3

I'm trying to determine the probability that three students cheated on a recent exam.

In a nutshell: What are the odds that three students sitting next to each other in a class of 22 could select (incorrectly) the same three colors (of 12 possible colors)?

Additional detail on the variables: Class size = 22, spread out in 2 rows of varying sizes. The three students in question got the same three questions wrong and the same 8 questions correct.

Thanks for your consideration.

Ollennjj
  • 31
  • 2
    Did they each have to select three colors or only one color? Also can you share how everyone else did? How many choices were there for the questions? – Patrick Oct 18 '19 at 15:05
  • They each had to select three colors as a response to one of the incorrect answers. Only a few got the color wheel question correct but nobody else chose the same three color combination. It was a humanities test so outside of the color question the answer possiblities were less fixed. – Ollennjj Oct 18 '19 at 15:06
  • Thanks for that. I added some other questions to my comment: Also can you share how everyone else did? How many choices were there for the questions? – Patrick Oct 18 '19 at 15:10
  • 1
    this is impossible to calculate exactly because the odds of selecting one of 12 colors are not 1/12 – Metariat Oct 18 '19 at 15:11
  • Most students performed better, Patrick. – Ollennjj Oct 18 '19 at 15:14
  • Exactness is not needed, Materiat, just want an approximation of the odds. – Ollennjj Oct 18 '19 at 15:16
  • The point Metariat raises is a good one and the problem is this: if one of those colors was a popular color that lot's of people picked then it might not be that unlikely. So one way to protect against this is to look at what colors the other students picked and see if they picked them roughly equally or not. – Patrick Oct 18 '19 at 15:24
  • 2
    Detecting 'cheating' on multiple-choice exams is a tricky business. Usual rules of independence for computing probabilities seldom apply. One often observes curious coincidences in real life.Of course, students do cheat. And armed with cell phones, they don't have to sit next to each other to do it. Avoid multiple choice tests, if possible. (MC question sort of implies there's a uniquely correct A, often not the case--even in a statistics course.) If you must have them, maybe make several versions with Qs and their A/s put in different orders. – BruceET Oct 18 '19 at 16:14
  • ...Maybe see this Q&A or google 'Birthday paradox'. If applicable: $1/{12 \choose 3} = 1/220.$ Also, you have ${22 \choose 3} = 1540$ possible triads of 'potential cheaters' in the class. 'Suspicious coincidence' and 'preponderance of evidence' are far apart. – BruceET Oct 18 '19 at 16:40
  • I have defended students and employees accused of cheating on standardized tests in disputes to be settled by courts, arbitrators, and university committees. In all cases one of the principal defenses was that some common characteristic of the defendants caused them to produce a larger than expected rate of IIRs (identical incorrect responses). The literature on this goes back 90 years. Statisticians have consistently pointed out that such probability calculations provide exaggerated evidence and they do not reflect the probability of cheating, which is a mistaken interpretation. – whuber Oct 19 '19 at 14:05
  • I may have created confusion in the way I described the scenario, this was not a multiple choice exam. The particular question that threw the red flag asked a color theory question that requires a response of a three color combination. There are 12 colors to select the combination. So to simplify, bare bones, without any additional factors or considerations, and assuming no bias towards any color, what are the chances of three students in a group of 22 to pick the same three of twelve color combination? I'm clear that this raw number inflates the probability of collaboration. – Ollennjj Oct 20 '19 at 02:48

0 Answers0