#OK to post homework
#Yuxuan Yang, Feb 17th, Assignment 8


#2
#When T>R>P>S,
#suppose player 1 uses probability [p1,1-p1] and player 2 uses probability [p2,1-p2], then 
#the expectation of player one is E1:=p1*p2*R+p1*(1-p2)*S+(1-p1)*p2*T+(1-p1)*(1-p2)*P.
#E1 gives maximum at p1=0 for any fixed p2, since T>R and P>S.
#Similarly, E2:=p1*p2*R+p1*(1-p2)*T+(1-p1)*p2*S+(1-p1)*(1-p2)*P gives maximum at p2=0 for any fixed p1,
#which comes from T>R and P>S.
#Note that until now all we need is "T>R and P>S" to give a unique NE. It is tragic because additionally we have R>P.
#It makes the NE payoff [P,P] is worse than another payoff [R,R].

#3
#[1,2] is NE because S=M[1][2][1]>M[2][2][1]=P and T=M[1][2][2]>M[1][1][2]=R,
#[2,1] is NE because T=M[2][1][1]>M[1][1][1]=R and S=M[2][1][2]>M[2][2][2]=P.
#For mixed NE, consider the derivative of E1:=p1*p2*R+p1*(1-p2)*S+(1-p1)*p2*T+(1-p1)*(1-p2)*P with respect of p1.
#It is a linear function of p2 and we let it equal to 0. Then p2=(P-S)/(R-S-T+P).
#Similarly we need p1=(P-S)/(R-T-S+P). 
#This is a mixed NE since each of two players has a fixed expectation when changing his own strategy.