Mathematical Conundrum or Not? Fun exercise. Here’s how I experienced it.
*read*
“Hm. Multiple choice random accuracy is 1/p for p choices, so the correct answer to this question is 1/p.”
“Oh, but 1/p shows up twice. More accurately, it’s 1/(degrees of freedom). If you know the full list of answers and you’ve picked 3, then you know what the 4th one is before you look at it, so it’s 1/(p-1)=1/3.”
“So the answer to this particular question is 1/3. But wait, no 1/3. This question’s correct answer is not listed, so there’s no chance of picking it. But wait, looks like 0% is actually the correct answer. Also not listed. The probability is still 0.”
“They tried to get me with the two 25%, tempting me to think the answer is 2*25%=50%, but that’s assuming the repeated answer choice doesn’t affect the odds. As an exercise in probability, it’s def 1/3, because we are picking at random and without assumptions about the meaning of the question. Strangely, I did have to know the meaning of the question in order to know the answer wasn’t listed, but now I’m just clarifying. My psychology does not affect the structure of the question.”
“If 0% were listed (once) alongside the two 25%, you would indeed go in circles. Finding no ‘1/3’, you revise to 0. Seeing 0 once, you revise again. Now thought processes divide. The meta view returns to 1/3, alternating with 0. The semantic view alternates between 1/2 and 1/4.”
“Does the fact that 25% is repeated change the odds? That’s double the risk of picking an answer that’s only 33% likely to be right. Well, no. The weighted average is still 33% because each hoice, being one of the three possible answers, has a 1/3 chance of winning.”
Very pleasant. Thank you.