### Terence Tao

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A probabilistic version of the liar paradox.
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Also interesting if the answer choices were: 25, 50, 50, 100.﻿

Will it still be a paradox if all outcomes are not equally likely?﻿

0% is a correct answer; how is this different from any other multiple-choice question where none of the listed answers are right?﻿

, it's different because if you pick an answer, you change the answer. So if 0% were an option, and you picked it, the answer would be 100%.﻿

One can also close this loophole by replacing 60% with 0% in answer C.﻿

, how would that close the loophole? If you replaced 60% with 0%, the other answers would still not be correct. And if 0% is the correct answer, and you choose 0%, then you are 100% correct, so 0% cannot be the right answer.﻿

I'm with . We can easily determine that none of the answers listed is correct. Therefore we know that if your method is to choose (at random) from the options given, you will never get the correct answer. So the probability is 0%. If I had a nickel for every time I had to answer a multiple choice question on a test where none of the answers was right.....﻿

you definitely need to close the loop hole, as the self referential question does have a fixed point (0%), also the fix of replacing choice C from 60% to 0% is more elegant, as currently 60% seems to be a junk number there without any purpose.﻿

, It's still a non-standard question because the answers (as a list) are self-referential.

I would prefer the phrasing "Which answer, if chosen randomly, is the probability of choosing itself?" since "correctness" does not seem to apply to situations where more than one answer works equally well (in math).

This is akin to the question: what is the longest a list of numbers (x_1, ... , x_n) can be such that P(choosing x_i) = x_i ? The same as count(x_i) = x_i * n; so we could ask for integers, count(y_i) = y_i (think y_i = x_i * n) ? We can choose something arbitrarily long: (1, 2, 2, 3, 3, 3, <4 4's>, etc).﻿

the answer is 42 as it is for everything... sheesh. piece of cake :P﻿

Here I'm trying to do a semi-formal proof. Please correct me if I'm wrong in any step.
[Assumption1] When identifying a paradox, one normally iterates through all possible assignment of a variable in the statement.
[Fact1] Notice that the number of correct choices can only take one of the 5 value in the set {0,1,2,3,4}
[Let] the number of correct choices be Y.
[Let] the answer to the question be X. (the percent chance of randomly answering the question correct.)
[Fact2] X = Y*25
[Fact3] the choices are (25,50,60,25)
[Try Y=0] X should be 0 by Fact2. and the answer 0 didn't appear in the list.
[Try Y=1] X should be 25 by Fact2. But the answer 25 appear 2 times, hence Y = 2. contradiction.
[Try Y=2] X should be 50 by Fact2. But the answer 50 appear 1 time, hence Y = 1. contradiction.
[Try Y=3] X should be 75 by Fact2. But the answer 75 appear 0 time, hence Y = 0. contradiction.
[Try Y=4] X should be 100 by Fact2. But the answer 100 appear 0 time, hence Y = 0. contradiction.
[Conclusion]
There is one possible answer, 0.
[Take home]
I think we can generalise the choice. Would someone else do it? thanks for you attention.﻿

Some thought of mine, using standard probability theory, gives answer 0%

P(Answer Correct | Answer randomly chosen from below) = P(Answer Correct, Answer randomly chosen from below) / P(Answer randomly chosen from below) = P(Answer Correct, Answer randomly chosen from below) / P(Answer randomly chosen from below) = P(A,B) / P(B) = P(B in A) / P(B in S) [where S = the set of all answers (everything in the universe)] = [can be intuitively shown by drawing some diagrams] P(S in A) = 0﻿

I spot a hidden assumption I made with the previous proof.
[Assumption2] There is ONE correct answer.
[Note] Sure the proof includes the possibility that there is NO correct answer too, in that case the proof will conclude a paradox.
The importance of this assumption becomes clear the proof is applied to choices (25,75,75,75)