A probability paradox?

david's picture
Submitted by david on Mon, 10/31/2011 - 08:49

I recently tweeted a link to this problem drawn on a blackboard, which got a lot of retweets.

Multiple Choice: 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%

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) (B) (C) (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.

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level 1

Comments

Mark Lewney's picture

Mark Lewney (not verified)

Mon, 10/31/2011 - 12:45

I consider it a feedback loop with no stopping citerion: 'Inputting' an answer of 50% on the first iteration makes the next iteration return an answer of 25%, which in turn puts the NEXT iteration at 50% etc. etc. One might as well have asked "Is the statement that "my previous statement was true" true or false?"
Tom's picture

Tom (not verified)

Fri, 11/25/2011 - 21:35

But what if none of the answers is correct?
Bilf's picture

Bilf (not verified)

Thu, 12/29/2011 - 09:59

If the answer is the probability of getting it right, it keeps going on and on so you never know if it's right and there is no way to fid out because you have to determine whether it's correct over and over and over again.
Dave Marsay's picture

Dave Marsay

Wed, 02/01/2012 - 21:56

David, I see no paradox. The answer is surely 0%. I give my reasoning at http://djmarsay.wordpress.com/2012/02/01/uncertainty-puzzle/ . I also give a variant that would be a probability if you thought that everything that looked like a well-formed question was, and that all probabilties are numeric. I think that it is well-formed, so it must be an example of a non-numeric probability. But you may want to ask a logician.
Dave Marsay's picture

Dave Marsay

Wed, 02/01/2012 - 22:00

I meant ... would be a paradox if ..., i.e. some things that linguistically seem to be probabilities aren't.
Blitz's picture

Blitz

Wed, 05/09/2012 - 03:23

I think that if you look at the question. It asks if you randomly choose, what are the chances YOU would be correct? Well then I say it is up to me if the answer is right. So there for it is right 100% of the time. Now then 100% is not one of the A B C D answers... But does it really tell me I have to pick from those 4?
efoula's picture

efoula

Thu, 05/10/2012 - 09:08

Basically, none of the answers. The probability of getting it right is 33.33333% because there are actually three percentages, since there is twice the number 25, so it's one out of three to get it right. That is my humble opinion.
Blitz's picture

Blitz

Sun, 05/13/2012 - 18:14

I love this question because it had so many layers. First thought was of course, what they said in school. "If you don't know the answer, you have a one in four chance if you guess." But then you see the other 25%... OK now it's 50%... but wait now there is a is a 0%... but that's not one of the answers ether. Take the answers out, and answer THIS question.

Multiple Choice: If you choose an answer to this question at random, what is the chance you will be correct? A B C D

or what if the answers were A)Blue B)Yellow C)Red D)Blue

This answer implies that there is another question. But the question states "this question"

So if I choose Yellow, Red, or ether Blue, I'm right... What if the implied question was whats my favorite color?

garybird's picture

garybird

Thu, 05/17/2012 - 15:49

The question doesn't imply that any of the answers are necessarily correct. So in that sense the actual answer to the question isn't tied to the multiple choice options. An equivalent question would be: What are the chances of answering this question "what is the capital of Spain" correct. A) Paris B) London C) Tokyo D) Wellington. You would have a zero percent chance of answering the question correctly. Every option in the probability example is also wrong, therefore the answer is zero percent chance of answering correctly.
pedroAbreu's picture

pedroAbreu

Mon, 10/15/2012 - 15:14

You always have an answer to the question :
1. How many answers are right from a set of four possible answers?
It is either 0, 1, 2, 3 or 4.

You also always have an answer to the question :
2. What is the probability of randomly picking the right answer from a set of four alternatives, given a fixed number of right answers?
It is either 0%, 25%, 50%, 75%, or 100%.

In this case they’re sort of asking you both questions simultaneously, since the content of the possible answers should be fixing and allowing you to discover both
a. the number of right answers
and
b. the right answer

Once you give an answer to either 1 or 2 you immediately get an answer to the other. In this case none of those answers is actually fixed, you should discover both at the same time: hence the (apparent) problem.
The trouble is that there are very there are some (actually very strict) constrains regarding the possible relation between a and b that are not being respected by the given possible answers: hence the incoherence.
Notice that for someone to be able to pick an answer correctly the following conditions must be met:
I) There is one correct answer ←→ There is one and only one option = 25%
II) There are two correct answers ←→ There should be two and only two options = 50%
III) There are three correct answers ←→ There should be three and only three options = 75%
IV) There are four correct answers ←→ There should be four and only four options = 100%
once you don’t respect this conditions you have an incompatibility between picking the right answer and picking the number of right answers, and so there is no way to answer the question and you should not be bogged by the appearance that there should be such an answer.