CoLoUr DoTZ Posted October 15, 2012 Share Posted October 15, 2012 If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25% If I were to choose, and only if I were to choose one of the below answers to a given question my chance to get it right would be 33% as there are only 3 diferent answers posible. Quote Link to comment Share on other sites More sharing options...
exotic Posted October 15, 2012 Share Posted October 15, 2012 I guess its only logical to assume that there is a second question, where did those %'s come from otherwise? The question only asks what is the chance of randomly selecting the right answer , but to which question are those answers answers to is what the question doesn't answer so we will have to assume its some phantom question . I would still stick to classic probability and not get swayed by the trickiness of the question . Also i don't get how i contradicted myself, i still stick to 33.3% Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 15, 2012 Share Posted October 15, 2012 I guess its only logical to assume that there is a second question, But there isn't. where did those %'s come from otherwise? The question only asks what is the chance of randomly selecting the right answer , but to which question are those answers answers to is what the question doesn't answer so we will have to assume its some phantom question . The question does tell us what the question is, with the word "this". The question refers to itself. I would still stick to classic probability and not get swayed by the trickiness of the question . There's nothing about my answer that contradicts classic probability. And the trickiness of the question is the whole point of the question. Also i don't get how i contradicted myself, i still stick to 33.3% This is how you contradict yourself: you say that the answer is 33.3%. Let's suppose that that is correct. Now let's suppose that you pick one of A, B, C and D at random, with any probabilities you like. If you choose A you get the answer wrong, since A says 25% but we're assuming that the answer is 33.3%. Similarly, if you choose B, C or D you get the the answer wrong, for the same reason. None of the four possible outcomes of choosing A, B, C or D at random will lead to you getting the right answer, namely 33.3%. Therefore the probability that you get the right answer is 0%. Hence the answer to the question "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25%" is 0%. But that contradicts our assumption that the answer was 33.3%. So that assumption must have been incorrect. Quote Link to comment Share on other sites More sharing options...
exotic Posted October 15, 2012 Share Posted October 15, 2012 Lets just agree to disagree on this , you are referring the question to itself - You are taking those choices itself to be the odds of selecting the right answer and when the actual chance of selecting correct randomly one of the choices isn't the same as one of the given answers you allege that it is in fact 0% or none of the above since it doesn't figure as one of the choices, while im assuming the correct answer lies in one of the choices and that there is a second question which those answers refer to. Apples and oranges. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 15, 2012 Share Posted October 15, 2012 Lets just agree to disagree on this , you are referring the question to itself - You are taking those choices itself to be the odds of selecting the right answer and when the actual chance of selecting correct randomly one of the choices isn't the same as one of the given answers you allege that it is in fact 0% or none of the above since it doesn't figure as one of the choices, That's not how my argument works. Rather, I show that each of the given answers, if correct, would lead to a contradiction and conclude that none of them is correct. Then I calculate that the correct answer is 0%. while im assuming the correct answer lies in one of the choices and that there is a second question which those answers refer to. Why would the author of the problem have used the words "this question" to refer to a different question without giving the different question? That's not how the word "this" is usually used. For that matter, if that was the author's intention then why would he or she have chosen a phantom question whose answers were probabilities, rather than colours or types of animal or something? Quote Link to comment Share on other sites More sharing options...
exotic Posted October 15, 2012 Share Posted October 15, 2012 That's not how my argument works. Rather, I show that each of the given answers, if correct, would lead to a contradiction and conclude that none of them is correct. Then I calculate that the correct answer is 0%.I was referring to your assumption argument and not to your original one. Why would the author of the problem have used the words "this question" to refer to a different question without giving the different question? That's not how the word "this" is usually used. For that matter, if that was the author's intention then why would he or she have chosen a phantom question whose answers were probabilities, rather than colours or types of animal or something?Thats a mystery.. i will have to get to the source of it and get back to you on this. Quote Link to comment Share on other sites More sharing options...
Ormion Posted October 15, 2012 Share Posted October 15, 2012 This is a famous riddle that has no answer. It's similar to the sentence below is false the sentence above is true Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 15, 2012 Share Posted October 15, 2012 This is a famous riddle that has no answer. It's similar to the sentence below is false the sentence above is true No, it isn't. The above two sentences (a variation of the liar paradox) are paradoxical because each sentence can neither be true nor fail to be true without leading to a contradiction - that is, if you suppose that the first sentence is true then that implies that it is not true, and conversely. exotic's problem isn't like that; there's a unique answer that doesn't lead to any contradiction, and that answer is 0%. It's easy enough to modify the problem so it becomes unanswerable, though; for example we could ask "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 0% D) 25%". Then there's no answer that fails to contradict itself. We could also ask something like "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 50%", in which case there are three different answers that are consistent (25%, 50% or 0%) and no way to choose between them. Quote Link to comment Share on other sites More sharing options...
