Mathematical Conundrum or Not?
Does this problem have a solution? Is it a paradox or other?
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%
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%
Comments (204)
If the correct answer is 50% or 60% then the correct answer is 25%
Therefore, the correct answer is 0% which is correct because it is not a choice on offer. If one could choose it, randomly or deliberately, it would be wrong. It's an amusing self-referential paradox, but I don't see anything special about it. It would have been more amusing, and neater, if one of the options (C) had been 0%
One expects there to be a right answer on offer, but there may not be.
2 + 2 = ?
A) 3
B) 203
C) 42
D) 8.73
Expect to get low marks in such tests, however hard you revise.
Quoting unenlightenedI agree, and the consistency of this makes it not a paradox. Choice C should probably have been 0% instead of 60%. Then it would be a paradox I think.
Thought of a variation of the original one:
If you choose an answer to this question at random, what is the chance you will be correct?
A) 25%
B) 50%
C) 100%
D) 50%
Each answer is correct (not necessarily in the same way), so are any of them really?
There are only 3 choices 25%, 50% and 60%.
Random selection would be a chance of 1/3=33.33%.
Guessing what correctly though? Guessing the answer yes, but... to what? The question... which is... about the odds of guessing correctly. It's a weird little circle.
The self-reference of the indexical 'this' hides the fact that there is no actual question to which a chance of answering corresponds.
The two 25%s are a trick, a distraction. The question is unanswerable from the very beginning.
Yes, circular, but that doesn't mean there's no actual question.
[Img]http://www.daviddarling.info/images/Penrose_stairway.gif[/img]
How many letters does the correct answer have?
There are four possible outcomes. That part never changes. The fact that two of the possible outcomes contain the same values does not change that there are four possible outcomes. We just increase our chances of being correct if the desired outcome is 25% since it is two of the four possible outcomes.
0% is not a possible outcome, which means there is a 0% chance of it being 0%.
The question concerns the answer to the question. That's a circle, or more formally, self-referential.
You ask about the odds of correctly guessing the answer giving the correct odds of correctly guessing the answer giving the correct odds of correctly guessing the answer giving the correct odds of correctly guessing the answer ....
Quoting JeremiahIf none of the choices is correct, then the correct answer is simply not among the choices. 0% is the answer since it cannot be chosen. This itself is not paradoxical. Hence my comment that (C) should have been 0%, not 60. That forces the paradox.
0% is not a possible outcome for the answer, therefore it cannot be chosen. Furthermore you are not wrong until after the chance event.
The idea that is it circular is just your perspective. There is a single chance event, one and only one. That is not a circle.
I considered the 0% version before making this thread, it really makes no difference at all as C is nothing but filler to give us four possible outcomes. The only reason to use the 0% version is to end moot arguments about how multiple choice questions work.
Also odds are not the same thing as chance.
There is only one purposed chance event. It does not loop back on itself, it is not a circle and, as long as you don't say C), then you are not wrong until after the chance event, only then can you be wrong.
You're wrong no matter what you choose. There is thus zero percent chance of choosing correctly, which is what the question asks. That the thing is self-referential doesn't seem to change that. It's not a paradox as written.
This too is unanswerable.
OK I'll try this one more time then that is it, as you don't seem to even comprehend what I am saying at all. Disagreeing is one thing, but not comprehending all together is quite another.
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. If you can't understand that simple concept then you are not worth anymore of my time.
And the only reason you saw this as a circle is the same reason humans tend to think of time as linear, that is how your brain is processing the information, but it is not a circle.
The main problem is that "this question" grammatically and conceptually must refer to the multiple choice group (without any given criterion pointing to "correctness") or else it is a circular question.
The correct solution is your response + 1. What is your response?: the self-referential or recursive nature of this question makes it impossible to answer correctly.
If we presume that the question is not self referential then it seems answerable. And it is 33%.
I like puzzles like these.
This is it.
I'm not so sure this works as a conundrum, since it is literally impossible to answer correctly, but not based on the problem itself, but because of the answers offered as a limited horizon.
A conundrum's difficulty is in it's riddle, in finding a satisfying answer despite difficult conditions.
In that case @TheMadFool and @VagabondSpectre are correct. :smile:
Here's why. This is not marble-picking. This is choosing a unique answer from among the choices of four. There are only 3 unique answers, as two are the same.
