I fully understand what you are saying, that no matter what you pick you'll be wrong, but this is not an essay question where you get to put in your own answers, this is a multiple-choice question and you can only select from the available options and 0% is not in those available options, so it cannot be selected. — Jeremiah

Then it's no different to asking for the capital city of France and not supplying "Paris" as a possible answer. There's no conundrum, just a bad question/answer set.

Then use your imagination and just imagine 0% is C. It does not really change anything.

Then use your imagination and just imagine 0% is C. It does not really change anything. — Jeremiah

It changes everything. If C is 0% then there isn't a correct answer, whereas if C is 60% then the correct answer is missing, and if C is 50% then there are two correct answers.

There are two correct answers you just can't pick them.

What?

The act of choosing 50% or 25% makes one of those answers correct, but they are not correct until you pick one of them. If you pick 50% then 25% is the correct answer, and if you choose 25% then 50% is the correct answer.

To be honest, I think the 60% version generates more discussion.

I don't understand your reasoning.

Maybe you should thinking about it longer than 3 or 4 minutes.

That's not a helpful response.

I think it is one of them most helpful post I have ever made.

Here's a better way to explain it.

An answer is correct if and only if its value matches the chance that an answer with that value will be selected. So, if there is a 10% chance that an answer with a value of 10% is selected then all and only answers with a value of 10% are correct.

So let's apply it to three different answer sets, where the percentage in brackets is the chance that an answer with that value will be selected:

1.

A. 25% (50%)
B. 50% (25%)
C. 60% (25%)
D. 25% (50%)

None of the answers are correct. If none of the answers are correct then there is a 0% chance of picking the correct answer. In this case, the correct answer just isn't provided as an option (à la my example of "Paris" not being provided as an answer to the question "what is the capital city of France?").

2.

A. 25% (50%)
B. 50% (50%)
C. 50% (50%)
D. 25% (50%)

Both B and C are correct.

3.

A. 25% (50%)
B. 50% (25%)
C. 0% (25%)
D. 25% (50%)

None of the answers are correct. If none of the answers are correct then there is a 0% chance of picking the correct answer. However, if there is a 0% chance of picking the correct answer then C is correct. We have a contradiction à la the Liar paradox, which would be a conundrum. Although earlier I said that this entails that there isn't a correct answer, I wonder if @StreetlightX had it right in saying that there is a correct answer (which would be 0%), albeit an answer to the meta-question and not the object question.

That does not follow. Green and blue are both non-red so you add their probability together. Hey you don't have to believe me as this is basic math, just look it up. — Jeremiah

We have to look into this further. There are only two possibilities.

Scenario 1. A)25% is the same as D)25%. If this is so then we have only 3 choices. 25%, 50% and 60%. The probability of choosing correctly is 1/3 = 33.33%

Scenario 2. A)25% is not the same as D)25%. Then there are 4 choices. That gives a probability of guessing right as 1/4 = 25%

As you can see I've exhausted all logically possible worlds. It can't be scenario 2 because the whole ''paradox'' depends on A)25% being same as D)25%. Therefore the correct answer must be in scenario 1 i.e. 33.33%

Another way to look at it is...

There are 4 options (A, B, C, D) but only 3 possibilities (25%, 50%, 60%) as A and D are same.

Remind yourself that the numbers 25%, 50% and 60% have no relevance in the calculation of the probability. They're just there to confuse you.

We can easily replace them with 2 red balls, 1 blue and 1 black.

One of these balls is the correct answer. What is the probability that, if you choose at random, you'll guess the correct answer?

We have to look at possibilities here and not options as you know choosing either A or D is the same thing. Same things in math get counted only once.

That means there are 3 possibilities - red, blue and black. The probability of guessing the right answer is 1/3 = 33.33%

We can easily replace them with 2 red balls, 1 blue and 1 black.

One of these balls is the correct answer. What is the probability that, if you choose at random, you'll guess the correct answer?

We have to look at possibilities here and not options as you know choosing either A or D is the same thing. Same things in math get counted only once.

That means there are 3 possibilities - red, blue and black. The probability of guessing the right answer is 1/3 = 33.33% — TheMadFool

If you pick a ball at random from a bag containing 2 red, 1 blue, and 1 black ball then there's a 50% chance that the ball is red, a 25% chance that the ball is blue, and a 25% chance that the ball is black.

But your post does highlight an ambiguity in the question. When we're asked to pick an answer at random, are we picking a letter (A, B, C, or D) at random – in which case there's a 1/4 chance of being picked – or are we picking a value (25%, 50%, or 60%) at random – in which case there's a 1/3 chance of being picked. Either method is viable without further clarification.

