#OK to post
#Rebecca Embar, 2-16-2022, Homework 8

#2
#The game is
#G := [[[10,10],[-5,15]],
#	   [[15,-5],[0 ,0 ]]]
#By running MNE22(G), we can see that the unique Nash Equilibria is [0,0]
#For a more mathematical proof, assume player 1 chooses to donate 
#with probability x and player 2 chooses to donate with probability y
#The payoff for player 1 is P(x,y) = 10*x*y-5*x*(1-y)+15*(1-x)*y+0*(1-x)*(1-y)
#Simplifying, we get P(x,y) = 15*y-5*x
#For fixed y in [0,1], P(x,y) is maximized when x is as small as possible, i.e.,
#when x = 0
#The same computations hold for player 2. Therefore, the unique Nash Equilibria
#is for both players to play strategy 2 with probability 1, yielding outcome [0,0]

#If we reiterate the game as,
#G := [[[R,R],[S,T]],
#	   [[T,S],[P,P]]]
#and let T=R+0.01, R=10000P, S=P-0.01
#If player 1 and player 2 play with the same probibilities as before,
#the payoff for player 1 is P(x,y) = R*x*y+S*x*(1-y)+T*(1-x)*y+P*(1-x)*(1-y)
#Since R+P=T+S, this simplifies to P(x,y) = (S-P)*x+(T-P)*y+P
#Since S-P is negative, for fixed y in [0,1], P(x,y) is maximized when x is as small
#as possible, i.e., when x = 0.

#3
#(Refuse,Pay) is a pure Nash equilibria since BR1(Pay) = Refuse and BR2(Refuse) = Pay
#Same for (Pay,Refuse)
#To find a mixed Nash equlibria, assume player 1 chooses "Pay" with probability x and 
#player 2 chooses "Pay" with probability y
#The payoff for player 1 is P(x,y) = R*x*y+S*x*(1-y)+T*(1-x)*y+P*(1-x)*(1-y)
#Simplifying, we get P(x,y) = (R-T+P-S)*x*y+(S-P)*x+(T-P)*y+P
#Setting y constant, we have P'(x) = (R-T+P-S)*y+(S-P)
#Thus, player 1 is indifferent in their choice for x when P'(x) = 0, i.e., when 
#y = (P-S)/(R-T+P-S)
#Notice that this value of y is inside of [0,1], since T>R>S>P implies that P-S and R-T+P-S
#are both negative (and therefore y > 0) and 1/y = (R-T)/(P-S)+1 > 1 since R-T, P-S are both
#negative (and therefore y < 1)
#By a similar computation, we can see that player 2 is indifferent in their choice for y when
#x = (P-S)/(R-T+P-S). Thus, a mixed Nash equlibria is [(P-S)/(R-T+P-S),(P-S)/(R-T+P-S)]