This is not your standard multiple choice question we are dealing with here. — Jeremiah

The question is irrelevant. We're discussing the logic of picking an answer at random. That logic doesn't depend at all on the question. A random selection is just a random selection.

I don't care about Paris, Michael, it is you trying to stack the deck.

I don't care about Paris, Michael, it is you trying to stack the deck. — Jeremiah

Whether or not you care is irrelevant.

You seem to want to have this conversation with me, so I would say it is very relevant.

You seem to want to have this conversation with me, so I would say it is very relevant. — Jeremiah

This makes no sense. Whether or not you care about Paris has no bearing on whether or not there are 2 or 4 answers to the question. There are only 2 answers, despite the repetitions. There are only 3 answers to your question, despite the repetitions. And as I'm asked to pick an answer at random, my sample space has 3 elements.

I don't think it is unreasonable of me to ask that we focus on the question in the OP and your aversion to do so is very suspicious. I also don't think these rapid fire replies do anything at all, I know from experience that they tend to lead nowhere.

Now, Michael, I have already said my part on this, and I do fully understand what you are saying, so unless you are going to bring a new argument to the table, then that I think this is it. I will continue to think you are just flat out wrong, that part will not change but unless you can provide some new perspective here, then there is no point for me to continue in this. The only thing that will happen is at some point I will insult you and I'll end up getting more post deleted and more messages in my inbox from mods.

There are two random events: the random selection of the actual correct answer from the list of 3 unique possibilities (ignoring that the right answer isn’t actually in there), and the random selection of the answer choice from a list of 4.

If you took 10000 questions of the same form, disregarding what they actually ask, this is how you would analyze them. You would predict random accuracy of 33%.

Jeremiah, if you really want to prove us wrong, set up a 10k question test and show us that it’s not 33%. Now, I don’t mean run 10k samples from an array of 4 elements. That’s only half the exercise. I mean 10k questions with 3 possible answers and 4 choices with 1 answer repeated. I did it already, results in my profile image.

I'm not sure if I've got the logic right, but I get a chance of 37.5%.

http://sandbox.onlinephpfunctions.com/code/ab2fb5e6371a618abe604072671d8a6c49878e21

If you don't know PHP, I've made a list of the OP's options and then run a million tests where I first pick a random element to be the "correct" answer and then a random element to be my guess. If they're the same then I add to the number of successes.

Strange, that’s exactly what I did, but I get a random variable centering on 33%. The logic must be different, but I don’t know PHP. I can work with R if you’ve got an equivalent. It sounded to me that your logic treats the two identical options somewhat differently. In your example of the classroom debate over the right answer, the teacher doesn’t acknowledge that A and D are the same, else there would be no debate. In any case, it sounded like your logic accounted for external factors, whereas my contention is purely about framing the random events. I will review your comments later today in case I missed something.

Javascript then?

https://jsfiddle.net/fw7tp0n8/

I don't know R, but if you show me your code I might be able to figure out what you're doing (as I did with Jeremiah's earlier).

Here it is in R:

Ah, that's the difference, thanks. Sorry, wasn't going to chime in until I could re-read the whole thread, but this settles it.

Here's what I'm saying:

possible_choice <- c("25", "50", "60", "25")
possible_correct_answer <- c("25", "50", "60")
s <- 0

for (i in 1:1000000)
{

if (sample(possible_choice, 1) == sample(possible_correct_answer, 1))
{
s = s + 1
}

}

print(s / 10000)


Looks like you're correctly weighting the selection from the answer choices A-D, but I would differ with you about how you're weighting the probability of each answer choice being the correct one. I'd argue the frequency of the answer's appearance in the multiple choice list should not affect the chance of the answer being correct. My profile pic now has our two scenarios compared.

I'd argue the frequency of the answer's appearance in the multiple choice list should not affect the chance of the answer being correct. — Fool

Ah, yes. That's a good point. Now I'm getting the 33%.

So we agree that we're dealing with two separate chance events. We've also precisely clarified the difference in our logic. Your scenario makes sense to me. If the teacher just picks one of A-D to be correct without considering that A and D are equally justified, then the chance that the content of the teacher's designated answer choice (A-D) is the semantically "correct" answer would need to be weighted by the frequency of its appearance in the list.

Also curious how mature and constructive this conversation becomes in the absence of some contributors.

Why don't multiple choice questions ever have repeated answers? If you were taking a math test and got to a question that had a repeated answer, you'd walk up to the teacher's desk and show him, and he'd make an announcement like, "Sorry, folks, Question 17 has a typo. Answer E should be '33%'."

Presented with a multiple-choice question, there are several methods you can use to choose an answer. Best method is knowing which answer is correct -- a word which here means "will be graded by the test preparer as correct". If you don't know, or don't think you know, you can go with your gut or choose randomly, and you can also eliminate answers you know are wrong before doing either of those to improve your (subjective) chances. There are some other methods, but the main point is that random choice is a fall-back when no better option is available.

Multiple choice tests are designed to test knowledge or reasoning, not luck. Tests are often designed with answers that will appear tempting if your reasoning is faulty, but not to foil a test taker who is lucky. That test takers have the option of making random choices is interesting, but test design needn't take this into account, and it's not perfectly clear that it can.

