Happy Halloween! Last week I described how @jbrownridge had sent me this lovely puzzle….

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

a) 25%

b) 50%

c) 60%

d) 25%

If you have not tried to solve it, have a go now. For everyone else, the answer (kinda) is after the break.

Well, its’ a tricky one. Essentially, there are three answers – 25%, 50% and 60%. So, if you choose one of these at random then you will have a 33% chance of being correct. However, this is not one of the answers, so you are in a somewhat difficult situation. My guess is that there isn’t an answer, but I can see how different linguistic assumptions make for different answers. So….what did you think?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

### Like this:

Like Loading...

*Related*

Oh…Does this mark the first time that the answer actually *wasn’t* after the break?

Yeah, next time can we go back to puzzles that have an answer please?

WTF?

I disagree. There are four answers, a, b, c and d. The fact that the text in the alternatives happen to be percentages is largely irrelevant, and the chances are still 25%

so does that mean u choose A or D then?

No, the choice is random, and the contents of the options is irrelevant. a) and d) say 25% but as far as this question is concerned, they could have said “sausage”

No. None of the answers can be correct because 25% and 50% contain contradictions. Therefore the answer is 0%, the chance the you will select the correct answer is zero. Imagine if the question was What is 2+2 and the answers were 1,3,6 and 2. None of those answers is correct but by your reasoning there is still a 25% chance of selecting the correct answer, there is a 25% chance that you will select any individual answer but if none of them are correct then you cannot select a correct answer.

Tort Says:

> No. None of the answers can be correct because 25% and 50% > contain contradictions.

What makes you think they are contradictions? What makes you think they are probabilities, or even the pool of possible answers? Which part of the question told you anything at all about the nature of the lines that started with letters?

You are right that 25% is wrong, I already changed that in a later post. The correct answer is not 0%, but asymptotically close to it

The answer is still 0%, as there is no correct answer given. The question asks about probability, which is why the fact that the answers are given in percentages implies that there is a correct answer. If 25% were given once, then that would be the correct answer, but since it is given twice, you have twice the opportunity to choose it, which makes the answer 50%, but then if the answer is 50%, you can only choose that one way, making it 25% again.

Well, to me the answer is 50%. The question is not “Which answer is right, a-b-c- or d?” The question is, “What is the chance that you will be correct?” So, usually, if you would answer randomly, you’d get 25% of chance to be right… but since there are two choices that says 25% (they could have said sausage like someone else said, the values in the choices are irrelevant, they are there to confuse us), therefore, the chance to be right is now 50%.

Exactly! You are the only one I have seen understanding this. The options aren’t 25%, 50%, 60% or 25%. They are A, B, C and D. There is no question asked. But there is another question of you chance of beeing correct. Don’t mix these two things together. We have to imagine a question with four different options. A is not the same as D. Even the ascii code differs and the don’t sound the same. They cannot be exchanged in words without the words changes, or lose their meaning. So forget about the percentage and consider the letters to be the four different values. Now, we are only given four options so we have a 25% chance.

Now comes the next part which is the trick and could confuse you again. The answer to the question in mind was 25%. Besides writing 25% down on a note, or just saying it, you can mark either line A or line D which both would indicate 25%. Now, do these two lines say the same thing, so it will be 50%. No no no, this is the mistake people make. A is not the same as D. Lets say that the “never asked” question was this:

If you choose an aswer to the question “which is the first letter in BORING” (not by random anymore), what is the chance that you will be correct? The answer in that case would be “B”. If you didn’t know which word it was, you would have had 25% chance which you can indicate by either mark A or D, but the answer could still be B for the actual question.

Great minds think alike! STRONGLY AGREE with Bocken Bruse!!At first i did this question mathematically, and the probability i got was 1/3, which gave me no right answer. Then when i looked back to the quesion, it said ” choose an answer at random”, which means GUESS! So, normally when we guess an answer for a multiple choice question, we assume there will be a right answer among the 4 options, and we don’t do any other probability calculations since we don’t know how to do it(that’s why we guess), also *KEY POINT* ignore the contents of every option(none of them make difference to our guess). Then,the answer of “choose an answer to any multiple choice question(4 options) at random” would be 25%, so both A and D are correct.

Bocken Bruse: Your explanation is based on the question about the percentages referring to the answers to a different, “never asked”, question — but that’s something you’ve added in, not in the original.

The question as posed by Richard clearly refers to itself, “this question”, not to a separate question. The answers have to be answers to that very question.

You accept that the purpose of the letters A–D is to label the possible answers, the percentages, in your final sentence where you say ‘A’ or ‘D’ could be use to indicate 25%. It’s nonsense to suggest using a second question in which the actual answers would be one of A–D and the percentages ignored.

If a second question is to be introduced, why would the letters A–D need to be involved at all? Indeed a separate question might have a different number of possible answers other than 4. For example, it could be ‘What colour is heliotrope?’ with the possible answers ‘purple’ and ‘black’ — in which case the chance of a randomly chosen answer being correct is 50%.

The only justification for calculating the chance based on picking from 4 possible answers is because of the 4 possible answers given, which are percentages. Ignoring those percentages as possible answers is ignoring part of the question as Richard posed it, and therefore not a solution to it.

you are incorrect as the assumption is that there is a correct answer: and there is not all can be eliminated, First this is the form of a multiple choice question, so it is understood that the answer must be selected from one of the answers offered, so the percentages are relevant as the question is asking for a percentage. the answer is there is no answer as all the selections can be eliminated. 25% can be eliminated because there is a 2/4 chance of that being selected randomly and 2/4 =50% and does not equal 25%. 50% can not be correct as there is a 25% chance of selecting B from A through D. 60% can likewise be eliminated as there is a 25% chance of randomly selecting C from A-D.

Multiple Choice questions are usually evaluated by a computer, in which case answer a) and d) are not the same. Chance you take the right one is therefore 25%.

Actually, you’ve got a 50% chance of picking “25%”, a 25% chance of picking “50%”, and a 25% chance of picking “60%”.

So presumably the answer is “0%”.

(it would be more amusing if you replaced “60%” with “0%” 🙂 )

this was the answer I came up with, and I believe is correct.

This is the correct answer.

0% as one of the answers would definetely make the question more interesting. That creates a paradox: The answer is than both right and wrong.

The interesting thing about this puzzle is that the answer influences the question.

yes! This is the correct answer!

More generally, this just comes down to the fact that it’s always possible to define a probability metric P on a discrete subset X of [0,1] where X is not {1} with the property P({x}) does not equal x for any x in X. Stated in this fashion, it seems much less paradoxical.

The problem as stated is solvable because when 0 is not an answer choice, it so happens that P({0})=0. When 0 is an answer choice but everything else stays the same, we have P({0})=.25 and hence lose the fixed point.

