was a "correct" choice.What was my choice?(b) 100%

You always have an answer to the question :

1. How many answers are right from a set of four possible answers?

It is either 0, 1, 2, 3 or 4.

You also always have an answer to the question :

2. What is the probability of randomly picking the right answer from a set of four alternatives, given a fixed number of right answers?

It is either 0%, 25%, 50%, 75%, or 100%.

In this case theyre sort of asking you both questions simultaneously, since the content of the possible answers should be fixing and allowing you to discover both

a. the number of right answers

and

b. the right answer

Once you give an answer to either 1 or 2 you immediately get an answer to the other. In this case none of those answers is actually fixed, you should discover both at the same time: hence the (apparent) problem.

The trouble is that there are very there are some (actually very strict) constrains regarding the possible relation between a and b that are not being respected by the given possible answers: hence the incoherence.

Notice that for someone to be able to pick an answer correctly the following conditions must be met:

I) There is one correct answer ←→ There is one and only one option = 25%

II) There are two correct answers ←→ There should be two and only two options = 50%

III) There are three correct answers ←→ There should be three and only three options = 75%

IV) There are four correct answers ←→ There should be four and only four options = 100%

once you dont respect this conditions you have an incompatibility between picking the right answer and picking the number of right answers, and so there is no way to answer the question and you should not be bogged by the appearance that there should be such an answer.

[bonanova]

Bon-chan brings up some good points, but thinking about them gives me a headache, so...

...and calculate the expected value

.

If every answer really corresponded to a probability of being correct when randomly chosen, then the expected value of being correct would be (1/4)*(.25+.5+1+.25)=.5, so I'm going with b)50%

Migraine averted

* Handing Y-San an Excedrin, just for emergencies ...