Pure and mixed strategy in Nash Equilibria with n player I got confused when I see the following problem:
There are n staffs and they want to raise their salary, if any one or more than one of these staffs suggest their boss they want raise salary, all staffs' salary will be raised by r, but whoever suggest the raise will be punished by cost c ( c < r ). If no one suggests their boss to raise salary, then salary end up the same.
Pure strategy
To find pure strategies Nash Equilibria, this is how I approach:
If more than one staff suggest raise, then those who didn't suggest get r, and those who suggest raise get r-c. So those who suggests raise can benefit by changing to not suggesting raise, and their result change from r-c to r as long as there's still one staff who is suggesting raise.
Now we come to a situation where there's one last person left who suggests raise and his result is r-c while every one else is r. But he can't get better result by not suggesting raise since if he doesn't suggest raise, there's no raise. (r - c > 0)
My question is: Does this mean this problem has n Nash equilibria since this hero staff can be any one of them?   And if r - c < 0, there's only one NE that is no one suggest raise?
Mixed Strategy
The problem remains the same, but each staff adapt mixed strategies, i.e. staff i suggest raise with probability  $p_i$.  How do I get Nash Equilibrium this time?
I think since the problem is symetric, so the mixed strategy NE should have $$p_i=p_1=p_2=...=p_n$$
And probability for a staff to get raised salary shall be
$$P(staff_i) = 1-(1-p)^n$$
But I'm not sure how to continue from here...Any help is appreciated.
 A: There is a related thread here: What is the pure strategy Nash Equilibria of asking your professor to cancel class?
1) Yes, there are $n$ Nash equilibria. Your reasoning is correct. Though I would also argue that suppose nobody asks for a raise. Then a single player can ask for a raise and better his outcome. Thus, a beneficial unilateral deviation is possible, so the outcome of nobody asking for a raise is not a Nash equilibrium. You have to argue all possible cases to prove that the Nash equilibria you have found are the only such equilibria.
2) So far, it looks good. Nice observation on the symmetry argument. The goal is to maximize the Expected value. So $r(1 - (1-p)^{n-1}) = r - c$. Then solve for $p$ to get $p = 1 - (\frac{c}{r})^{\frac{1}{n-1}}$.
So the mixed strategies Nash equilibrium would be for player $i$ to ask with proportion $p$ and not ask with proportion $1-p$. Because you have solved for the optimal value of $p$, a player cannot unilaterally increase his payoff by deviating from this strategy.