Yes, exactly. It’s interesting that this is not obvious to more people (where by “obvious”, I mean, evident after lots of thought, of course!).

It’s a funny joke, though!

replacing “60%” with “0%” would not change anything, this would only eliminate the possible solution of 0%, leading to no possible correct answer…

For me this riddle is just a case of “catch 22…”

First of all, we need to be clear what the question is. Some seem to miss this part. So…

1) “If you choose an answer to this question at random, what is the chance that you will be correct?” The question is what the chance is of answering the question itself correctly. Exactly as it says. This also clearly creates the loop, which can lead to ‘puzzling’ results.

2) Secondly, four answers possible .. so the answer seems to be 25% at first. Making both of the answers “25%” correct.

3) Since both the 25% answers are correct, that makes the chance 50%. For which there is only 1 option.

4) But if there is only one option, the answer would be 25% again. So there is no right answer. Whatever you pick from these four, it is wrong.

The suggestion was that replacing 60% with 0% would make it more interesting. Is it? This is basically exactly the same situation as before. However, it does create a weird situation.

A) Whatever choice you would pick, it would be the wrong answer. Even if it were 0%, the answer would be wrong.

B) In this case every answer would always be the wrong answer. And so the percentage would still be 0%, as there is no good option to pick from. Yet, the option “0%” is not good.

ASSUMPTION 1

HOWEVER … there is one an important assumption we are making here. In that we have to pick one of the four options as a correct answer. So we may pick any answer at random. Doesn’t necessarily have to be the first number that comes to mind either. It could even be 2125,2139001% … Doesn’t have to make sense or anything. Even better? It may not even have to be a number or percentage .. but I’ll save that for last :-p

ASSUMPTION 2

Another assumption? That it is a theoretical and purely logical problem. Could very well be interpreted as an experimental problem. In this case we could try to guess the answer a few times at random .. and after having done this a number of times check how often we got each percentage .. and see if there is a match there. Example: you may have picked 50% 12 out of 24 times ..

ASSUMPTION 3

Many people here seem to have found out some weird exceptions .. and probably all of us have instantly picked an answer, and then found out that it could not be right… One of the assumptions many seem to make is that their new found answer is correct. You’ve just had this great new insight .. and this newly acquired view must surely be the right answer. Right? 😉 Until you found out you’re wrong .. again! Don’t believe me …?

So, is there an actual good answer? Try asking this question to people at random .. especially try it out on people who don’t like puzzles, think they suck at puzzles .. or just don’t care to bother. Good chance they’ll answer with: “I don’t know”. And that would be correct. If you don’t think about it .. or are not too sure of yourself, that is the only correct answer to give. And immediately you would be correct. And since there is a chance of getting this answer .. 0% clearly seems to be wrong also.

Ultimately, there are various interpretations and solutions imaginable. And although there may not be a single ‘right answer’, there certainly are wrong answers .. as well as lines of reasoning. Have fun! 🙂

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

a) 25%

b) 50%

c) 60%

d) 25%

“If you choose an answer to this “question” at random (by chance)”,

the only question being ask is…

” what is the chance that you will be correct?”…

It did not tell that, you have to choose among the four choices provided (If you choose an answer “among the four choices” to this “question” at random (by chance)”. “If you choose an answer” was spoken generally, means if you choose an answer at any thoughts that you have right now to answer the question.

for any question being ask, there are only two possibilities (chance), you either give a “correct” or “wrong” answer.

If that is the case, if it ask you – “what is the chance that you will be either right or wrong?” means there are two possible outcome to your answer, it’s either “correct” or “wrong” answer.

Now the chance that you are providing a correct answer is always 100%, because at any question you will be either providing correct or wrong answer. An answer will always be 100% correct if it is “correct”, while 100% wrong if it is “not correct”.There are no in between.

If you are “correct” then you are 100% correct and 0% wrong,

If you are “wrong” then you are 100% wrong and 0% correct.

Now since it only ask “what is the chance that you are “correct”?”

Then the answer is “100% correct”. Likewise at any exam, you will either get a “check” or “x” mark, that means your answer will either be counted or not counted in your score.

… 100% wrong and 0% correct…

its not the question how correct the answer is – its about how likely your pick of the 4 answeres is the right one!

since 2 answeres are the same you can pick one and still have 2 other possible answeres – that makes 33,3% possibility

but this answer isnt available – so your chance being right must be 0%

but who makes questions which answer-possibilities depends on the real answer?

There’s 3 answers but 4 choices. If a computer were to randomly choose 1, there’s a 50% chance it’ll choose 25% and a 50% chance it’ll choose either 50% or 60%.

I’m sure there’s a mathematical way to figure out the probability of a computer choosing the right one. I’ll figure it out and post it tomorrow if I remember.

Reminds me of this, alhough not quite the same

This might be even funnier if they were Canadian, eh.

I’m with repton on this. I think RW’s analysis is a bit off. If you randomly pick a) or d) you have to ask yourself if it’s the right answer. So is there a 25% chance of picking 25%? No, there’s a 50% chance. So a) and d) are not correct. Apply the same to b) – is there a 50% chance of choosing that answer? No. Same applies to c). As none of the options will give you the chance of having chosen that answer, then none is correct, so there is a 0% chance.

But 0% isn’t one of the options to pick so…

Mick, there is no reason to assume that one of the options has to be correct.

Josh, yes I know. My point is that 0% isn’t one of the options. There is no correct anwser to this.

It’s a multiple choice question with 4 wrong answers. No one ever said that the question needs to have a right answer included in the 4 options. The answer stil exists, it’s just not one of the options.

Therefore if you select an answer randomly, you have a 0% chance of getting the right answer. 0% is also the answer to the question.

The puzzle is solvable, and the right answer is 0% chance. As far as I can tell it’s the only answer that won’t produce a loop or a paradox as well.

0% couldn’t be included as one of the four options though, or it would stop being the correct choice.

It would still be solvable. Your answer wouldn’t be a choice of the given ones but rather, your own answer, a new 0%! You can also justify with words your trying to avoid the paradox! Anyway, since there is no statement of this question being a multiple choice, the answer is 0%.

Starcore

The old version of that question included “0%” already (instead of “60%)

As Sohvan said, if “0%” is included in the 4 options, then won’t a correct answer though. And that’s because if it’s correct, then it can be randomly picked (with a chance of “25%”), and therefore the “asked chance” won’t be “0%” anymore. Got the point!?