You could in fact help the random chooser by repeating correct answers, and people do this sort of thing for comic effect. What are the three most important principles of retail? Location, location, location. There are only two rules for working here: 1 is "Do what I tell you" and 2 is "Do what I tell you".

So here's the answer to the question at the top of this post: multiple choice questions never (deliberately) have repeated answers because a duplicated incorrect answer will not foil the random chooser, while a duplicated correct answer will help him. And the test preparer has no reason to help the random chooser.

The question that remains is whether a duplicated incorrect answer also helps the random chooser by changing his chance of getting the right answer from, say, 1 in 4 to 1 in 3. That would seem to depend entirely on the test taker -- that is, on whether he reads the question at all or just bubbles in something on the answer sheet.

And that opens up the possibility of a gap -- noted by several people in this thread -- between the chance of my picking the answer that is correct, on the one hand, and the chance of the answer I pick being correct. In the usual case, with no duplicated answers, these are identical, by design. But if there are duplicated wrong answers, will the random chooser who reads the questions out-perform the random chooser who doesn't?

If I did the simulation right, I get 16.6162 for the test taker who reads the questions, and 12.4914 for the test taker who doesn't.

The question that remains is whether a duplicated incorrect answer also helps the random chooser by changing his chance of getting the right answer from, say, 1 in 4 to 1 in 3. — Srap Tasmaner

This reminds me of the Monty Hall problem.

Dude, I didn't even think of that!

Not sure if I did the simulation right, and I'm at work now. :-(

(I had right answer being chosen from {b, c} and two students: one chooses from {a, b, c, a} and one from {a, b, c}. I don't understand the result though, so must've muffed it.)

Also, I got something wrong about the duplicated correct answer: that will help the student who doesn't read the questions even more than it helps the one who does. So whether the duplication helps the reader or non-reader more flips depending on whether it's a right duplicated or a wrong. Cool.

In any case, it still seems that duplication can only help random choosing, even if it helps in a differentiated way.

What I wanted to get back to eventually was how this puzzle conflicts with our expectations about how tests work...

(Pointless anecdote: I had a professor in college who didn't give multiple choice questions because he said they helped poor students and hurt good ones.)

My values are halves of the 1 in 3, and 1 in 4 totals, duh. The only reason to look at it by cases as I was -- and I should have done the other two -- is to see if there's any point in a test designer choosing one of these options, and I don't see any such reason, or any reason to follow some other strategy like randomly duplicating, etc.

Perhaps just to cause confusion as an additional layer if rigor.

In high school one my teachers set all the answers to a multiple choice test as option B (a statistics test, har har :( ) and so the class spent the hour in frustration, assuming every question was a trick and second guessing ourselves. Kind of stupid of him now that I think about it...

So it would be weird if I felt a deep sense of kinship with this man ...

Anywho, this sort of gamesmanship is practically built into the multiple choice test and it would be easy to (ahem) multiply examples.

None of these strategies look like a counter to the random chooser though. (The "all B" sequence messes with our faulty intuitions.)

All of which points to something weird in Jeremiah's puzzle.

The conversation would have never got this far without my input. Don't like me fine, but no one has actually proven me wrong and my argument is a very valid take. I am an ass, yes, but I am also smart and you'd be fooling yourself thinking otherwise.

Notice that I never called you unintelligent, and I actually thanked you for your entertaining post.

The conversation would have never got this far without my input. — Jeremiah

In the sense that an obstacle, once removed, makes the road seems so much easier.

Feel free to actually address my argument, as far as I can tell you only joined this conversation to poke at me.

I'll keep my thoughts to myself.

no one has actually proven me wrong — Jeremiah

Wrong in what? So far you have made the following two assertions - The question should be interpreted as having a sample space of four variables with a single random choice event and a single random choice event in a sample space of four variables produces the vacillating probabilities you claim (25%, 50%).

You've invited people to prove you wrong by showing you a chance other than 25% from a sample space of four variables with a single random choice event. As no-one has done so, that means no-one has proven your second assertion wrong.

So what would constitute a proof that your first assertion is wrong, what kind of proof would we need to present, and what standards would be assessing that proof by to see if it held?

I actually made several points, but the one that got everyone hopping is that with the sample space in the OP you get a distribution which favors 25%. The entire conversation is recorded if you wish to review it and if I thought it was incorrect then I won't assert it as correct, would I?

So what would constitute a proof that your first assertion is wrong, what kind of proof would we need to present, and what standards would be assessing that proof by to see if it held? — Pseudonym

A simple example to the contrary suffices in proving wrong an assertion that all A is B, or in this case, the only valid interpretation (A) is one of a sample space of 4 (B). Many of us have produced that alternate interpretation of a sample space of three (~B).
That leaves it up to Jerimiah to demonstrate that either this sample space is not 3 in number, or that one has no way of randomly selecting from this sample space. The latter would be the case if the answers were hidden, but the OP concerned an open multiple choice question, not drawing of hidden names from a bag.

Taking things out of context, doesn't prove you are right, it just proves that you can't read.

Yes, that's what I thought, but it sounded very much as if @Jeremiah was still looking for some other sort of refutation that I couldn't think of. I thought I'd ask. It seems from the reply above that his first assertion is supported by the evidence that... he asserted it. Not a type of logical proof I'm familiar with.