Actually, there are four choices A, B, C and D. I get what they are driving at but the chance event is for four slots, which mean if we just look at the three values then 25 is 1/2 and 50 is 1/4 and 60 is 1/4. Saying they each have a 33% chance is suggesting they all have the same probability of being chosen by the chance event, which is not true; 25 has double the chance of being selected than the other values.
Furthermore just as 0% is not a selectable option neither is 33%.
*read*
“Hm. Multiple choice random accuracy is 1/p for p choices, so the correct answer to this question is 1/p.”
“Oh, but 1/p shows up twice. More accurately, it’s 1/(degrees of freedom). If you know the full list of answers and you’ve picked 3, then you know what the 4th one is before you look at it, so it’s 1/(p-1)=1/3.”
“So the answer to this particular question is 1/3. But wait, no 1/3. This question’s correct answer is not listed, so there’s no chance of picking it. But wait, looks like 0% is actually the correct answer. Also not listed. The probability is still 0.”
“They tried to get me with the two 25%, tempting me to think the answer is 2*25%=50%, but that’s assuming the repeated answer choice doesn’t affect the odds. As an exercise in probability, it’s def 1/3, because we are picking at random and without assumptions about the meaning of the question. Strangely, I did have to know the meaning of the question in order to know the answer wasn’t listed, but now I’m just clarifying. My psychology does not affect the structure of the question.”
“If 0% were listed (once) alongside the two 25%, you would indeed go in circles. Finding no ‘1/3’, you revise to 0. Seeing 0 once, you revise again. Now thought processes divide. The meta view returns to 1/3, alternating with 0. The semantic view alternates between 1/2 and 1/4.”
“Does the fact that 25% is repeated change the odds? That’s double the risk of picking an answer that’s only 33% likely to be right. Well, no. The weighted average is still 33% because each hoice, being one of the three possible answers, has a 1/3 chance of winning.”
Very pleasant. Thank you.
Thanks for clarifying. What I think is...
When we calculate the probability and say we have 50% chance of it being 25% we are assuming A)25% is the same as B)25%. But when we say if the correct answer is 50% or 60% then the probability is 25% we're making a distinction between A)25% and D)25%.
Seems like the fallacy of equivocation to me.
An example might help to clarify...
Scenario 1: we have 4 balls: 1 red, 1 green, 1 blue and 1 pink. I've chosen one ball.
The probability of you guessing my ball is a uniform 1/4 = 25%.
Scenario 2: there are 4 balls: 2 red, 1 green and 1 blue.
I've chosen a ball. What is the probability that you'll guess correctly.
You'll say 50% chance if it's red and 25% chance if it's non-red. Notice however that when yoy say 50% chance if it's red you're saying the two red balls are same. However when you say 25% chance that if it's non-red you're making a distinction between the two red balls. This is a fallacy.
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.
Let me give a clear example of what I mean.
You have 5 beads 4 of them are blue and one of them is green. If one is selected at random then you have a 4/5 chance of getting a blue and a 1/5 chance of getting a green. You can't group all the blue beads into one bead and claim you have a 1/2 chance of getting a blue and a 1/2 chance of getting a green as then you are no longer accounting for the fact that there are 4 blue and only one green.
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.
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.
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.
Maybe you should thinking about it longer than 3 or 4 minutes.
I think it is one of them most helpful post I have ever made.
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.
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%
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.
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.
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.
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.
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.
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.
Quoting 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.
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:
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.
It is more than clear that you are "only half paying attention".
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.
I'm suggesting that this is a viable way to select an answer at random:
Which gives:
https://ibb.co/n0joBT
Michael's right that c) should be 0% to really make it a paradox through and through.
But Jeremiah is right that when you are given 4 options to choose AT RANDOM (your selection then can't be based on knowing that A and D are the same) the probability would never be 33%.
That being said, I'm pretty sure none of these misunderstandings mean you need to get snarky, Jeremiah. It's just math and logic.
What is the capital city of France?
A. Paris
B. Paris
C. Paris
D. Paris
You say there are 4 answers (A, B, C, and D). I say there is 1 answer (Paris) repeated three times.
Only if one of the answers actually is correct, but none of them are.
