In a class of three students, the teacher says an astrochemechanicalculus exam will be given next week. He warns that after the tests are graded, he will swap the highest and lowest scores. For example, a set of 0, 50, & 100 will result in the perfect scorer receiving a zero while the bum gets a 100.

Each student has a job that pays $1 per hour (they work at underground child labour sweatshops). Meanwhile, their parents will give them $1 for every point they score on the test.

Every hour the student studies, they would earn 1 point on the test. If the money is the goal, what is the optimum strategy on how long one student should study?

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TopNewestThe answer is study zero hours and work all the hundred hours.

Let the hours studied by student \( i \) be \( x_i \) and hours earning money be \( y_i \). Then \( x_i + y_i = 100 \). Now, consider the person who has the highest marks.(If two people have the highest marks, choose the one whose marks gets exchanged.) WLOG , let it be Student 1.

Now, he is at an obvious disadvantage, since his marks get exchanged for lower ones(WLOG Student 2). His earnings = \( x_1 + y_2 < 100 \). Whereas Student 2 is at an advantage as his earnings = \( x_2 + y_1 > 100 \).

Therefore, those students who study more are at a disadvantage and those who study less are at an advantage. Thus, the strategy of each student should be to study the least. i.e. 0 hours.

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Very thorough! It is the absolute Nash and anybody who studies more gets cheated out of money. Kind of an ironic way to make all students get a 0. Try to attempt the revised question. +1

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I still think the Nash equilibrium is for all of them to study 100 hours since making $100 from doing so is very much easier than 100 hours in a sweatshop.

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I disagree with this outcome although it is a possible outcome if all of the students are irrational thinkers and only act selfishly for their own good. The best outcome is for all of them to study 100 hours and earn a score of 100 each. That way no one gets cheated from their studying time.

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If money is the prime motivator, say making a $100 either from working 100 hours or studying 100 hours( getting 100 points and getting $100 reward for doing so), one is better off studying since that entails less sweat. Since all 3 students know that each know that whoever studied the least will ultimately gets the highest score, everyone of them will study just as hard, to max out their score to 100.That way, the equilibrium point is all 3 students will earn 100 marks and $100.

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Good, I agree half way although it is not a Nash Equilibrium. What if the other 2 get a 100 while 1 gets a 0? In this situation, however, everybody still gets the max net if everyone cooperates, like the farmer and the villagers game. +1

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For this to work, ALL of them has to study 100 hours, i.e. to earn a $100 from their parent. ( Assuming studying 100 hours earn you 100 marks ) . Any slight inclination of one of them ( or two or 3) to NOT follow this plan will result in the action- ALL of them will prefer not to study since the only person who chose to do so will only end up getting nothing. So it is a clear choice for all of them to stick to this plan of maximizing their study hours. It is quite clear that making a $100 from studying is much easier than doing hard labor.

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+1 again for the following situations: What is the optimal solution if there are 6 students, and this time the parents will pay double (id est $200 for a perfect score)?

I'm planning on turning it into a problem, but I need to make sure I've got it right first.

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Same strategy. It doesn't change with the number of students or the money earned by studying.

Formally, the only Nash equilibrium is when everyone doesn't study.Infact, the number of students and the factor of increase don't matter.

Proof:- Suppose there does exist a Nash equilibrium where someone studies.Let the hours studied by Student \(i \) be \(x_i \) and WLOG, \(x_1 \geq x_2 \geq \ldots \geq x_n \). and \( x_1 > 0 \).

Case 1 :- \( x_1 > x_n \). Student 1's earnings = \( (100 - x_1) + kx_n \). If Student 1 changes their strategy to studying marginally less than Student 2, their earnings = \( (100 - x_2) + kx_2 \)

This is greater as \( 100 - x_2 \geq 100 - x_1 \) and \( kx_2 \geq kx_6 \). Equality occurs when \( x_1 = x_2 = x_n \) which is not possible since \( x_1 > x_n \). Therefore, His new earnings is strictly greater than his original earnings, which is not possible, since we assumed this is a Nash equilibrium.

Case 2 :- \( x_1 = x_n \).Therefore \(x_1 = x_2 = \ldots = x_n > 0 \).

Student 1's original earnings = \( (100 - x_1) + kx_n = 100 + x_1 \). By studying 0 hours, his earnings = \( (100 - 0 + kx_2) = 100 + kx_1 \) Since \( x_1 > 0 \).we have \( 100 + x_1 < 100 + kx_1 \) Therefore, His new earnings is strictly greater than his original earnings, which is not possible, since we assumed this is a Nash equilibrium.

Thus there doesn't exist a Nash equilibrium with \( x_1 > 0 \).

Note:- I proved this for "n" students of a increase factor of "k".

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I still think making a $100 from studying is much easier than from hard labor. So, it is only rational for them to cooperate and study 100 hours so they ALL can earn a max of $100. Any slight inclination of not following this strategy ( i.e. studying less than 100 hrs or not studying at all) means the other 2 students will follow suit as well, meaning all of them would have to earn their keep the hard way!! So the best strategy is to cooperate and max out their study hours.

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They know that they must get a lowest score to get the most money, so they'll just spend 100 hrs doing the job. This will be the best solution for individuals. But I didn't say that doing job for 100 hrs is the only way.

If they can get the same points, they also get a same amount of money. So they have to talk together about this at some time I don't care.

The another weird way is do whatever they want. The total money they get is always $300. Then share some money together at some time I also don't care.

PS: I honestly don't know that what the "optimum strategy" means. Sorry about that. : <

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