The reason Richard is wrong is that the chance of getting any of these answers is not 33%. Imagine the following: What colour are zugs? a) yellow, b) blue, c) yellow d) red. If zugs are yellow, you have a 50% chance of being right. If zugs are blue or red you have a 25% chance of being right. The ambiguity in the original question is not because 33% is missing; it’s because the chance of getting 25% is 50% and vice versa.

I’m with all of you saying that 33% is wrong. There’s a fifty percent chance of randomly choosing 25% and a 25% chance of randomly choosing either of the other answers. I’ll accept that 0% is correct even though it isn’t in the answers, because none of the given answers can ever be correct, as soon as you identify one as being correct, it becomes wrong.

Zug being the German word for train, and trains in Germany predominantly being red, I’d have to say d) red.

Starcore

The old version of that question included “0%” already (instead of “60%)

As Sohvan said, if “0%” is included in the 4 options, then won’t a correct answer though. And that’s because if it’s correct, then it can be randomly picked (with a chance of “25%”), and therefore the “asked chance” won’t be “0%” anymore. Got the point!?

Gus Snarp

33.3% is not wrong (if we assume there is a correct answer of the given options), and here is the proof:

First, let’s have an introduction: assume that we have 4 different answers (no repeated 25%, e.g., we have 40% instead of one of them), Ok?

So, we have: any answer can be chosen out of 4 (i.e., 1/4) , and any answer can be the correct one is also one out of 4,(i.e., 1/4)

Then, the whole set of possibilities is 16 (not 4), and so:

The probability (chance) to randomly choose one answer out of four, and the chosen answer be correct = 1/16 + 1/16 * 1/16 + 1/16 = 4/16 = 1/4 = 25%

Now, come to our question:

we have: any answer can be chosen out of 4 (i.e., 1/4) , and any answer can be the correct one is also one out of 3,(i.e., 1/3)

The whole set of possibilities is also = 3*4= 12, and so:

The probability (chance) to randomly choose one answer out of four, and the chosen answer be correct = 2/12 (for the case of 25%) + 1/12 + 1/12 =4/12 ((because when you consider selecting one of the two 25%, there will be a doubled possibility of being correct)

And that is 33.3% (which is not available on the given choices!)

((By the way, if you draw the all possibilities on a sketch, you will find they are really 4 out of 12)

What do you think?

having two answers the same does not make the chances of picking the right one 33%. There is no correct answer because if you choose 25% there is a 50% chance of picking it, but if you choose 50% there is a 25% chance of picking it.

There is a way to make both ‘a’ and ‘d’ be correct answers. If an eight sided die had ‘a’ and ‘d’ on one side each, then the answer “25%” has a 25% chance of being selected even though ‘a’ and ‘d’ each have a 12.5% chance of being selected.

I’ll go with the interpretation that “choose an answer to this question at random” means that choosing among the four option tags, not as choosing among the distinct values.

Thinking about variants, in that case, if the 4 values are all distinct, then the probability of picking any is 25%, so if one of them is 25%, that is the answer.

If one of the values repeats (and none of the other distinct values are 25%), then if the repeating value is 50%, then that is the answer; rather either of the two tags are the answers.

similarly 75% occurring three times, or 100% occurring all four times have valid solutions. Other variants dont.

I got 33.3%

There’s a 33.3% chance the answer will be 25%, and a 50% chance you’ll choose 25%

There’s a 33.3% chance you’ll choose 50% and a 25% chance you’ll choose 50% – and the same for 60%.

So 50% x 33.3% + 25% x 33.3% + 25% x 33.3% = 33.3%

Bit annoying there’s ‘no answer’. I thought I had it wrong all weekend. Chances are I still have it wrong, but I’d have liked to have been put out of my misery at least.

OOps – replied to the wrong post.

But 33% isn’t one of the four options, so you have no chance of selecting it as the answer randomly.

The puzzle is solvable, so Richard is wrong about this one. The answer is 0%

I gave B as the answer. Ignoring the question, in any multiple-choice question you assume that one of the answers is going to be correct. As there are 4 possible answers, that means you have a 1/4 chance (25%) of getting it correct, regardless of the question. As there are two answers here that give the 25% option, that means you have twice as many chances of getting it correct i.e. 50%. So the answer is B, no?

a) 25%

b) 50%

c) 60%

d) 25%

No: because when you say that there is a 50% chance that you answer the question right, then the answers a, c and d are wrong.

The point is that the answers make each other incorrect. Just see it this way:

When you pick a, it changes the answer into “there is a 25% chance that you have this answer right”. But that is incorrect because there is a 50% chance that you pick 25%.

When you pick b, it says “there is a 50% chance that you have this answer right”. But that is incorrect because there is a 25% chance that you pick 50%.

When you pick c, it says “there is a 60% chance that you have this answer right”. But that is incorrect because there is a 25% chance that you pick 60%.

d is the same as a.

So all answers are incorrect making it 0%.

If the answers were this:

a) 25%

b) 50%

c) 0%

d) 25%

And you randomly choose c, it would say: “there is a 0% chance that you have this answer right”. But that is incorrect because there is a 25% chance that you pick 0%.

Again, making 0% the right answer. But according to the question 0% cannot be right, but it is. If you were not confused, then you are now.

The answer is that you don’t have an answer? Uninspiring to say the least.

I agree with Repton above, it’s 0%.

Artificially restrictin the possible answers to a question to some small subset of logically possible answers doesn’t mean the correct answer will always be part of that subset. In this case it’s not, which can be easily shown using proof by contradiction:

Assume a) is correct: d) must necessarily also be correct, but then you’d have a 50% chance so they’re both false.

Assume b) is correct: answer is 50%, but only 25% of the answers are this.

Same logic for c) as for b)

The above is assuming that ‘random’ means a flat distribution, rather than some weighted function giving different chances of picking each answer. Anything else would be entirelu unfair if not specified in the puzzle, since you could play with the function to arbitrarily force any answer to be correct.

This isn’t really a puzzle, it’s just a question and four wrong answers.

Chris,

The old version of the question had “0%” instead of “60%”, so I wonder what would you say about that case!?

It depends which way you read the question. If you read the question as the answer being 25% chance of picking a random letter, then your chances of picking the 25% as the answer would be 50%. But if you did not take into consideration that 25% is the answer but any of them could have been the answer to something different, the right answer then it would be 30%. Does that make sense? I am confusing myself now. I thought it was 50% anyway.

This very puzzle was brougth up in my mathematics class on friday.

First I reasoned as follows:

* If you choose 25 %, your answer is wrong, since the probability of chosing 25 % at random is 50%:

* If you choose 50 %, your answer is wrong, since the probability of chosing 50 % at random is 25%:

* If you choose 60 %, your answer is wrong, since the probability of chosing 60 % at random is 25%:

* Therefore there is no right answer and the probablity is 0%.