None of the answers provided in the original question are correct, so the chance of being right is 0%.
Yeah, thought we were all on board with that.
A, B, C, D is not the sample space I used in my code. You remember that?
My sample space was: 25, 50, 60, 25. We'll call this sample space 1.
You remember that? Or do I need to quote myself? It sure the heck was not A, B, C, D.
Now when you used the unique function you changed that sample space to: 25, 50, 60. We'll call this sample space 2.
So when I ran the sample command it sampled from sample space 1.
When you ran the sample command it sampled from sample space 2.
That is what happened, and it is in black and white.
Now the question of interest is: 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%
It has a sample space of: 25, 50, 60, 25. Now wait, that looks a lot like my sample space, sample space 1. While sample space 2, your sample space, is missing a value. Hmm... that can't be, as the random event for the question is for the sample space of the question. How could this possibly happen? Oh I know, you are doing it wrong.
Why don't you run the code correctly this time, with the correct sample space and stop trying to save face. I am sorry, but this is just beyond sad.
You presented this as a conundrum. By definition, conundrum are hard to answer satisfyingly, possibly impossible. You don't get to pretend you have a clearly obvious answer afterward.
We have at least a 1/4th chance of being right, and on top of that we have the scenarios of guessing A and the answer being D (making us correct) and the scenario of guessing D and the answer being A (also making us correct).
The odds of choosing A but the answer turning up D is (1/4*1/4) (in other words, 1/16 possible outcomes fits this win condition) and the same holds true for the scenario of guessing D and the answer being A
So the full odds are (1/4+1/16+1/16) which works out to (3/8) or 37.5%
If you don't believe me run the experiment yourself (it's Processing.js)
[hide][/hide]
Since 37.5% isn't a possible answer, 0% of the time the correct answer will be guessed.
Since the correct answers to questions are normally not determined by random dice rolls which include redundant options (weighting the dice) we would normally calculate 33% as the outcome because we don't ascribe causal value to the existence of redundant options in multiple choice questions. (we also assume one of the answers is correct)
I have the same attitude everywhere, it has nothing to do with "here".
@VagabondSpectre I hadn't considered how to structure the iterative game. That's an interesting take.
Do you even have an answer for me?
Blithering about us not getting it isn't helpful. As I've just shown mathematically and proven experimentally (with your interpretation of sampling), the odds of guessing the correct answer to that question are 37.5%. That's the product of your "conundrum". If we assume that the question is self-referential (you seem to maintain that it is not) then it becomes unanswerable because 37.5% (or 33% under a sensical interpretation of what multiple choice questions are) is not an option.
do you have a different solution in mind?
Are you able to explain the understanding that we lack instead of just telling us we're stupid?
Imagine selecting option D on a test but your teacher/professor arguing that A was the correct choice; since they're the same they would have no choice but to mark it correct. If you made the 1/4 guess you would be right 37.5 percent of the time.
I have explaind it well enough, and if you cannot grasps it by now then that is your problem not mine.
It's evident you have not explained it well at all. By continuously saying this instead of responding properly you make it amply clear.
It's not us, it's your inability to explain.
You make me glad I am me.
Good luck with your conundrum.
A: wise
B: foolish
C: foolish
D: foolish
E: foolish
F: foolish
G: foolish
H: foolish
I: foolish
I got your goat, didn't I.
If by "got my goat" you mean to say you've frightened it and every other goat away with your jackassery, then yes.
If you are trying to shame me, it won't work.
Your answer to this question will go far toward clearing things up.
I’m thinking of a number. It’s either 1, 2 or 3. If you randomly choose a number from the list below, what are the odds that it’ll be the one I have in mind?
1, 2, 3, 1
If your answer is not 33%, I must question your mathematical literacy. If it is 33%, then we’re not arguing about the same thing.
You really live up to your name.
There he is. Good timing, was just checking one last time. I am a fool, but I know intelligence when I see it. I will keep my thoughts about you to myself. Good day, sir.
Ya, I don't really care what you think about me.
Person A ask : "Conundrum : What can swallow a man, and be swallowed by him"?
Person B answers : euh, I don't know, a piranah?
Person A replies : You fool, the only answer is "pride".
Person B object : yeah, I mean, classically perhaps, but you asked me a question, and I gave you an answer that fitted the parameters.
