Wow, my little quiz received a lot more attention than I thought it would. Surprisingly, of all the answers that I have read so far (over 150 between the comments on my blog and Google+), I only spotted three that match exactly my reasoning, which in itself, has to be a statistical anomaly.
Anyway, enough waiting, here is my interpretation of the puzzler.
This is a meta quiz: a question about a question. The first step to figure it out is to solve the “inner” question, which is determining what “right answer” or means. In other words, we need to answer “If you have a choice between four options and you pick randomly, what are the odds you’ll get the right answer?”.
This is a very general question that can be answered easily: 1 in 4. 25%.
Armed with this knowledge, we now focus on the “outer” question, which we can rephrase as follows:
If you choose an answer to this question at random, what is the chance you will pick 25%?
Looking at the choices, we notice that 25% appears twice among four choices, so you will be right 50% of the time.
Therefore, the answer is 50%, right?
It’s B! You’re asked to pick a choice between A, B, C and D, not a percentage. Some teachers will actually fail you if your answer doesn’t typecheck
I think this is the answer that the creator of this quiz expected, however, the more I thought about it, the more I started to realize that there was a crack in the reasoning. Can you spot it?
It’s the answer to the question above:
If you have a choice between four options and you pick randomly, what are the odds you’ll get the right answer?
The answer will be 25% only if the answer is present exactly once among the options. Not zero, not two, three or four.
Do you think the meta quiz obeys this constraint? After all, 25% appears twice in the options, right? Yes, except that the correct answer is not 25% but 50%, which appears exactly once, so the quiz seems to be consistent with this caveat.
Exercise left to the reader: can you rephrase the original quiz to close this loophole?