Then someone told me there was a sublte error in my premises, so I reasoned as follows:

* The prior probability of chosing any of the options at random is 25%

* The probability of chosing 25 % at random is 50 % since 2/4=0,5

* Therefore the right answer is option B 50 %

This was the consensus until somebody pointed out that this means 50 % IS your answer to the question, which means that WHAT IS BEING ASKED FOR is the probability of THIS answer.

So now I’m inclined to think my first response was correct.

Your first response is correct.

Because if 50% is right, because you have 2 answers of 25%, which can both be right, then 50% is also right. That adds up to 75%. That answer is not there. So: 0%.

Your first response is correct – the probability of choosing ‘25%’ is 50%, because that is two of the four answers. But the question asks for the probability of randomly selecting the correct answer (i.e. the answer ‘x%’, for some ‘x’ which you have an x% chance of randomly selecting), not the probability of selecting 25% – as you have a 50% chance of selecting 25%, 25% cannot be correct.

Well, I thought there wasn’t any answer to the question, since there is no question regarding the options.

“If you choose an answer to this question at random, what is the chance that you will be correct?” What is the question “this” is referring to? There is no previous question, so there is no answer.

Why do you need a previous question? “This” usually refers to the current thing/place/time/whatever.

For example:

This sentence has seven words in it.

The “this” in that sentence refers to itself, just as the “this” in the question refers to the question.

Well, for me the answer was 60%; maybe different for others if they chose different answer at random. I can only think it’s pretending to throw us, when actually it’s just asking a very simple question, and asking us to say out loud what we picked? If we play the puzzle logically, then we subvert the puzzle as we’re no longer picking an answer at random??

Any good?

Bjarte, so are you saying that if you are being asked what the probability of you picking 50% is then wouldn’t that then be 25%?

I say the correct answer is B, 50%. You have a 50% chance of choosing 25%, which is the correct answer.

Note, that if do you answer B, your answer really is wrong. But, nobody said that the answer you choose should be the correct one!

If one does require that the answer you give is also the correct one, then there is no solution to the problem. But, as said, this is not what is required in the formulation of the question, as far as I can see. I also think Bjarte above me was correct in his latter reasoning, and is being unnecessarily swayed by the popular vote.

Though I bet many will disagree with me on this one, I’m sure (and there is always the possibility that I’m wrong — failure is always an option for me ;-))!

Wait, no, darn. That doesn’t feel right. I gotta think this again.

Well of course the answer has to be correct. Otherwise it’s entirely meaningless to ask. You may as well state the the answer is ‘cheesecake’, and question why anybody was expecting you to look for the correct answer.

You are only answering “what is the probability that you choose ‘25%’?”, which is not what the question asks for: assuming 25% to be correct leads to a contradiction (you do not have a 25% chance of choosing it), and thus 25% cannot be correct.

It is interesting to see the number of people who think the text of the alternatives have a bearing on the probabilities of the question.

The classic riddle in this regard is “This sentence is false”. The solution, of course, is that the sentence is meaningless and has no truth value. It is just a sequence of letters.

Similarly, the options here were never stated to contain the possible probabilities to the question. The question was “what are the chances” and because the options contain something that looks like probabilities, people immediately assume that they must be. “This probability is false”.

Very good explanation, Anders. I also believe there really is no question to answer.

Actually, thinking a bit more about this, I’m guilty of an unstated assumption as well. The question never actually said that the choice was between the four letters, that was an assumption on my part based on years of conditioning.

The range of possible answers then is infinite, and the odds of hitting the right one when choosing at random is not quite 0 but infinitely close to it

AnnieB: Yes, but then the probability of randomly picking that answer (the very thing that is being asked for) is not right according to the anwer you actually picked. :-S

A fun variant:

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

a) 25%

b) 50%

c) 50%

d) 0%

A, B and C could be correct answers. D is always false.

If A is the correct answer, you have 25% of choosing the right one.

If B or C is the correct answer, you have 50% of choosing the right one.

The chance you pick the right one is 25% * 25% + 2 * (50% * 50%) = 9/16 (= 56,25%), but that answer is not there. So the chance you will choose the correct answer is 0%, which is answer D, and for that you will have 25% of guessing. But that implies 0% = 25%.

After initial thought that it’s 50% I realized that there’s no right answers therefore the chance to choose the right answer is 0%

That’s why I commented it as sneaky on Friday 😛

Assuming “correct” answer is also random, with probabilities of P(25%) = 1/2, P(50%) = 1/4, P(60%) = 1/4, then the probability for choosing the correct answer at random is 50%.

If you now allow your answer to retroactively change the “correct” answer, then the question becomes unanswerable, however, why would you do this? That makes no sense.

This is actually an example of a Russel Paradox

The probability of choosing “an” answer is actually 25% since there are 4 possible options. However the likelyhood of picking an answer that lists as 25% is 50% (because the answer 25% is listed twice).

Now you would think that the answer would be 50%, but the chances of you picking that answer is only 25%. Which brings us back to the front 😉

I can’t remember where but I’ve seen this puzzle before where this answer was given to the puzzle.

It a trick question, so the chance of answering correctly is 0%

Chris wrote: ‘Artificially restrictin the possible answers to a question to some small subset of logically possible answers doesn’t mean the correct answer will always be part of that subset. In this case it’s not, which can be easily shown using proof by contradiction:’

I’m with Chris on this. It’s a bit of an Alice-in-Wonderland-logic problem, but it seems to me that the answer must be 25%, irrespective of what each ‘answer’ reads as.

Or it’s 0%.

There – that helps the debate, dunnit 🙂

Okay, so let’s rephrase the question. If you chose an answer to this question at random, what are the chances of you being correct:

a) Apple

b) Pear

c) Banana

d) Apple

Oh Christ, now I’m starting to see the logic of 33.33333% (as mentioned above).

Thats a good idea. I was also experementing in this way. But there is something to clarify at all: Does “chose at random” mean that we close our eyes and tip with a stylo on one of the four answers or does it mean that we read the answers an then make a well-considered choice ?

How about this version:

Four envelopes contain the answers: 25%, 50%, 50%, 100%

I ask AnnieB to pick one envelope at random. What is the probability that AnnieB has picked the correct answer to this question?

This is quite a stretch, but flip a coin to choose whether or not you answer the question. Then flip the coin twice to choose an answer. Then ‘a’ and ‘d’ could each be correct answers.

Not answering would have to count as not answering correctly.

I feel Cheated by this. we have spent all weekend pondering what the answer would be only to find out there isnt one.

Rubbish!!!

That’s not true, there is one, it’s 0%.

I think AnnieB has decoupled the answers from the question. If the question is “What are the chances of you being correct?”, the answer cannot be “Apple”, or “Pear” or “Banana”. Therefore you would have a 0% chance of answering correctly.