Person A : What are you, defective?
repeat
Actually, the thread has been very useful to me. I learned a lot about the people participating in it.
How many cities are in this list?
A. Paris
B. Paris
C. Paris
D. London
4 or 2? I say 2. So if I'm asked to pick a city at random from that list, I think it reasonable to flip a coin and select Paris if it's heads or London if it's tails. The question is ambiguous as to the sample space.
The question is not how many are in the list. You people are confusing yoursleves because you all keep changing the parameters of the question with your nonsense examples. It is the dumbest thing I have ever seen.
If we took a random sample from your sample space it would be Paris 75% of the time and London 25%. If I was a betting man I'd bet on Paris.
You are not paying attention to the sample space that the actual random sample is taken from. It is not a 50/50 chance. They would have to be equally distributed for it to be 50. So your chance of being correct if you choose Paris is 75%. As the question in the OP is not how many are in the list, it ask what are your chances of being correct when one is selected at random.
Stop changing the question, stop changing the sample space and learn how to read.
I'm not changing the sample space. I'm saying that there is room to disagree over what is the sample space. You say that the sample space is the rows in the list (of which there are 4); I say that the sample space is the cities (of which there are 2).
I think you just can't admit that you were wrong.
One what? One answer? But what counts as an answer? If I were to ask you for the capital city of France, and then list A) Paris, B) Paris, C) Paris, and D) London, you might say that there are 4 answers (A, B, C, or D) but I'm saying that there are only 2 answers (Paris or London).
Repetitions aren't different answers. They're all one answer. Paris repeated 99 times is still just one answer, not 100 answers.
When figuring probability repeated values are very important, if you remove them then you will misrepresent the distribution. Repetition is not a valid reason to remove a datum. What you are doing is changing the sample space and it should be a clue to you, that in order for your numbers to come out that you have to change the data.
Oi oi, look in the 'math and motives' thread and see what you think.
It is in some contexts.
Not in this context.
That's up to interpretation.
I clearly disagree with that.
I do this.
No, I'm disagreeing over what is the sample space.
There are two cities in this list, not four:
A. Paris
B. Paris
C. Paris
D. London
If I'm asked what the capital city of France is, and given the list above as choices, I only have two possible answers (Paris or London), not four. A, B, and C are the same answer. So if I'm asked to pick an answer at random, and if there are only two answers (which there are: Paris or London) then I have a 50% chance of picking Paris.
You were not asked what the capital of France is. That is an entirely different question. You changed the question.
I think there have been too many redefined examples already and I don't think they are helping at all. In fact I think that's what is mixing you up, I would suggest you focus on the actual question. Furthermore I already gave you my comments on this example of yours, they have not changed and this post did not convince me otherwise, so I don't see any reason to repeat myself, and I'll refer you to my prior post.
The principle behind there being 2 answers, not 4, is the same. In your question there are 3 answers, not 4. So if I'm asked to pick an answer at random my sample space has 3 elements, not 4. The actual question is irrelevant, so your responses are just deflections.
I have understood what you are doing since TheMadFool made his first post, it was incorrect then and it incorrect now.
And if you are so right, then whey do you need an alternative example to prove it? What is wrong with using the question in the OP? I think your examples is you deflecting.
Because you don't seem to understand it with your question, so I'm providing an alternative that should make the reasoning clearer. There are only two answers to the question, even though Paris appears three times, so an answer picked at random has a 50% chance of being Paris. There are only three answers to your question, even though 25% appears twice, so an answer picked at random has a 33.33% chance of being 25%.
This is not your standard multiple choice question we are dealing with here. Most multiple choice questions do not involve a random sample.
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.
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.
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.
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.
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.
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.
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.
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.
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.)
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.
In the sense that an obstacle, once removed, makes the road seems so much easier.
I'll keep my thoughts to myself.
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?
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.
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.
It's exactly the same as everyone else here. The question asks for the selection of an "answer" at random, there are three possible "answers" (25%, 50% and 60%) one of which, it is implied by the format, is correct. The fact that I've been provided with two different ways of indicating to the questioner that I've chosen 25% from the three available options, is odd, but irrelevant to this interpretation.
I'll ask again, what would define a valid reason? Everyone here has given you what they believe to be a reason. I can guarantee you that this forum is made up at least from some of your epistemic peers (other people of your intellect) . If they're wrong about their reasons being valid then it follows that it must be possible for someone of your intellect to be wrong about this issue. So how do you know it isn't you?
That has observational bias all over it. That is the difference that you are not seeing, I am not reinterpreting the question.
No one's "re-interpreting" the question more than any other. We're all interpreting it one way on our first pass, maybe seeing another interpretation on second reading. What makes your first interpretation more right than anyone else's first, second or third attempt?
As I said, your argument "my argument is sound... because it is", is not a style of logical analysis I'm familiar with.
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%
Answer: ...
You have to write on the dotted line the answer. How many different things can we write on the dotted line? I say just three: 25%, 50%, or 60%. If there are only three possible answers then my sample space for the random selection has only three elements.
I already know your argument, Michael. Unless you are repeating yourself for someone else?
My position has not changed, despite your renewed effort.
Given a multiple choice test with a fixed format, say, every question having four possible answers, there are two ways to choose randomly:
By design these are equivalent on multiple choice tests because answers are never deliberately duplicated. If use method (2), it won't even matter to you that this question directs you to choose randomly -- that's what you're doing already. People tend to use method (1) in part so they can throw out "(a) a fish" as a possible answer to "What is 6 x 7?" and then randomly select among the other three.
I claim all this is relevant because you rely on our expectations about how multiple choice tests work but then screw with them. It is literally a trick question, just not the sort that's typical on these tests.
Suppose this was a real test, and the duplication was a mistake. Then the instructor would discover that this question is broken. Admittedly it's broken in a really unusual way, but the result is that no student can give a correct answer. As far as that goes, it's no different from a typo in which the correct answer was supposed to be "(d) 33%" but was printed as "(d) 55%".
And again, all of this is lost on the student who just randomly bubbled in A or B or C or D without even reading the broken question. To him, the question is just
Blah blah blah
A something
B something
C something
D something
Do you have a proof that he is Doing It Wrong™?
The second 25% is clearly intentional and not a typo.
The second 25% is obviously a typo that leaves no correct answer as an option.
To the student who just bubbles away, there's no difference between a broken question and a question he just didn't luckily answer correctly.
To the test designer, a duplicated answer is always a mistake that can only help such students.
This is not a test and clearly the second 25% is meant to be there.
Can we say Occam's Razor?
You might say the two-event analysis looks more baroque, but it’s also more adequate.
By marking all answers as wrong?
Yeah. I didn’t say it would be fair. It would be coherent. The test also wouldn’t have any pedagogical use, but that’s a practical problem.
That's the definition of a broken question.
In that case, broken questions can still be coherent.
But not answered correctly. And the question is about answering correctly, so it's not like I'm forcing my own preconceptions on it.
I see two possibilities:
But we're not asked what the chance is that we've succeeded in answering; we're asked what the chance is that we've answered correctly.
I mean, I agree that it’s a defective prompt in the pedagogical sense. I would protest as a student. But I’m gonna carry on with the analysis.
Btw, I accidentally clicked smthg on your post when I went to reply. Dunno what it was or what it did. If I somehow flagged you or smthg, I am profusely sorry.
Nobody got flagged. Just checked.
Thank you, sir/madam.
See, there could be a point to that. Suppose it were done this way:
16. ...
17. Do not select an answer to this question.
18. ...
Then the instructor could assess a heavy penalty on anyone who bubbled in any answer to (17). It would be a check on students answering randomly.
That is interesting.
I see you creating an entire narrative, a fantasy world, to try and facilitate your position. It is on a test, it is being graded, there is an instructor, there are students, etc. . .
Not only are you making a ton of assumptions about the nature of all these fictional story elements, but you are taking the OP out of context.
What is see is people doing what people tend to do: Unnecessarily over complicate things. It is why the Law of Parsimony exist, to help mitigate such observational bias.
I would call that more of a guideline, second to the guideline that your theory should be empirically adequate. That said, I don’t consider this a question of theory choice. It’s a straightforward exercise in chance with a self-referential but non-circular twist. I quite enjoyed it.
By the way, I appreciate your dialing down the vitriol. I am happy to debate this as friends.
