# Please do not post
# Daniela Elizondo
# February 20, 2022
# Assignment 8

read "C7.txt":
# From hw 2:
# IsNash(G,a1,a2) returns true if [a1,a2] is a Nash Equilibrium, and false otherwise

IsNash:= proc(G,a1,a2): member(a1,BR1(G,a2)) and member(a2,BR2(G,a1)): end:

# PureNashEqui(G) returns a (possibly empty) set of pairs [a1,a2] that are pure Nash Equilibria

PureNashEqui:= proc(G) local a1,a2, NE: 
NE:={}: 
for a1 from 1 to nops(G) do
 for a2 from 1 to nops(G[1]) do 
  if IsNash(G,a1,a2) then 
   NE:= NE union {[a1,a2]}: 
 fi: 
 od: 
 od: 
 NE: 
end:

# End hw 2

# 2. Prove that (Defect,Defect) is always the unique Nash Equilibria, (including mixed one) of the version of Prisoner's dilemma (called the "donation game" in Sigmund's book) Show that if T=R+0.01, R=10000P, S=P-0.01, In (1.1)) how tragic it is. 

DonationGame := [[[R,R], [S,T]], [[T,S], [P,P]]]:

# Under these assumptions, DonationGame is the Prisoner's Dilemma:
assume(T>R, R>P, P>S);

PureNashEqui(DonationGame);
#                             {[2, 2]}
MNE22(DonationGame);
#                             {[0, 0]}

# Defect is strategy 2 for both players, so this tells us that (Defect, Defect) is always the unique Nash equilibrium in the Prisoner's dilemma. 

# If T=R+0.01, R=10000P, S=P-0.01, then both players would still have a payoff of P because [P,P] is the payoff in the Nash equilibrium. But, if they both cooperated they could both have a payoff of R=1000P. Unfortunately they both choose not to cooperate in order to avoid the payoff P-0.01 which is barely less than their payoffs in the equilibrium.


# 3. Prove that in the version of the Snowdrift game in section 1.4, with T>R>S>P, (Refuse,Pay) and (Pay,Refuse) are two pure Nash equilibria. Either by hand, or using Maple, prove that in addition there is a mixed Nash equilibrium. Find an explicit expression for that additional Nash equilibrium, in terms of the parameters T,R,S,P, and check it for random numerical values using MNE(G). 

# We switch the assumptions on T, R, P, and S to those of the Snowdrift game:

assume(T>R, R>S, S>P):
PureNashEqui(DonationGame);
#                           {[1, 2], [2, 1]}
# This confirms that the two pure Nash equilibria are (Refuse, Pay) and (Pay, Refuse).

MNE22(DonationGame);
# {[0, 1], [1, 0], [(P - S)/(P + R - S - T), (P - S)/(P + R - S - T)]}
# This confirms that there is an additional mixed Nash equilibrium. We save the additional equilibrium below:

snowdriftMixed := (P,R,S,T) -> [(P - S)/(P + R - S - T), (P - S)/(P + R - S - T)]:

# We now find this equilibrium for different values of P,R,S, and T.

snowdriftMixed(10,8,6,4);
#                       [1/2, 1/2]


snowdriftMixed(10,9,3,1);
#                       [7/15, 7/15]


snowdriftMixed(10,7,6,0); 
#                       [4/11, 4/11]