In the taxonomy of word games, I would be inclined to class this as a Riddle rather than a Puzzle. Or maybe it would be better termed a Linguistic Illusion; a bit like Chomsky’s “Colourless green dreams sleep furiously”.

Multiple choice questions are very common in the school system, so we’ve been conditioned to expect that one of the choices will always be a correct one. Think about a question like this one for example:

What is 2+2?

a) 1

b) 2

c) 3

d) 5

The question is still valid and answerable, even if none of the given options are right. Of course it’s not a fair question, but no one ever said it would be. You just expect it to be fair because that’s how it’s always been in the thousands of multiple choice questions you’ve completed in your life.

The question in the friday puzzle too is solvable and has a correct answer, which is 0%. It is not the equivalent of “this statement is a lie.” which is a true paradox.

It is the equivalent of the liar’s paradox in that people ascribe meaning and truth value where none is warranted

But the liar’s paradox doesn’t have a valid answer, while this question does. Just because you are prevented from giving the right answer, doesn’t mean that a question becomes unanswerable.

Of course you can get into semantics about what is the true meaning of words like chance, random, question, choose, answer or is. Human communication relies largely on certain unspoken assumptions. If you peel back enough layers, you can make any communication impossible.

The liar’s paradox does have a valid answer, and I gave it above. The correct answer to the liar’s paradox is that a sentence is just a string of letters, and to assume there is meaning or truth value is unwarranted.

(of course the answer to the real, original liar’s paradox, is that all Cretans are liars, is that not all Cretans are liars and the speaker is a liar – it’s not even a paradox, really)

You are correct that in normal conversation, we have to be able to rely on assumptions, but when those assumptions turn out to be wrong, we make mistakes, misunderstand, get things wrong, perhaps get into fights. That is the real point of riddles like the liar’s paradox. Challenging assumptions, understanding when we can, understanding when we have to.

In riddles though, assumptions are nearly always wrong. The riddle format is designed to exploit such things (the surgeon is a woman)

By that same reasoning we can declare half the riddles and puzzles in the universe to be unanswerable, because after all “a sentence is just a string of letters, and to assume there is meaning or truth value is unwarranted.”

Suppose a question asks you what is the shortest amount of time in which John and Erica can complete the chores, which was a recent weekly puzzle. Do you then give the answer that time is relative, and we need to know how fast the observer is moving or whether they’re near any highly massive objects? If you stretch enough, you can make almost anything become an answer to any puzzle, but what is the fun in that?

While assumptions are indeed dangerous, all of communication and intelligent thinking becomes worthless if no assumptions at all are made. You can always derail anything with the equivalent of “But what is Truth reallly?” A lot of the beauty in riddles comes from knowing which assumptions are appropriate for the situation. A puzzle can have multiple answers depending on which assumptions are made to begin with.

Not really, you are being overly dramatic here. The assumptions in a riddle are stated in the question. This is always true, and in 99% of riddles, this is the entire point of the whole thing, and the riddle leaves something unsaid which you assume, and this throws you off.

By challenging assumptions and going strictly by what is in the riddle (or indeed exam question), we do not “declare half the riddles and puzzles in the universe to be unanswerable”, we in fact answer them correctly. The people who stay with their assumptions never understand the point of riddles and questions of this nature.

So, assume only that which is given in the question, nothing else. That works, assuming the riddler (or exam setter) is competent.

In this case, the question only asked for the odds of answering the question correctly. Nothing else. In every exam I have ever been in, and in every book of puzzles, there is *always* the statement that the answer is to be chosen from the alternatives listed. If there is no such statement, the question is open-ended and any answer may be given. Here, there were no such statement

The puzzle you refer to did leave many aspects open, which led to a healthy discussion about possible alternative answers allowed by the phrasing of the question. This is not wrong, it is the whole point.

By the way, maybe you’ve missed this, but the people who claim there is no answer are the ones who go with the conventional assumptions (the percentages are probabilities, the answer is one of the 4 etc.). They say it can’t be answered. I actually gave an answer, with justification, and I would defend it to any mathematics professor. So please don’t claim that *I* would leave half the puzzles without an answer.

My original comment was not aimed at you specifically, but rather at people calling the puzzle unanswerable, a waste of time, or even not a puzzle at all. I’d be interested in knowing what you did think of the puzzle overall?

I agree that an answer should be chosen from the alternatives given in a question like this. Are you saying that it needs to specifically stated in every single puzzle, or we can assume that any answer can be chosen?

The point was the the lack of a correct answering option does not make the question itself invalid. The existence of a solution is separate from your ability to give it as an answer. So 0% is still a valid answer to the question presented, even if its not included in the options. In fact it would seize being a valid option if it was included. That’s part of the beauty of the puzzle. To me the whole point of the puzzle was challenging those assumptions people make about the nature of multiple choice questions.

I think it needs to be stated. At the very least something like “which of the following…” or similar. The version I just saw here in a comment, which rewrote it to have four envelopes and ask to pick one envelope at random would also be acceptable, and comes close to being the question most people here have answered.

I do believe however that the question becomes more interesting when you read it the way I did. Then it becomes about challenging assumptions, and not answering something that was never actually asked, and all the rest that I have argued above and elsewhere – as opposed to a not-really-answerable probability calculus exercise. It would be something I’d expect from a psychology professor, and more in line with the events, books and videos that have made Richard Wiseman famous. All these semi-skilled maths problems on this blog are so far removed from the image I have of him

Nope. This question, unlike your example, explicitly demands you “choose one” (assumedly one of those provided).

The question tells us that there is a correct answer by asking us what the chances of picking it are. The only way the answer can be 0% is if there isn’t a right answer. So 0% can’t be the right answer.

The question asks “What is the chance you will be correct?” It does not ask what is the chance of picking the correct answer. So it doesn’t tell us anything about the existence or non-existence of a correct answer.

Just because it tells you there’s a correct answer doesn’t mean there is one; the question might be wrong. Among the answers given, there’s certainly no correct one, and you absolutely have a 0% chance of the correct answer if you’re restricted to the selection of a, b, c, or d.

If you decide you’re not restricted to the answer’s suggested, then it becomes impossible to really quantify. My best answer if you’re *not* restricted to just a, b, c, or d is:

You then have an infinite choice of possible answers. If the selection is truly random, each is equally likely, and thus each has a probability of 1/(infinity), which is undefined but is smaller than 1/x for any positive x. And so whichever answer is correct, your chance of picking it basically end up at zero again (or more strictly, infinitesimal), and in this case I can’t see any knock-on effects where the definition of that answer as correct would lead to a contradiction.

