0
$\begingroup$

Each student in a class has the option to remain silent or ask the professor to cancel class. If any students asks to cancel class, all students get a payoff of $r$. However, the student that asks incurs a cost of $c < r$ so that his payoff if $r-c$. If none of the students ask, they all get a payoff of $0$.

  1. If the students adopt pure strategies, what is the pure strategy Nash Equilibria for the game?

  2. If all students adopt mixed strategies (i.e. student $i$ asks with probability $p_i$ for $i = 1,...,n$), is there a mixed strategy Nash Equilibria for the game?

I would appreciate if someone could briefly explain each part. I'm not sure how to deal with the case of arbitrary $n$. Every example of NE I've seen has a table with underlined values, and choosing the one with all values underlined represents a Nash Equilibria.

$\endgroup$
  • $\begingroup$ What have you tried? Using the basic definition of a Nash equilibrium, you should be able to tackle part 1. How far have you got, even with the case of $n = 2$? $\endgroup$ – Simon S Dec 8 '14 at 4:18
0
$\begingroup$

a) There are exactly $n$ Nash Equilibria, where $n$ denotes the number of students. The equilibria are for student $i$ to ask the professor to cancel, and the remaining $-i$ students to stay silent.

So suppose a second student asked the professor to cancel class. Then exactly one of these two students could unilaterally change his action (to staying silent) and eliminate the cost, thus benefiting. So no more than one student can ask the professor to cancel class. Similarly, if nobody asks the professor to cancel class, a single student can unilaterally deviate and gain profit $r - c > 0$.

b) We seek to determine the probability of asking the professor to cancel class. So each pure strategy of player $i$, $s_{i} \in S_{i}$, must have equal payoff against $\sigma^{*}_{-i}$. So the probability only player $i$ calls is $(1 - (1-p)^{n-1})$ (that is, $1$ minus the probability that no other player calls). The game is symmetric, so the probability is the same for all players. So the equation of interest is: $r(1 - (1-p)^{n-1}) = r - c$. This yields $p = 1 - (\frac{c}{r})^{\frac{1}{n-1}}$. So player $i$ asks with probability $p$ and does not ask with probability $1-p$.

$\endgroup$
  • $\begingroup$ How do you go about determining if there is a mixed Nash Equilibria from these probabilities in part (b), or does the fact that we have this probability show that the a mixed Nash Equilibria exists? $\endgroup$ – Carley Dec 8 '14 at 22:13
  • $\begingroup$ The mixed-strategies Nash Equilibrium is to ask the professor with proportion $p$, and not ask with proportion $1-p$. By symmetry, everyone else will be adhering to the same strategy. The derivation of $p$ maximizes the expected payoff, and so one cannot unilaterally benefit by changing strategy. $\endgroup$ – ml0105 Dec 8 '14 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.