Ormion Posted October 16, 2012 Share Posted October 16, 2012 I'm pretty sure the original question asks for the chance, but also requires to choose one of the following answers and neither of them is 0%, so it's a paradox. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 16, 2012 Share Posted October 16, 2012 I don't know about the original, but the version exotic posted was If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25% Quote Link to comment Share on other sites More sharing options...
Ormion Posted October 16, 2012 Share Posted October 16, 2012 Hmm... 0% is right, but I believe that the original question requires that the answer must be one of 25%,50%,75%. I've seen this riddle many times before and each time most people agreed it's a paradox. The way exotic put it (which can be the right one and I'm the one who don't remember correctly) then 0% is indeed right. Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 16, 2012 Share Posted October 16, 2012 A quick Google reveals there are a load of variants floating around the tubes. The only link I found where I can vouch for the fact that a good proportion of posters know what they're talking about was this one. The top-rated answer points out that the version given by the OP is paradoxical but that a variation similar to that given by exotic is not. Quote Link to comment Share on other sites More sharing options...
JISNEGRO Posted October 16, 2012 Share Posted October 16, 2012 A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say? Quote Link to comment Share on other sites More sharing options...
exotic Posted October 16, 2012 Share Posted October 16, 2012 It's easy enough to modify the problem so it becomes unanswerable, though; for example we could ask "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 0% D) 25%". Then there's no answer that fails to contradict itself. We could also ask something like "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 50%", in which case there are three different answers that are consistent (25%, 50% or 0%) and no way to choose between them. 0% fails to contradict itself , using your contradiction argument : Lets say you say that the answer is 0% . Let's suppose that that is correct. Now let's suppose that you pick one of A, B, C and D at random, with any probabilities you like. If you choose A you get the answer wrong, since A says 25% but we're assuming that the answer is 0 %. But if you choose C you get the answer right. One of the four possible outcomes of choosing A, B, C or D at random, in this case C will lead to you getting the right answer, namely 0%. Therefore the probability that you get the right answer is 0%. Hence the answer to the question "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 0% D) 25%" is 0%. And that doesn't contradict our assumption that the answer is 0% So that assumption must have been correct. So basically in the same vein all the given options can be correct which obviously isn't the case. Quote Link to comment Share on other sites More sharing options...
exotic Posted October 16, 2012 Share Posted October 16, 2012 A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say? He said you will sentence me to six years in prison. Quote Link to comment Share on other sites More sharing options...
Lemmiwinks Posted October 16, 2012 Author Share Posted October 16, 2012 What is the only sport where all the players agree? Quote Link to comment Share on other sites More sharing options...
Rotwang Posted October 16, 2012 Share Posted October 16, 2012 0% fails to contradict itself , using your contradiction argument : No, wrong. Lets say you say that the answer is 0% . Let's suppose that that is correct. Now let's suppose that you pick one of A, B, C and D at random, with any probabilities you like. If you choose A you get the answer wrong, since A says 25% but we're assuming that the answer is 0 %. But if you choose C you get the answer right. One of the four possible outcomes of choosing A, B, C or D at random, in this case C will lead to you getting the right answer, namely 0%. OK so far... Therefore the probability that you get the right answer is 0%. No. If C, and only C, is the right answer then the probability that you get the right answer is equal to the probability that you choose C. If you pick one of A, B, C or D with equal probability - which is what the question meant - then the probability that you choose C is 25%, not 0%. In the argument of mine you're trying to imitate (in which I assumed that the correct answer to your original problem was 33.3% and derived a contradiction), the probability of getting the right answer was 0% because none of the four possible answers that you could choose at random was equal to the correct answer. That is no longer the case in your version of the argument. Hence the answer to the question "If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 0% D) 25%" is 0%. And that doesn't contradict our assumption that the answer is 0% So that assumption must have been correct. And this is nonsense too. The (non-)fact that you can't derive a contradiction from an assumption does not imply that the assumption must be true (as I already pointed out, with the second unanswerable modification I proposed). The reason I concluded that 0% was the answer to your original problem is because every possible answer except 0% leads to a contradiction. Quote Link to comment Share on other sites More sharing options...
JISNEGRO Posted October 16, 2012 Share Posted October 16, 2012 He said you will sentence me to six years in prison. Correct! that was easy Quote Link to comment Share on other sites More sharing options...
Lemmiwinks Posted November 6, 2012 Author Share Posted November 6, 2012 Quote Link to comment Share on other sites More sharing options...
Rotwang Posted November 6, 2012 Share Posted November 6, 2012 Mary Quote Link to comment Share on other sites More sharing options...