Of course, to suggest we could truly pick at random from an infinite set is absurd.

Does the question state there are only 4 answers? Or just the probability to pick the correct answer?

0% is the answer. If the answer is 25%, there is a 50% chance of picking it => contradiction.

If the answer is 50%, there is a 25% chance of picking it=> contradiction, similarly with 60%. There is a 0% chance of picking 0%, and this is the only consistent possibility.

I agree that the correct answer is 0%, but it might be helpful to articulate the difference between the 25% chance of picking a given answer as cited by many correspondents and the 33% chance cited by RW.

Put simply, it depends on the algorithm selected by the chooser. Faced with the given list of options and told to pick one at random, there are two equally reasonable procedures which could be adopted. One is to ignore the content of the four options and simply roll a four-sided die and thus pick an answer at random. Only once a, b, c or d had been chosen would you pay any attention to the answer selected.

The other option, which RW must have had in mind, would be to assess the number of different answers first, ignoring how the options were labelled. There are three different answers, so to choose randomly amongst these, one throws a three sided die (or more practically a six-sided die) and thus picks an answer at random.

As far as I can see, both procedures are reasonable and with no other information there is no particular reason to prefer one to the other. This means that if the options were

a. 33%

b. 50%

c. 33%

d. 25%

Then 33% would arguably be a correct answer (to the nearest whole number).

Wait a minute….. Richard gives us a problem he doesn’t think there is an answer to, leaving us to squabble amongst ourselves as to what the answer may be. Is this some kind of warped psychological experiment?

That would be funny! I like the comments of the one’s who feel all indignant, as though they’re actually owed an explanation. I would like to see it resolved as I have the small problem that I shared this puzzle with my FB friends and they are all waiting for the answer. Jokes on me! ROTFL

Next Friday’s Puzzle:

“Will the number of indignant replies on Monday be greatest if I posted the correct answer to this question, if I posted the wrong answer to this question, or I posted no answer?”

What a complete waste of time!!!

What a complete waste of time!!!

It’s a paradox in that a question that doesn’t appear to have an answer, does have an answer (that is, 0%), but only because we’re applying the conventional rules of MCQ. Given that two choices are identical indicates that this isn’t a typical MCQ. This is a hint that since MCQ rules no longer apply, the question can be reinterpreted as: “If four envelopes contain answers: 25%, 25%, 50%, 60% and I ask someone to pick one at random, what is the probability that she has chosen the correct answer to this question?”

Since the question is no longer in MCQ form, the paradox is resolved — until I replace the 60% envelope with one conatining 0%, that is.

“If you roll a pair of dice at random, there are essentially 11 different outcomes, so the probability of each is 1/11.” We know this is wrong, because the randomisation procedure has been specified. The problem of the present question is that the randomisation procedure has not been specified. Are we to choose one of a, b, c, d at random, like wagdog? That is certainly the most common interpretation of the kind of words used here.

The question does not ask you to choose an answer from the choices below it. It only asks “what is the chance that you will be correct?” Therefore, the ABCD answers below the question are unrelated to the question itself, and the answer is 25%!

I’m not convinced humans are capable of making a random selection.

i suggest it’s a basic 1-in-4 chance and someone has changed the numbers from the original puzzle and failed to realise the impact it would have outcome. If you assume the normal multiple choice format (1 question & 4 possible answers and only one is the correct) then it’s always a 1-in- 4 chance of being randomly being correct answer

But that’s assuming it’s still out of four possible choices – which one can’t assume if you’re going to discard the provided choices.

I spent a good few minutes vacillating between 25% & 50% then smoke started coming out of my ears & I decided to wait for the answer. Strangely relieved that there isn’t one. 🙂

It took me some time but I think the right answer is 42.

I think this is a question about linguistics, the chance of you doing something is always the same, you either will or you won’t. When you have four answers, it is “probability” that will determine which you will pick or which is right.

I think its kind of like the Russel paradox. If the answer is 25% then beacause it appears 2 times you have 50% chance of choosing it. If the answer is 50% then there are only 1 value which appears 2 times therefore the answer is 25%. So the answer is 50% if and only if the aswer is 25% and vica versa.

Exactly.

Has anyone consulted Kurt Godel or Douglas Hofstader?

the answer is

e) 20%

as clever as that pi puzzle

but more acceptable

No, the answer is f) 16 2/3 %

Wait, no, it’s g) 14 2/7 %

…No, wait, it’s….

Of course the puzzle did not say that the answer had to be in the list, so 33% is perfectly acceptable.

I am thinking like this:

If the answer is 25%, the chance is 50% – contradiction

If the answer is 50%, the chance is 25% – contradiction

If the answer is 60%, the chance is 25% – contradiction

Since none of those answers is correct, the chance is 0%, which is correct, because 0% is not one of the listed options.

This is my reasoning, too. Tricky puzzle.

Outright cheating! Best expressed by person who said “WTF?”

The answer is ‘no’

a) I was right in thinking that I was wrong.

b) I was wrong in thinking I was right.

Any multiple choice question must always allow for an (n+1)th response of “none of the above”. That’s simply because in even the best prepped quiz misteaks can creep in.

Thereforewise, is five possible answers: a, b, c, d, e, none,

If I guess one of those with equal randomness, then in one occassion in five (1 in 5) I will guess none.

As none of the answers in Richard’s quiz is right, then none is right.

Means I am guessed right 1 in 5. Which is (when multiplied using simple school arithmetic) 4 in 20 or 20%

Thereas, 20% is the rightestmost answer in this case.

And here I rest that case.

Considering the many-worlds interpretation, it’s you have a 100% chance to choosing the right answer somwhere. 🙂

The question does not speciifcally limit the correct answer to the choices listed below. In fact, it suggets quite the opposite. A legitimate answer might – in fact – be “elephant.”

Suppose that “random” does not refer to a uniform distribution. Then I can pick (a) and (d) with 1/8 probability each, and spread out the remaining 3/4 on alternatives (b) and (c) equally. Then I will be correct with 25% probability.

You can do the same to justify answer (b) and (c) too.

Another solution suggested to me as that one may choose two distinct alternatives (as is common in many multiple choice problems). Then there is a 25% chance of choosing (a,d).

At first I broke it down into 33% for each actual value.. but even if 33% was an option it would change the actual true answer to 100%.. I think the key to the puzzle is the word RANDOM.. if all the question asks is for one to RANDOMLY choose an answer then none of the choices can be wrong

The answer is 33%.

Reasons: The question isn’t defined, just need to work out the ‘chance’ that one of those answers is the correct answer. The chance you will pick the correct answer at random is 33%

Forget the actual choices, they are irrelevant and only there to confuse. Another example:

Pick an answer at random, what is the chance that it is correct:

A) Sausage

B) Tree

C) House

D) Sausage

The chance is 33%, the question isn’t even important, because the answer is random.

