If you were to pick an answer at random, what are the chances you would pick the correct answer?
a) 20%
b) 40%
c) 25%
d) 0%
e) 20%
My answer in the next post.
This should be a poll.
Very clever. The correct answer, is of course Not 20% – Not 40% – but 0% – no that makes it 20%
When I first heard this yesterday, my initial thought was “this is stupid.” There’s no real question being asked and so any one of these answers could be correct or incorrect. But 100% wasn’t an option, so the correct answer is probably 0% depending on the inclination of the person asking the question.
But since 0% is an actual choice, then choosing that number automatically invalidates the correctness of that choice.
So then, on a random basis, you have a 1/5 chance of choosing the correct number. That’s 20%. But since there’s two choices for 20%, the odds are really 2/5 or 40%. But 40% is only listed once and if that’s the correct answer then the answer goes back to 20%.
It’s possible to combine the 20% and 20% and eliminate one choice altogether, thus getting 25% or 1/4 as the correct answer. But that’s really not how odds work and so we can eliminate that option.
So that leaves the two 20% and the 40% as both being right and not right simultaneously. Similarly, but for a different reason, 0% is right and not right simultaneously.
Since we can eliminate the 25% as obviously wrong, that means we have a 4/5 chance of choosing both the correct and not correct answer. That brings us back to an average of 40%.
So I choose 40% by the sheer law of averages. And my final thought on this logic problem is “this is stupid.”
To make this more competitive, anyone whose final score is in the bottom half should be given compensatory points sufficient to place them in the top 10%.
I think this question is an elaborate version of “this statement is false”. In fact choosing (d) seems very like that statement. Choosing (d) doesn’t make it correct. Therefore it is the correct answer. Therefore it is wrong. Etc. But the other answers except 25% are self-contradictory in similar ways.
I think this is unanswerable myself.
There is a 20% chance of choosing 0%
a 20% chance of choosing 40%
a 20% chance of choosing 25%
a 40% chance of choosing 20%
Logically the answer has to be 20% or 40%,
so if we remove 0 + 25
you will pick 20% 40% of the time, thus invalidating it as the answer, and 40% 20% of the time, thus invalidating it as the answer.
Does this sound right to anyone? My head hurts.
First, try this simple example:
If you were to pick an answer at random, what are
the chances you would pick the correct answer?
a) 100%
b) 100%
c) 100%
d) 100%
e) 100%
Well, that’s easy: the answer is 100%, and there is no self contradiction. It’s a bit like:
A. Statement B is true.
B. Statement A is true.
But then try this:
If you were to pick an answer at random, what are
the chances you would pick the correct answer?
a) 0%
b) 0%
c) 0%
d) 0%
e) 0%
That needs some thinking. You have a 100% chance of choosing an answer saying “0%”, but that’s the wrong answer unless it’s the right answer. So that’s like:
A. Statement B is false.
B. Statement A is false.
In other words, if you create a system with circular references to truth, or probability of truth, it’s easy to set up internal self-contradiction. The system in the OP is just a more complex way of doing it.