#OK to post homework
#Victoria Chayes, 2/16/2022, Assignment 8

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#2 Prove that (Defect,Defect) is always the unique Nash Equilbria, (including mixed one) of the version of Prisoner's dillema (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.

# We write the bi-matrix in terms of T,R,P, and S as 
#  [ R, R ]  [ S, T ]
#  [ T, S ]  [ P, P ]
# and as designed, T>R>P>S. Then [R,R] cannot be pure Nash Eq., because defecting strictly increases payoff for each player; [S,T] cannot be pure Nash Eq., because defecting strictly increases payoff for Player Row; and [T,S] cannot be pure Nash Eq., because defecting strictly increases payoff for Player Column. Similarly, even if Player Row chooses a random strategy, Player Column always has larger payoff choosing defect, and vice versa, so even introducing randomness does not change this. 

# This is especially tragic for T=R+0.01, R=10000P, S=P-0.01 because if the players cooperated, they would be 10000x richer than the [Defect, Defect] state, and being the "sucker" while your partner chooses to defect is almost no decrease from that, and choosing the temptation to defect when your partner cooperates has almost no noticeable increase, and yet the Equilibrium value is still that both players lose. 

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#3 Prove that in the version of the Snwodrift 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).

# The same bi-matrix 
#  [ R, R ]  [ S, T ]
#  [ T, S ]  [ P, P ]
# represents the outcomes, but now we have T>R>S>P. [R, R] and [P, P] are clearly not equilibrium because at least one player increases their payoff via changing their answer; [S, T] and [T, S] are pure equilibrium because in this state, neither player can increase their payoff by by switching. 

# For the mixed Nash equilibrium, the joint payoffs for player 1 choosing cooperate with probability p1 and player 2 cooperate with probability p2 is
# [P*p1*p2 + R*p1*p2 - S*p1*p2 - T*p1*p2 - P*p1 - P*p2 + S*p1 + T*p2 + P, P*p1*p2 + R*p1*p2 - S*p1*p2 - T*p1*p2 - P*p1 - P*p2 + S*p2 + T*p1 + P]
# We can then take the p1 derivative of the first and p2 derivative of the second and set each to 0, to determine for an arbitrary probability that the other player chooses, what probability maximizes payoff for the other player:
# diff(P*p1*p2 + R*p1*p2 - S*p1*p2 - T*p1*p2 - P*p1 - P*p2 + S*p1 + T*p2 + P,p1)
#               P p2 + R p2 - S p2 - T p2 - P + S
# diff(P*p1*p2 + R*p1*p2 - S*p1*p2 - T*p1*p2 - P*p1 - P*p2 + S*p2 + T*p1 + P, p2)
#               P p1 + R p1 - S p1 - T p1 - P + S
# solve({p1=P*p2 + R*p2 - S*p2 - T*p2 - P + S, p2=P*p1 + R*p1 - S*p1 - T*p1 - P + S},{p1,p2})
#                   P - S                     P - S       
#        p1 = -----------------,   p2 = ----------------- 
#             P + R - S - T - 1         P + R - S - T - 1
#
# Note that as P>S and T>R, both p1 and p2 are in [0,1]

# The following programs are useful for testing: 
 
Snowdrift:=proc(T,R,S,P)
[[[ R, R ] , [ S, T ]],[[ T, S ] , [ P, P ]]]:
end:

RandSnowdrift:=proc(pmax) local ra,L:
ra:=rand(0..pmax):
L:=sort([seq(ra(),i=1..4)]):
[[[ L[3], L[3] ], [ L[2], L[4] ]],[[ L[4], L[2] ] , [ L[1], L[1] ]]]:
end: