Greg posted this question

The answer seemed obvious to me but I may be missing some deeper point. My answer under the fold.

.25*.5+.25*.25+.25*.25+.25*.5 = .375 = 37.5%

If I had to guess it seems to me that the confusion results from two sources

1) There are percentage signs in the answers

2) The fact that 50% of the answers are “25%” and 25% of the answers are “50%”

However what if the question were reworded:

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

A) 250

B) 500

C) 600

D) 250

Here the answer is clearly 37.5% and the “lesson” is that there being three possible right answers does not make your chance of picking the right answer at random 33%, because one of the potential right answers has two “balls” associated with it.

However, the question does not say that the answer must among the listed options. It only asks what is the probability that you will be correct if you answer at random. If there were no percentages signs, no one would assume that the question was self-referential.

Why assume it just because there are?

**UPDATE:** I still think part of the “cuteness” of this problem is that it is not explicitly self-referential but my commenters are right that they do not specify the set to which the “correct” answer belongs. I simply assumed it was {“25%”, “50%”, “60%}

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Wednesday ~ November 2nd, 2011 at 6:01 am

foosionIt’s clearly meant as a self-referential question, or at least a question that requires consideration of that interpretation.

The odds of a random answer are 25%, but there’s a 50% chance of picking 25%, so 25% isn’t correct. There’s a 25% chance of picking 50%, so that isn’t right either.

Wednesday ~ November 2nd, 2011 at 6:03 am

foosionBTW “Why assume it just because there are?”

Because there are. To conclude otherwise would require the question to be reworded, as you appear to acknowledge.

Wednesday ~ November 2nd, 2011 at 6:24 am

marco1) “If you choose an answer to THIS question at random”. That’s self-referential, no way out in my view (250 is NOT an answer to a question in the form: “what is the probability…?”.

2) As you pick an answer outside the set, as you do, there is no reason to limit randomization to the set. Thus as the set of possible answer is unbounded, uniform randomization would give a probability of 0. Well I’m not so convinced about that, but sounds reasonable…

3) You assume uniform randomization, but that is what is really unspecified in the question.

My view is that any of the answer may be true (or none can), given a suitable randomization.

Wednesday ~ November 2nd, 2011 at 6:40 am

Richard WilliamsonI think the point is that in order for there to be a ‘correct’ answer there has to be something that makes it correct. For the reason pointed out above, if the question is supposed to be self-referential then there can be no ‘correct’ answer. Similarly, there is no correct answer to the question Karl posed, unless Karl arbitrarily decided which answer is correct. But things get more complicated if we consider the subjective probability that my random guess is correct – Karl could decide that a number was the correct answer, or that only one of A, B, C or D is the correct answer even if two of them are the same number (remember, we’re being arbitrary here!). The subjective bayesian probability of guessing correctly is therefore not 37.5%, but 31.25% (assuming probability of Karl choosing each strategy is 50%)

Wednesday ~ November 2nd, 2011 at 6:46 am

AxelAren’t you confusing average answer with probability of being right? If you assume uniform randomization (which is usually implicit with the ‘at random’ expression), I would think the answer is simply 0% proba of being correct (as marco pointed out) with an average response of 37.5%.

To see this just assume the correct answer is 25%, then you have 50% proba to pick it up which is hence absurd. In the same manner 50% cannot be the correct answer as you have only 25% to pick it up. 60% is obv not correct with 25% proba as well. So none of the potential answers can be the correct one, then you would surely fail this test were it at random.

Is it related to any economics ?

Wednesday ~ November 2nd, 2011 at 6:50 am

anonThis is a trick question: there is no “right answer” amongst the choices offered so there’s zero probability you can answer it correctly.

Wednesday ~ November 2nd, 2011 at 6:54 am

anonAlso note that the question did not require you to provide an answer from the choices offered. The answer is 0%.

(If the question required that then in would be a poorly formed question, with no answer – a paradox.)

Wednesday ~ November 2nd, 2011 at 7:05 am

Nick RoweThe question works better if you replace the 60% in C with 0%.

Then, if you say “there is no right answer” that means 0% chance of getting it right at random, but that means there is one right answer (C), which is a contradiction.

Wednesday ~ November 2nd, 2011 at 5:39 pm

Dale Sheldon-HessYes; the question would be even more devious if the provided answers were something like “25%, 25%, 50%, None of the above”, because even then there isn’t a right answer. Perhaps I should mock that one up and see if it can go as viral as this one has…

Wednesday ~ November 2nd, 2011 at 8:16 am

Lukas Daalder37.5% is not an option, that’s the point.

