Metagame it. Instead of ensuring that the largest number of students get credit, ensure that none do for fairness. Bell curve adjustment of grades will nullify the question.
I was thinking since it is the average not median of the group in question, somebody could try and game it by just reporting a reaaaaly large number. they wont be accurate but will be closest to the final answer. now if everyone thinks this...
They won't be the closest. If there are 5 people, 4 write 10 and 1 writes 510 then the average is 550/5 or 110. Which is much closer to 10 than it is to 510.
right, I see how I missed that. but the silver lining is that one outlier would have made everyone else seem very close to the final answer (% wise). this means if the professor is grading on a curve the whole class gets a 'C' except the outlier. sort of like 'Jesus died for your sins' ;-)
My initial hunch was just "write down 10, don't overthink it", and was pleasantly surprised to find that I would pass the exam if plover was the one evaluating it.
The truly strategic player would do research on the final exam ahead of time, pool together class resources and just hire someone to sign up for the course to take the loss.
Unfortunately, I think this is not an equilibrium strategy. An equilibrium strategy profile is one where, given everyone is using that strategy, no single individual cannot profitably deviate. But if I know everyone else in the class is going to use this strategy, and I draw the short straw (and assuming for simplicity that nobody else can draw the short straw), then I know everybody else will answer 10, and I maximize my chance of winning by choosing x such that ((n-1)10 + x) / n) + 10 = x.
But I wonder if this or a similar strategy would provide the maximum expected reward for the group*. In that case, then there's a situation that is somewhat like a multi-player prisoner's dilemma, in that if everybody always "cooperated", they'd all be better off in the long run. Yet whoever draws the short straw is better off not cooperating. Therefore nobody cooperates.
In the case of repeated one-on-one games (where you play the same game with the same player multiple times), there is a solution to the prisoner's dilemma: strategies like "tit for tat" where you can essentially punish people for non-cooperation (as described in fantastic classic "The Evolution of Cooperation" by Robert Axelrod). But since there are multiple people in the class, this is more like a a multi-player prisoners dilemma (as described by Thomas Schelling in "Micromotives and Macrobehavior"), or a "collective action" problem. These can also be solved if you go "outside" the game and have systems for punishing people who don't cooperate (e.g. union members beating up scabs).
Another thing about Schelling points: a "strategy" can be a Schelling point. It is unlikely that the whole class would independently do the same analysis and come up with the same strategy (even if it was an equilibrium strategy). But if you stood up before the class and explained the strategy before the exam, it would become a Schelling point.
Maybe a lot of people will think most will use similar strategies like 30 because it’s 10 more than 20 so you would go and say 40 if you actually don’t respect your peers intelligence, but if you do you may want to go around 50
An unethical life pro would make a class group to discuss the consensus & the make a separate smaller class group of 'inside people' who'd use previous conclusion to adjust their answers .. then .. turtles all the way down with diminishing returns.
but everyone has to cooperate here, if a few guys go crazy and say some super outlandish number to shift the average then all bets are off. Hence my most common pet-peeve with data .. median not average.
If everyone follows this strategy, then I will win if exactly one person draws the short straw and if that one person isn't me. The former has a probability that rapidly approaches 1/e≈36.8% as n increases
Why is "if exactly one person draws the short straw" probabilistic at all? I assume that's what "the former" means?
Each person samples a private random variable to decide if they have drawn the short straw. To use a shared sample would violate the no-communication assumption.
Because you're in an exam, and you cannot really draw a straw. So you have to choose a number randomly, and pretend you chose the short straw if it's a 0. And then hope you're the only one.
The article also discussed how it'll work if you prepare beforehand and actually can draw a straw. Spoiler: you might want to cheat if you chose the short straw :)
Using the phrase "draw the short straw" is a little confusing here, because there are 0-n short straws in this scenario. He is describing the probability that there is exactly one short straw (which isn't just 1/n+1 because the straws are not all "present" at once... it's just not a perfect analogy).
Suppose there are two students. Each has a 50% chance of drawing the short straw (by picking either 0 or 1). Therefore, there is a 25% chance that neither will draw the short straw, 25% chance that both will draw the short straw, and 50% chance that only one will draw the short straw.
That is what the former means. It's probabilistic because we assume (or at least hope) everyone is doing as I do, rolling an n-sided die to decide if they will write -10(n-1) instead of writing 10.
