Probability Question for David Sklansky

[RJT]

Quick hijack. The only question she answered differently is #11. She got this one correct and he got it wrong I assume as she got a higher score by one. She answered it as C.

Like they say when in doubt answer C. - Scary, isn't it? [img]/images/graemlins/wink.gif[/img]

[bluesbassman]

How was this test proctored?

Also, how many (if any) of the problems were the same/similar as problems on previous tests to which the students reasonably had access? If they studied together by working previous tests, it's more likely they would answer similar questions the same way. I know when I was an undergrad, I would sometimes study that way with classmates, and we would often (without cheating) achieve similar grades and tend to answer the same questions the same way.

[thylacine]

Looks like cheating/copying (may be unilateral) to me.

Check out your institution's policies. Where I am most of the suggestions in this thread are absolutely forbidden. Here you must either grade at face value, or make a formal accusation.

Talk to the appropriate person in your department, and see what they suggest.

Also, watch them like a hawk in the final exam. Repeated `coincidences' seal the deal.

[Enrique]

Cheating for sure.
I don't understand how some of the posters said "60-70%". My guess is that they have never graded exams. I'd say the probability that cheating occured is 100%, not 99.9, 100%.
I know the probabilty is < 1, but it is certainly higher than .999, probably higher than .99999, so for all practical matters it is 1. People have already given compelling reasons why it is super high (the argument for at least 1/800). I think you could argue (1/256)(1/800) by combining the arguments of PairTheBoard and the guy who said 1/800.

[vhawk01]

[ QUOTE ]
Cheating for sure.
I don't understand how some of the posters said "60-70%". My guess is that they have never graded exams. I'd say the probability that cheating occured is 100%, not 99.9, 100%.
I know the probabilty is < 1, but it is certainly higher than .999, probably higher than .99999, so for all practical matters it is 1. People have already given compelling reasons why it is super high (the argument for at least 1/800). I think you could argue (1/256)(1/800) by combining the arguments of PairTheBoard and the guy who said 1/800.

[/ QUOTE ]

Higher than .9999? That seems absurd. Heck, any reasonable interpretation of a random-guessing strategy is going to give this more than .0001% of the time.

[RoundGuy]

[ QUOTE ]
Higher than .9999? That seems absurd. Heck, any reasonable interpretation of a random-guessing strategy is going to give this more than .0001% of the time.

[/ QUOTE ]
We're talking two separate individuals, at the exact same time. Are you saying the probability of them both random guessing the exact same answers, at the exact same time, is >.0001? Or if only one of them random guessed, what is the probability the guessing person would have the exact same answers as the non-guessing person -- still >.0001?

I don't think so. But I'm no probability expert.

[vhawk01]

[ QUOTE ]
[ QUOTE ]
Higher than .9999? That seems absurd. Heck, any reasonable interpretation of a random-guessing strategy is going to give this more than .0001% of the time.

[/ QUOTE ]
We're talking two separate individuals, at the exact same time. Are you saying the probability of them both random guessing the exact same answers, at the exact same time, is >.0001? Or if only one of them random guessed, what is the probability the guessing person would have the exact same answers as the non-guessing person -- still >.0001?

I don't think so. But I'm no probability expert.

[/ QUOTE ]

No, thats why I added the caveat 'any reasonable interpretation.' Of course, if they randomly guessed for ALL of the questions, this would be far more unlikely than .0001. But it isn't unreasonable at all to grant them all (or most) of the correct answers, since those aren't going to be guesses. Even the incorrect answers aren't going to be entirely guesses, since some answers are more obviously wrong, and we don't know how many choices there were to select from.

[Borodog]

Update. I spoke to the guy briefly since I do not suspect him of copying off of her paper (he is an easy A student).

1) Her attractiveness is not an issue with him, if you get my drift.
2) They know each other, they were on the cheerleading team together at some point in the past, but do not "hang out" or study together.

Also, there are no previous tests available. I write all my tests from scratch. What can I say; I'm a pervert for writing physics problems and drawing overly elaborate diagrams.

If you were to run a computer simulation where each student had a 75% chance of getting a question right (a good assumption given the 2 grades), and if they got it wrong each incorrect answer had a 25% chance of being chosen, I think the probability of those two students independently creating identical tests (which was the state before the erasure on number 11) would be very, very low. Even if certain incorrect answers are more likely to be chosen than others on certain problems, this might increase the probability, by it would still be incredibly low.

In fact, there should be a straitforward way to calculate this, right? The chances that both students will get the same problem either right or wrong is 10/16. There are 19 problems. So the chances of them getting all problems right or wrong in lockstep is about (10/16)^19 ~ 1.3x10^-4, or almost 7,555:1 against.

This neglects the need to get the incorrect problems incorrect in the correct way, but I'm too lazy to figure that out right now.

[arahant]

[ QUOTE ]
Update. I spoke to the guy briefly since I do not suspect him of copying off of her paper (he is an easy A student).

1) Her attractiveness is not an issue with him, if you get my drift.



[/ QUOTE ]
I don't, actually.

[ QUOTE ]

2) They know each other, they were on the cheerleading team together at some point in the past



[/ QUOTE ]

OOOhhhhh...NOW I get it!

[Borodog]

This is not a rigorous way to calculate it, but what the hell. Assuming 5 incorrect responses, there is a 1/4 chance that they will get each wrong answer wrong in the same way. Hence there is approximately a (1/4)^5 chance that they will get all 5 wrong in the same way, or 1/1024.

That brings the total odds against to something in the neighborhood of 7.7 million to 1 against.

If I did that (approximately) correctly.