Although the solution in either case is the same; the correct answer (0%) isn't provided as an option.

The probability of choosing correctly is 1/3 = 33.33% — TheMadFool

Only if one of the answers is correct. But for the answer to be correct the probability of choosing it must equal its value. Given that 33.33% isn't an option, none of the answers are correct.

33% is not a valid answer no matter how you slice it. It completely ingores the distribution. I am sorry but that is just incorrect math, there is no point to aruge on that front because it is just wrong.

The random event is for four slots, which means 25 is weighted more than the others. It cannot have an equal chance as the others no matter how you scale it.

The people saying 33% are demonstrating a fundamental misunderstanding of basic probability.

33% is not a valid answer no matter how you slice it. It completely ingores the distribution. I am sorry but that is just incorrect math, there is no point to aruge on that front because it is just wrong.

The random event is for four slots, which means 25 is weighted more than the others. It cannot have an equal chance as the others no matter how you scale it. — Jeremiah

You are playing a game of rock paper scissors. Your opponent selects scissors. If you select one of the below at random, what is the chance that you will win?

A. Rock
B. Paper
C. Scissors
D. Paper

You say that because there are 4 options I only have a 25% chance of winning (if I pick A). I say that I will ignore the letters and just pick one of the three possible actions (rock, paper, or scissors) at random. I have a 1/3 chance of winning.

The same principle goes for the question in the OP. It isn't specified that I have to pick A, B, C, or D at random. Just "an answer". I might interpret that as to mean "unique value", and so pick one of 25%, 50%, or 60% at random.

The people saying 33% are demonstrating a fundamental misunderstanding of basic probability. — Jeremiah

I'd like to know how I misunderstand. Thanks.

My simple understanding of the matter:

There are 3 possible answers 25%, 50% and 60%.

There are 4 options A(25%), B(50%), C(60%) and D(25%).

If I were to make a random selection then...
It could be A)25% with a 25% chance
It could be B)50% with a 25% chance
It could be C)60% with a 25% chance
It could be D)25% with a 25% chance

We need to go a little further because we have two options that express the same possibility viz. A and B.

Scenario 1. If the correct answer is 25% then we have a 50% chance of getting the answer right. 2/4=50%

Scenario 2. If the correct answer is 50% or 60% then we have a 25% chance of getting the right answer.

We don't have a single answer like that if all options were different (I think we all agree that 25% would be the answer).

So, it depends on which is the correct answer.

Would the 33% geniuses care to explain how 25% goes from having a greater probability to having an equal probability; from having one to one odds then going to a one to two odds?

Spin it however you want, but the math doesn't lie. 25% has a greater probability and it cannot just magically go to having an equal probability. If you try and shove it into three then you are not accounting for the second 25%, which is what gives it a greater likelihood.

I sure would love to gamble with some of you.

33% is not a valid answer no matter how you slice it. It completely ingores the distribution. — Jeremiah

Not so. You just refuse to slice it the way some others are.

The people saying 33% are demonstrating a fundamental misunderstanding of basic probability. — Jeremiah

This misunderstanding comes from an imprecisely wording of the problem, and you're assuming everybody interprets the problem the way you are.

So back to the red, yellow, and two blue balls example. I have one of those colors in mind, and I ask the odds of you guessing correctly. The odds are 25%, 33%, or 50% depending on the choice and the interpretation. 25 and 50 come from blindly reaching into the bag and selecting a ball. 33% comes from looking at the available colors, randomly selecting one of the valid options, and then choosing a ball that matches that choice.

Similarly, your original problem had three unique answers to choose from, and there is a 33% chance of each of those if I choose from them randomly. But there is a 25% of each of A,B,C,D and I don't need to look at the answers to choose those randomly.

Am I choosing randomly from the 4 bullets, or from the three unique answers? Never mind that in both interpretations, all the answers provided are wrong.

Since some of you seem to be struggling with this. Also see my profile pic if that's easier.

First Glance Schematization
Answer Choices Chance Correct Chance Selected Chance Correct & Selected
AC 1 33% 50% 17%
AC 2 33% 25% 8%
AC 3 33% 25% 8%
Total (also the correct answer) 33%


Revision, Seeing 33% Not Listed
Answer Choices Chance Correct Chance Selected Chance Correct & Selected
AC 1 0% 50% 0%
AC 2 0% 25% 0%
AC 3 0% 25% 0%
Total (also the correct answer) 0%


Revision, Seeing 0% Not Listed
Answer Choices Chance Correct Chance Selected Chance Correct & Selected
AC 1 0% 50% 0%
AC 2 0% 25% 0%
AC 3 0% 25% 0%
Total (also the correct answer) 0%

What is your username?