Just to clarify again, the question isn’t important, the value’s of the 4 answers isn’t important either.

You chose either A, B, C or D. Because 2 of them are the same, the probability you pick the right answer is 33%.

The answer to this puzzle doesn’t have to be one of the value’s of ABCD. My above example doesn’t have a “Sausage Percent” of being correct lol.

What question? There is no answer because there is no question. I may not always be correct, but I’m never wrong. 😉

Formatting is off after pasting, but this leads you through the logic. The chance you will be correct is 0%.

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%

Guess: A B C D

% of choosing: 25% 25% 25% 25%

Answer Value: 25% 50% 60% 25%

chance of choosing answer value: 50% 25% 25% 50%

Answer that turns out being correct: b) 50% a) and d) 25% a) and d) 25% b) 50%

Did I choose the correct answer? No No No No

0 out of 4 = 0%

Not sure what the awnser is at this point, but I thought of this riddle the complete weekend. I could come up with the next awnser:

Lets assume the right awnser is in there. Then its either A and D (since they contain the same awnser) or B or C.

Theorem 1: Awnser A and D are incorrect.

If so, it must be either B or C. Both having a chance of 25 percent to be randomly choosen. This makes A and D correct, so Theorem 1 is incorrect and A and D must be true.

Theorem 2: Awnser A and D are correct.

If A and D are both correct there is a 50 percent chance of randomly choose the awnser correct. This, as many people above pointed out, leads to anwser B being correct as well and since I’m not looking at the values of A and D yet now there is a 75 percent chance of awnsering this question at random correctly.

Theorem 2 is therefore also incorrect since A and D point out 25 percent.

Therefore the assumption that the right awnser is in there is incorrect and the chance automatically is 0.

But to calculate the actual value of the right awnser one could aproach it maybe like this:

Imagine a pie divided in two halves. One halve is also devided in a halve. So we have three area’s now, all different in taste. The mission now is throwing a spoon randomly in the pie and hit the pie peace with the correct taste.

Since we don’t know what the right taste is we are facing a problem. The chance to hit the biggest peace is 1/2, but the chance its the right taste is 1/3. For the smaller peaces, the chance to hit it is 1/4, but the chance they are the right taste is still 1/3. So the chance to hit the peace with the correct taste:

(1/2)(1/3) + (1/4)(1/3) + (1/4)(1/3) = 1/3…..?

Not so sure about this calculation though

[…] Answer to the Friday puzzle…. « Richard WisemanOct 31, 2011 … You are right that 25% is wrong, I already changed that in a later post. The correct answer is not 0%, but asymptotically close to it. Chris V. says: … […]

Let us assume that the correct unswer is the one that given in “Answers” section (can be one of the four choices: A, B, C, or D). If it is A or D (25%) then I would not object.

(in other words if we use A and D as different “input” events we don’t need to merge them in the “output”)

We may also find “None of the above” in the “Answers” section, who knows …

_____________CORRECT ANSWER___________________

EVERY SINGLE ANSWER THAT HAS BEEN GIVEN THUS FAR IS INCORRECT.

THE CORRECT ANSWER IS 17%…. Read more to find out how

I believe the author of this problem intended it to be a probability question, not a trick question. So we have to calculate the chance we have to be correct if we choose A-D of an unknown question.

Now, we do not know which one (A-D) is the correct answer. So we have one of two possible outcomes.

IF a) and d) is the correct answer to this unknown question then the chance that we will choose the correct answer would be 50% (Because 2 of the possible 4 is correct)

However, IF a) and d) is the incorrect answer to this unknown question then the chance that we have of choosing the correct answer is 25% (Because 1 of the possible 4 is correct).

I believe we can take it a step further…

a) and d) has a 33% chance to be the CORRECT answer to the UNKNOWN question.

This means that a) and d) has a 66% chance to be the INCORRECT answer to the UNKNOWN question.

This means that there is a 33% possibility that you have a 50% chance to choose the correct answer, and a 66% possibility that you have a 25% chance to choose the correct answer.

The possibility of a) and d) being correct AND that you chose a) or d) (Thus having the CORRECT answer to the UNKNOWN question) is 16.5%

The possibility b) OR c) being correct AND you choosing b) OR c) is 16.7%

The difference in decimal places has to do with the initial 33 and 67 percentages having repeating decimals.

In conclusion you have about 17% chance of choosing the correct answer.

Yours sincerely,

A HIGH SCHOOL STUDENT.

This was good to read! Not because it was interesting, but because your answer to the question was so abysmally stupid that now I NEVER have to read your books to know what a terrible writer you are!

For everyone who has agreed on 0% I first have to say that if all the alternatives were wrong it would not be a trick question. Therefore lets consider one of the answers to be correct. We have to realize that the answer must be more than 25% since there are only 3 possible answers… 25, 50 or 60. But with 50 % change to pick 25. The answer must also be less than 33% since it is a greater chance to pick one of the alternatives. good luck! 🙂

isnt it B 50%?

assume b is the correct answer. then 25% is the probability of choosing b as the answer.

when you randomly choose the answer from the choices, there is a 50% possibility of choosing a or d.

i think the misconception comes from the fact that chance of choosing the answer does not have to be the same as the answer itself.

probability of choosing (B) as the answer=25%

when randomly choosing->getting 25% as the answer=50%->(B)

[…] richardwiseman.wordpress.com […]

Ha! I’m pretty bad about over thinking things but this is nuts. I like Simon’s answer above and it seems the best choice from a statistical perspective. I’d also like to add the following:

We have no information about what constitutes a correct answer. Therefore the we can likely consider the whole thing invalid. If not invalid we can assume that any chosen answer is wrong (because there is no correct answer to a question that is not asked) or we can consider every answer right because we have no way to definitively declare an answer wrong.

It is a paradox. The answer would be 25%, but since there are two of them, it is 50%. Since the right answer is 50%, and that is only one possible answer, it is 25%. Repeat forever.

The answers to the questions are irrelevant. The answers could be apple, orange, banana, and apple. Or anything else. The fact that two of the answers are the same percentage is irrelevant. There are still four distinct answers. Given that it’s multiple choice, we must assume one of the answers if correct when asked to pick. If we pick one at random, we have a 25% chance of picking the “right” one. Whether you agree that 25% is correct or not doesn’t matter. If I tell you the correct answer is C then you pick at ransom and get C, you picked the correct answer.

What is the question????

The puzzle can be easily solved, but only before choosing the answer. As soon as you pick one answer the it becomes a paradox…you can picture it as something fluctuating ad infinitum between 25%, 50%, 25% …..

