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OSBot :: 2007 OSRS Botting

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

If you choose an answer to this question at random, what is the chance that you will be correct? 16 members have voted

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

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Featured Replies

Use the poll system.

 

Do not open the spoiler before you voted!

 

 This is a fun question whose paradoxical, self-referential nature quickly reveals itself – A) seems to be fine until one realizes the D) option is also 25%.

A quick search reveals hundreds of discussion contributions of this problem, for example here and here and from a year ago. People often appear very confident that their answer is the only possible solution.

I am no logician and so unqualified to place this within the grand structures of mathematical paradoxes. I have not waded through all the discussions and so there may be something I have missed, but in among all the arguments there seem to be four conclusions that could be considered as 'correct'. These are my personal comments:

1) There can be no solution, since the ambiguity of ‘correct’ makes the question ill-posed.

It's true the question is ambiguous, but this still seems a bit of a cop-out. 

2) There is no solution. 

This seems to take this interpretation of the question. 

Which answer (or set of answers) of “p%”, is such that the statement ‘the probability of picking such an answer is p%’ is true? 

Then this appears to be a well-posed question, but there is no solution.

3) 0%.

Consider a different interpretation of the question. 

Is there a p%, such that the statement ‘the probability of picking an answer “p%” is p%’ is true?

Then this appears a well-posed question and has the solution p = 0, even though this is not one of the answers. Of course if answer C) were changed to “0%” (as it is in this 2007 version of the question ), then this would also have no solution.

4) We can produce any answer we want by changing the probability distribution for the choice.

Why should ‘random’ mean an equally likely chance of picking the 4 answers? If we, say, assume the probabilities of choosing (A) (cool.png © (D) to be (10%, 20%, 60%, 10%) then the answer to either formulation (2) and (3) is now “60%”. But if we make the distribution (12.5%, 15%, 60%, 12.5%) then we seem to back to square one again, since there is now both a 25% chance of picking “25%”, and a 60% chance of picking “60%”.

I like conclusion 3) best, ie 0%

Maybe the main lesson is: ambiguity and paradox are often the basis for a good joke. 

 

 

*Spoiler copied from another forum (not writte by myself)

No 21%? :doge:

Edited by FahQ

When in doubt go with C is my motto and it's done me well on tests :)

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