A: Jeremiah
B: Zarathustra
C: Epicurus
D: Jeremiah

This is where 33% chance of being right comes from (Michael explains this in detail).

If we randomly select a letter, 50% of the time we will choose A or D, and 25%-25% for B and C respectively. BUT, the odds of choosing one of four letters are not the same as choosing the response that happens to be correct because if A is correct then D is also correct, and vice versa, so we can actually eliminate D entirely from our list of possible unique outcomes.

You could put 1000 additional selections, all the name Jeremiah, but if to us there is an equal possibility of you being named Zarathustra as there is of you being named Jeremiah, then it doesn't matter. The odds of making random selections between duplicated options is not the same as the possibility of unique options being true in the end. Even if most of the time we end up with the name Jeremiah as a selection, if one out of three times that turns out to be correct then we will still be correct 33.3% of the time.

Like the above, normally the correct answer to a question does not recursively alter itself when you actually make a selection (your question uses our selection to revise the criteria, which thwarts attempts to solve it) so this is why the correct answer to your question is actually 0% (because the odds of selecting the correct answer are 0%, as 33.3% (which would not recursively alter itself) is not an option). It's a circular-false trilemma (a false dilemma with an unmentioned fourth option).

Alternatively I could just pen in an E) and assign it anything but 25%, and then A or D become correct answers. (If option E was also 25%, since it would not constitute a new unique option, the odds of guessing correctly would still be 33%, which isn't an option, making the real answer 0%).

The fun part of your question comes only from its recursive or self-referential nature, the doubled multiple choice is a separate issue entirely. Here's your question with that bit removed:

If you randomly select an answer to the following multiple choice question, what are the odds of selecting correctly?

A: 0/1
B: 1/3
C: 1/1

A recursively refutes itself. If we have 0% chance of guessing correctly because the correct option is not available, then if we select 0%, we're correct, therefore our chance of being correct must be greater than 0%, making us incorrect again. Much of the fun of your example is illustrated by option A.

B seems very enticing. If we assume one of the three answers is correct then B seems to hold with theoretical odds, so let's eliminate it as a possible option to induce more enjoyable confusion:

If you randomly select an answer to the following multiple choice question, what are the odds of selecting correctly?

A: 0%
B: 100%

The funny thing about A is that it can simultaneously be correct and incorrect (it paradoxically flits back and fourth as we chart our way around it's loop).

So if we select A, in a sense, we would be correct (because the correct answer of 50% is missing).

If we select B: 100%, it could actually also be called correct, because the only other option is A, which is technically correct, and so if B is also correct then B recursively verifies itself as correct because both A and B are technically correct. Both answers are simultaneously correct and incorrect because their criterion for being correct are self-contained and circular.

If to determine the current position of a coin (heads or tails) we had to flip it over, by doing so we change the truth of it's current position, rendering our result incorrect.

I think this better captures the peculiar nature of your original example. Perhaps another interlocutor could explain this with greater clarity.

This is our question:

Multiple Choice: 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%

This is our sample space: 25, 50, 60, 25

One of those will be choose at random.

OK, following along now? Everyone got that much?

Now using R I took 10,000 random samples of that sample space and put that into a bar graph.

Here is the graph:

https://ibb.co/iaE5rT

Here is the code:

x <- c("25", "50", "60", "25")
x2 <- sample(x, 10000, replace = TRUE, prob = NULL)
library(ggplot2)
x2 <- as.data.frame(x2)
ggplot(x2, aes(x=x2)) + geom_bar(fill="blue") + xlab('Values') + ylab('Count')


Now I'll do same with some of your examples:

Michael said rock, paper, scissors, let's try that:

The sample space:

A. Rock
B. Paper
C. Scissors
D. Paper

https://ibb.co/ifBtBT

TheMadFool,

Sample space:

2 red balls, 1 blue and 1 black.

https://ibb.co/ceOC5o

Do I need to keep going?

Where is the equal 33% between the three choices? 10,00 samples and it is not there. If it was an equal 1/3 chance the bars should be even, but they are not.

Only half paying attention here, but it looks like you’re confusing the probability of selecting a unique answer choice with the probability of selecting a correct answer choice. We’re looking at the compound event.

It is more than clear that you are "only half paying attention".

Just because you decided there were three choices that does not mean they have the same likelihood of occurring.

They don’t. I didn’t see anyone say that. Answer choice ‘25%’ has a 50% chance , obviously. We noticed that.

The only way you could have a 33% chance is if each outcome has an equal chance of being selected. Clearly, after 10000 samples, they don't.