There is no paradox here, and there is a straightforward answer which may become clearer to some if we first look at a related puzzle.

A fifth grade class is asked which year the Declaration of Independence was signed in: . a)1066, b)1215, or c)1492. If they choose an answer at random, what is the probability of their being correct?

The chance is zero, and surely no one will argue with that. None of the answers offered is correct.

The puzzle we’re looking at is no different. Assuming that “choose an answer” means “choose from the following four answers,” there is no correct answer among the four and therefore no chance of choosing a correct answer and the answer is zero, no paradox at all. It isn’t one of the answers offered, but so what? Nothing in the wording of the puzzle reduces the logical range of answers to those on offer.

Wish I could edit my previous. A better, simpler last line: “Nothing in the wording of the puzzle requires the correct answer to be one of the four offered.”

I disagree the very form of the question is that it is a multiple choice, therefore it is expected that an answer must be select from A-D, in either case there is no correct answer, and since 0% is not a selection the answer is not 0% there is no answer, the question itself is invalid.

No, the question is in the subjunctive. IF you were to select from A-D, what is the probability of being correct. You are asked to make a judgment about what would happen IF you chose one of the four. The answer is just as straightforward as in the Declaration of Independence question above, which is also multiple choice. It is the solver who is lured into committing a category error in imagining that zero is out of bounds.

Do you believe that the Declaration of Independence question is also a paradox? Why can’t its answer be zero?

Let’s make it even simpler. You are shown two doors.

A) a door leading to a beautiful woman and nothing else

B) a door leading to a tiger and nothing else.

IF you were to choose A or B, what is the probability it would lead to a shiny new car?

Do you see this as a paradox, a category error, or simply a question whose answer is zero? And why is the Friday puzzle any different?

The question is a not a paradox but a categorical error.

Making no assumptions, the first step to determine is how many answers are there to this question. Having had multiple choice questions where any, all and none of the answers could be correct, we have to include those possibilities. Hence, that would be expressed as…

((4!/(4-4)!)+(4!/(4-3)!)+(4!/(4-2)!)+(4!/(4-1)!))+1

Which equals 24+24+12+4+1 or 65 possible answers.

With no assumptions, you have a 1/65 chance of answering this (considering we don’t know what “this” is) correctly or roughly 1.54%.

You can’t assume that the two 25% answers are the same because the question could easily have an answer that is one of the 25% choices and not the other.

So, 1.54%.

The total of any one, all, or none among 4 is only 6

0000

000x

00×0

0x00

x000

xxxx

And the probability of picking one of six is 16.66…%

Of course, as you probably intended, “any” could mean any combination. But there’s still something wrong with your math, as any combination including all or none still totals only 16 possibilities:

0000

000x

00×0

00xx

0x00

0x0x

0xx0

0xxx

x000

x00x

x0x0

x0xx

xx00

xx0x

xxx0

xxxx

The probability of getting the only correct one (0000, or “none”) would therefore be 1/16, or 6.25%.

But in any case, you are making a big and unwarrantable assumption: that “all or none or any combination” can be implied for any multiple choice question. A fair test always makes these options explicit if they are permitted, and standardized tests have been invalidated for failing to do so. If you pick one of the four actual choices, you cannot get a correct answer, so the probability that you can remains zero.

It’s also true that the wording of the puzzle does not explicitly say that the four answers following the word “choice” are the choices referred to, but that seems like a reasonable assumption. And if we cast it aside and permit any conceivable answer in the universe (including 17.124, 17.125, 17.126, “France,” and “a spot of jam,”) that’s an infinite number, and one divided by infinity is as close to zero as makes no difference.

You are assuming that if the answer is AB it is also BA which is not explicit either. You have to account for order of choices to which is how I got to 65.

I initially came up with 16 combinations but then figured in the possibility it could be order contingent which boosts the number to 65.

I do appreciate learning the rationale for your numbers, but I really can’t see any meaningful distinction between “I choose 50%, _then_ also 60%. _then_ also 25%” and “I choose 25%, _then_ also 60%. _then_ also 50%.”–even if those would be considered sensible answers for a multiple choice test, which of course they wouldn’t. (The only sensible combination is A and also D, and I defy you to tell me that D and then A would be meaningfully different.) Unless the implication, is “I choose 50%, no 60% — no, no, wait! I really choose 25%. Honestly..” Which would be indicated by some crossing out, but even then the order of choice would be incompletely known, and the answer simply recorded as 25%.

[…] a lot of discussion at Richard Wiseman’s blog and more at Lifehacker, where I first saw […]

Does the question state there are only 4 answers? Or just the probability to pick the correct answer? Another way to look at it????

Either way… I’m not sure if this a paradox? Just the way the question is worded…

how can i change my billing info (or am i being stupid, here?)

Ans is 50% .

( I have answered same answer in an interview)

Now this is the reason:

The question is tricky . First I will prove my point than eliminate other options.

Question has asked what is probability that answer is correct …. Note this.

Every answer to any question not only this has two probability viz

it may be right

it may be wrong.

So if one choose answer to this question at random its chance to be correct or wrong is 50%.

Now eliminating other option.

Every option has equal chance to be correct. Hence every option has 25 % chance. But if you select an option and it will be correct or not has 50 50 chance .

So eliminate A and D.

Option C is there to create confusion.

( Reasons for C that it creates confusion :

It is all a different option and according to human psychology our brain prefers such thing…

Moreover , mcq solver has habit of marking option C more … This is according to surveys done on the answer sheet of examination.

From the first part of the question:

If you choose an answer to this question at random.

1- the question is not mentioned, so we actually do not know the correct answer.

2-the person who is going to answer randomly is the reader of the question ( got that from *If you choose*)

From the second part of the question:

what is the chance that you will be correct?

and he gave four options:

a) 25%

b) 50%

c) 60%

d) 25%

Ask your selves, to a question you don not know the answer to, and you have these options above.

what are the percentages you will thinking of to answer the question?

for sure it is 25%,50% and 60%.

if you would be wondering what is correct choice A or choice D, then you to get your self checked.

finally the chance of choosing the correct answer from 3 choices is 33.33%

let us try another method.

you have 4 choices:

a) 25%

b) 50%

c) 60%

d) 25%

and if you try to answer it 100 times,

if the answer is 25% then 50 out of 100 it will be right>>> 50%

if the answer is 50% then 25 out of 100 it will be right>>>25%

if the answer is 60% then 25 out of 100 it will be right>>>25%

50%+25%+25% = 100

How to Find the Mean

The mean is the average of the numbers.

It is easy to calculate: add up all the numbers, then divide by how many numbers there are.

so 50+25+25= 100

100/3= 33.33%