There is simply no solution. If you choose an answer as “correct” it is shown to be not correct, (not a 25% chance to choose 25%, not a 50% chance to choose 50%, not a 0% chance to choose 0%) showing there is no correct answer.
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Triumph@fedia.io 6 hours ago
All right, let's break this down.
For this question, for these available answers, choosing at random:
First glance says the correct percentage is 25%. There are four answers, you get to pick one.
However, two of the answers are 25%. This means you have a 50% chance of picking an answer that's right.
Which makes the correct answer C: 50%.
But there's only one answer that's "50%", so you have a 25% chance of picking that one.
Which makes the correct answer "25%", so you have a 50% chance of picking that one.
If we consider that "two answer" equation, we can then consider the correct percentage to be "37.5%" -- halfway between 25% and 50%. That makes the correct answer from the available answers to be B: 0%.
And you have a 25% chance of picking that one at random. So we go back to the beginning, where the correct percentage is 25%.
I think we need to get Matt Parker on this one.
TowardsTheFuture@lemmy.zip 4 hours ago
Triumph@fedia.io 4 hours ago
There's always a solution. Even if it's "empty set".
TowardsTheFuture@lemmy.zip 4 hours ago
Sure same thing.
Soup@lemmy.world 5 hours ago
There’s also, I think, the weird fucky option were 75% sorta works because the 25% applies to choosing 50% and 50% applies to choosing 25% which means that as long as you don’t choose 0% you’re good?
Artisian@lemmy.world 6 hours ago
They don’t say that the random answer is chosen uniformly (though that is the norm in the field). If we relax that, then we’re putting a distribution on these where we want:
P(correct with distribution (a,b,c,d)) = some value shown on A,B,C,D
I don’t see any assumption that we will pick using that distribution, so I think this avoids the recursion.
Unfortunately this has too many solutions. If you put a total of 0.25 weight on A and D, then the rest does not matter. If you put 0.5 weight on C, again the rest is irrelevant.
Triumph@fedia.io 6 hours ago
You've added details that aren't in the question. It's like asking what are the odds of rolling a "one" on a 1d4, and then saying "Well, if it's not a fair 1d4, then ..."
Artisian@lemmy.world 6 hours ago
… I… I literally talked about this. It’s the first words.
What more was needed?
Triumph@fedia.io 5 hours ago
The question is as posed. We have no indication that we should assume a different distribution of "random".