Wednesday ~ November 2nd, 2011 at 10:10 am

PsychohistorianThe answer should be 0%. Unless you assume that the correct answer choice need not correctly answer the question, and that two identical answer choices must either both be wrong or both be right, and that two different answer choices cannot both be right. Then you get the 37.5%.

But because B is self-defeating, A & D are self-defeating, and C makes no sense, there is a 0% chance your answer choice is correct.

Wednesday ~ November 2nd, 2011 at 10:13 am

RobertWrong. If you choose an answer from those given, you cannot be right, so a random choice of A,B,C, or D gives you a zero chance of being right(given your comments on self-reference). If you answer at random fron all total possible answers you have an infinetessimal chance of being right as you might just as easily blurt out the answer: “Krugman!” as you might anything else.

Wednesday ~ November 2nd, 2011 at 10:14 am

Joe Colucci (@wonkinakilt)I’m with Marco: a question that asks about “this question” is pretty clearly self-referential. If the question referred to “a four-option multiple-choice question, with two identical answers,” then it might have a different answer, but the difference between “this” and “a” is hugely important.

As presented in the photo, it’s answerable correctly as a free-response question (non-multiple choice): the probability of choosing randomly is 0%. If you replace 60% with 0% in option C, it becomes a wonderfully circular paradox.

Wednesday ~ November 2nd, 2011 at 11:27 am

CraigOf course it’s self-referential. And self-destructive. It’s a fancy version of “This sentence is false:” selecting any response invalidates that response. This is exactly the kind of logical paradox that drove Bertrand Russell to an early grave.

Karl attemts to escape self-referentiality (as Bertrand Russell did in his day) by simply forbidding it. He wants the question to say: “Here is a set of four possible answers to a problem, in the customary ‘multiple choice’ format. I will not state the actual problem associated with these answers, but you are allowed to assume that there is a correct answer among the choices given. Now, if you were to pick one of them at random, what would be your chance of answering correctly? ”

That is what he means by replacing the probabilities with integers in his rewording. He could have as easily replaced the answers with “blue,” “red,” “green,” and “blue.”

Nice try, but it ammounts to nothing more than sticking your fingers in your ear and humming loudly. Or publshing six jillion pages of the “Principia Mathematica” in an attempt to outlaw paradox. The question is not what Karl wishes it were. A question on a chalkboard with four answers labeled A through D is a multiple-choice question.

(Or else, why even assume that the question is written in English? Maybe it’s a secret code I just invented! Maybe, by a convention of my code, any question that begins with the words “If you choose” should always be answered “A!”)

(What makes you think you have the right to go one step–but no further–down the rabbit hole? If you revoke one convention of our shared reality, you’re back with Decartes, wondering if your entire world is a dream, or you are deceived by a malicious demon. Be very careful about opening that box.)

Every sentence in this comment is false.

Wednesday ~ November 2nd, 2011 at 7:09 pm

RafaelIn my opinion the answer is 50%. You can assume that the probability to answer correctly is more than 25%. So you can exclude the options A and D because there are wrong. So you have only two possible answers and a 50% chance to be right.

Thursday ~ November 3rd, 2011 at 1:54 am

anonWikipedia “Russel’s Paradox”, this is essentially a restatement of that. Ideas of this form are what led to the construction of Zermelo-Fraenkel set theory and rigorous first order logic. The long and short of it is that this is an unacceptable question to ask in first order logic; which is basically a consequence of the fact that ZF set theory doesn’t let you include a set as a member of itself.

Friday ~ November 4th, 2011 at 4:54 pm

JovfIt’s interesting that everybody thinks of a uniform distribution when they hear the word random.

Thursday ~ November 29th, 2012 at 3:01 am

ElianIt’s a tricky question.First of all ,U must understand that it’s not multiple choice question and A B C D is related to our assumption not the answer.Then we have to exclude A&D because in the test,if two choice are the same,they are wrong. so B C remain,the chance is 50%.

Wednesday ~ May 22nd, 2013 at 8:58 am

azathothI want to understand the formula which finds %37.5. i did use that on different examples i made up and it worked and it seems logical but i am not sure why.

Saturday ~ September 28th, 2013 at 4:48 pm

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