Some want an answer, even though the question, which does have an answer, can't be answered. That is where the 33% comes in, they need it to have a best answer. It is human nature, to try and put everything in its place even if it has no place; we are the diligent organizers of an endless chaos.
However, to get to those points they must redefine the question; they must change the question. As if they are trying to avoid the original question, because it messes with their human nature.
The question as it is in the OP is the best version if it, as it is nonsensical chaos on more levels than one, which is the point. It is a reminder that these systems we have built, like mathematics, are not perfect and they only work up to a point. It is a reminder that for all our great insights, our understanding of even our own concepts is incomplete.
Here are two more versions.
1. 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%
Only when you go to the answer sheet, you see (I'm putting the letters where there would be bubbles in labeled columns)
1. A. B. C
Wait, so is D part of the sample space or not, since it turns out I can't "choose" it?
2. 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)
Now what? There's "D", but can I answer "D"? If I assert that D is the answer, what would I be asserting?
Which brings me to this point: a question and answer pair should be recastible as an argument or at least an assertion. If it can't, then it's either inconsistent or a contradiction.
I get the point you're making about mathematics, I think. If I drag in some context, it's too show that mathematics doesn't stand naked on its own, but relies on a broader conception of rationality. That conception can be seen at work in the way we talk to each other, and in the way we make tests. It's those conventions your question violates, and that's why it's a conundrum.
I actually like the puzzle, else I wouldn't spend my time trying to figure out how it works and why it resists solution.
No it isn't. The answer to the version in the OP is 0%, which just isn't offered as an option. You might as well ask:
Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct?
A) 1%
B) 2%
C) 3%
D) 4%
Or:
What is the capital city of France?
A) London
B) Washington D.C.
C) Berlin
D) Ottawa
The paradox works best when 0% is offered as an option.
If 0% was in the OP we would have lost that entire part of the discussion.
What is the right answer to this question?
Is it...
A) b
B) c
C) d
D) a
If you revise the parameters to say the teacher picks one and only one of A or D to be true regardless of the content of the answer, then I will accept your analysis. Then the answer is simply 25% bc, even though it’s listed twice, it doesn’t matter what the answer choices actually say. We can observe that 50% of the students get the semantically correct answer, but we don’t go in circles because we stipulated that the criterion of correctness has nothing to do with the actual question, so the actual frequency of correct answers is irrelevant to the teacher’s designated answer choice. I appreciate the conundrum that, on some interpretations, the meaning of the question is part of the format of the question.
Quoting Jeremiah
Sorry if I'm still trying to understand this problem, you seem to have moved on to another thread.
I think (as other participants here are seeing also), the "chance of being correct " in the first quote above, and the "probability of being chosen [by chance event]" imply two different commands. I say "imply" although it is clearly obvious. In the second quote from your post, you are referring to 'what's a chance that each choice will be randomly selected. This is a good question if you're doing it in a blind selection. But the choices are out there listed as multiple choice. So, the question as I see it is, how many times a 25% occurs? That's 50% of the 4 choices. (As you correctly identified).
Then, there is a second question -- what is the chance you will be correct?
So, let's say 25% is the correct answer -- and it occurs twice in the multiple choice, giving it 50% occurrence rate, does it also mean that the chance of us being correct is 50%?
I'm choosing from 3 unique answers, not 4.
My question is really for this:
Quoting Fool
This is what's at stake.
I think that formulation is incorrect, because if this truth condition yields "true" for more than one value, the chance to be correct overall is greater than for any of the individual values.
Take for example:
a) 25 %
b) 50 %
c) 50 %
d) 60 %
A has a value with a probability of 25 % to be chosen, so it's correct. B and C both have both have a value that has a chance of 50 % to be chosen, so they're correct, too. But that would render the chance to be correct overall at 75 % and according to the problem's formulation, none of them would be correct. But if none of them is correct, then the way we arrived at the correctness of the individual values isn't valid, as it doesn't address the problem.
I don't even know how to formulate this problem in mathematical terms. I don't understand the truth condition.
Good point.
True.
Quoting Jeremiah
Quoting Jeremiah
This popped up this morning:
https://www.gocomics.com/fminus/2018/05/23
Thought it expressed my takeaway on this topic.
And I am sure that was your only take away.