Rotwang Posted November 6, 2012 Share Posted November 6, 2012 You are presented with 12 marbles. You are told that 11 of them are identical, but one of them is either lighter or heavier than the others. You have a pair of scales - the old-fashioned sort that will only tell you whether the items placed on one end are heavier than, lighter than or the same mass as the items placed on the other end. You are allowed to use the scales three times, and no more. How do you identify which marble is the odd one out, and whether it is lighter or heavier than the others? Oh, I forgot about this. Here's one possible solution (there are others): Suppose we label the marbles with numbers 1 to 12. For the first measurement, weigh marbles 1-4 against marbles 5-8. What happens next depends on the outcome: If 1-4 weigh the same as 5-8, that means that the odd one out is in 9-12. Weigh 1-3 against 9-11. If they weigh the same then the odd one out is 12; with your third go on the scales you can weigh 1 against 12 to learn whether 12 is lighter or heavier than the others. On the other hand, if 1-3 is lighter or heavier than 9-11, you know that the odd one out is among 9-11, and also whether it is lighter or heavier than the other marbles. With your third go on the scales, weigh 9 against 10; if they are different then you know that one of them is the odd one out, and you also know which one because you already know whether the odd one out is lighter or heavier. If 9 and 10 weigh the same then 11 is the odd one out. That covers the case where 1-4 and 5-8 weigh the same. Suppose instead they are different, and that e.g. 1-4 weigh more than 5-8 (the case where 1-4 weigh less is dealt with similarly). With your second go on the scales, weigh 1-3 and 5-6 against 9-12 and 4. If they weigh the same then the odd one out must be 7 or 8. Since 5-8 weighed less than 1-4, the odd one out must be lighter than the others, so with your final go on the scales you can weigh 7 against 8 to learn which is the odd one out. If 1-3 and 5-6 weigh more than 9-12 and 4 then the odd one out must be among 1-3 (for we already know that 9-12 are all normal; if the odd one out were 4 then it would be heavier than the rest and 1-3 + 5-6 would be lighter than 9-12 + 4, while if the odd one out were 5 or 6 then it would be lighter than the rest and 1-3 + 5-6 would be lighter than 9-12 + 4) and, since 1-4 was heavier than 5-8, it follows that the odd one out is heavier than the rest. So, with your final go on the scales weigh 1 against 2; if one is heavier than the other then that one is the odd one out, otherwise 3 is the odd one out. The final case to consider is the case where 1-3 and 5-6 weigh less than 9-12 and 4. There are two possibilities: either 4 is the odd one out, and heavier, or one of 5 or 6 is the odd one out, and lighter. Weight 5 against 6 to learn which: if they're the same then 4 is heavier, otherwise you will find out which of 5 or 6 is ligher. Quote Link to comment Share on other sites More sharing options...
Lemmiwinks Posted November 7, 2012 Author Share Posted November 7, 2012 Mary Quote Link to comment Share on other sites More sharing options...
ouroboros Posted November 17, 2012 Share Posted November 17, 2012 the probability of guessing the correct answer to any question is either 100% or 0%. it doesnt matter how many options of answers you are given to choose from. there are only 2 possibilities...the guessed answer is either correct or it isnt. the percentage is an unknown variable untill the guess is made. Quote Link to comment Share on other sites More sharing options...
ouroboros Posted November 17, 2012 Share Posted November 17, 2012 If you were to choose one of the below answers to this question at random, what would be the chance that you got it right? A) 25% B ) 50% C) 75% D) 25% ok let me try. there is no answer for this question. initialy there is a 25% chance of randomly picking a b c or d. so one might say since there are 4 options (and we are assuming one is correct) then the answer is 25%. seeing as how we are being asked what is the percentage and not which option (a b c or d) is correct, and there are 2 options for 25%...that changes the "correct" anser to 50%...2 out of 4. once that happens there is only one option for 50% which brings the actual chances down to 25%...which then makes it 50%...which makes it 25%....ad infinitum. Quote Link to comment Share on other sites More sharing options...
ouroboros Posted November 18, 2012 Share Posted November 18, 2012 ok let me try. there is no answer for this question. initialy there is a 25% chance of randomly picking a b c or d. so one might say since there are 4 options (and we are assuming one is correct) then the answer is 25%. seeing as how we are being asked what is the percentage and not which option (a b c or d) is correct, and there are 2 options for 25%...that changes the "correct" anser to 50%...2 out of 4. once that happens there is only one option for 50% which brings the actual chances down to 25%...which then makes it 50%...which makes it 25%....ad infinitum. nevermind....im my attempts to not be pursueded by other answers i didnt pay attention to the initial question....the answer is 0%. rotwang is right. the question states "if you were to choose" it doesnt say that you actually have to choose one of those. doh! Quote Link to comment Share on other sites More sharing options...
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