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By way of Raymond Johnson, the best statistics multiple choice question ever written on a chalkboard. Try not to think too hard.
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%
To see it on the blackboard go to best-statistics-question-ever .
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Sunday, February 12, 2012
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50%
ReplyDeleteI saw this posted a while back and thought 50% too. I didn't think about it too hard until you posted this, Dr. Bell. Now that I have read the forums and others' logic, I am totally confused. I will think about it harder, but I hope you give your analysis on it at a later date.
ReplyDeleteI love it!
ReplyDeleteThe question is “mind boggling” because we commonly make assumptions AND when we make one that is seemingly intuitively obvious, but in fact wrong the simple answer evades us.
Since we are to make our choice randomly I choose to make mine (symbolically) blind folded. To me the values provided in the list are irrelevant; as I the fact that 2 of the answers are duplicated. So my chance of randomly picking the correct answer from the list (or any list) is equal to the percentage of correct answers in the list.
I think it's 0%.
ReplyDeleteAssume 25% is correct, but there are two options with 25%. Which make a 50% chance of picking it. That does not equal.
Assume 50% is correct, but there is only one option with 50% so you have a 25% chance of picking it. That does not equal.
Assume 60% is correct, but there is only one option with 60% so you have 25% chance of picking it. That does not equal.
This does assume that the answer must be within the multiple choice selection though. COME ON DR. BELL. Tell us what you think!
DeleteI think if you assumed that the answer did not have to be in the multiple choice then the equation would be different.
ReplyDeleteAssuming that each of the 3 choices (25%,50%,60%) had an equal opportunity to be chosen as the right answer, then I think the equation would look like this.
Also, let's assume that each value has an equal chance of being chosen as the correct answer, which is 33%.
x = The chance of selecting the value .25 out of the multiple choice which is 50%.
y = The chance of selecting the value .5 out of the multiple choice is 25%.
z = The chance of selecting the value .6 out of the multiple choice is 25%.
.33*(x) + .33*(y) + .33*(z)
Then I think you need to find the average of the averages. So...
(.33*(.5) + .33*(.25) + .33*(.25))/3 = 11%
Unknown, I must admit I did not take that possibility into consideration. Very good. The way the question is worded the correct answer is not necessarily a,b,c or d. I think you may be on to something.
ReplyDeleteHummm!
ReplyDeleteUnknown’s last post does introduce an interesting possibility. That is the assumption that each “value’ has an equal chance of being chosen as the correct answer whereas I would have assumed that the correct answer was indeed the correct answer and not a selection (variable). Interesting twist!
I often tell students that on a multiple choice exam with 4 possible answers for each question they start off with a free 25% (assuming there is no penalty for a wrong answer). If there were some questions that had more that 1 answer that would/may be counted as correct then I would suggest that their free percentage would go up. I have difficulty with the 11% suggestion.
Just because you have 3 choices doesn't mean that they are all equally likely to be chosen. The probability of choosing (randomly)25% is indeed 50%.
ReplyDeleteUnknown's second post is, I think, correct. However, once an academic always an academic so here is my extended answer.
The short answer is: Just because you throw some words together, doesn’t mean they make sense.
The medium length answer is :
It is a bit like saying: “Everything I am saying is false.” Is that true or false? If it is true then it is false. (contradiction) If it is false then it is true. (contradiction) Therefore it is neither true nor false. But - one might say – every statement is either true or false. How do we get out of that? Elementary, my dear Watson. “Everything I say is false.” is not a statement.
The long answer is: I think that our probability question is a special type of what is call an “ill posed problem”. The “question” has the “answer” as part of the question which leads you in a circle.
So let’s begin with: What exactly is the question? It is to determine the probability of guessing the probability of ... what exactly? … guessing the probability of ... guessing the probability of … .
Let’s rephrase the question: A is going to pick one of the answers at random. Leave out the distracting word “correct”. We can now talk about two numbers: The number A picked and the probability that A would pick that number. The question then is: What is the probability that: the number A picked is the same as the Probability that A would pick that number? I will list them as pairs: (the number that A picked, the probability that A would pick that number). The pairs then look like (25, 50), (50, 25), (60,25). Then the question is: What is the probability that the second number is the same as the first? The answer is zero.
That’s my story and I’m stickin’ to it.
An ill posed problem... is that like when a woman asks "Do these jeans make me look fat?"
ReplyDeleteNo, a mathematician would call that a dangerous question.
Delete