Give the general definition for a Prisoners dilemma.
Bookwork: https://vknight.org/gt/chapters/09/#Prisoners-Dilemma
Justify if the following games are Prisoners dilemmas or not:
$ A = \begin{pmatrix} 3 & 0\\ 5 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 3 & 5\\ 0 & 1 \end{pmatrix} $
This is a Prisoners Dilemma: $(R, S, T, P) = (3, 1, 5, 0)$.
$ A = \begin{pmatrix} 1 & -1\\ 2 & 0 \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 2\\ -1 & 0 \end{pmatrix} $
This is a Prisoners Dilemma: $(R, S, T, P) = (1, -1, 2, 0)$: $2>1>0>-1$ and $2\times 1 > 2 - 1$.
$ A = \begin{pmatrix} 1 & -1\\ 2 & 0 \end{pmatrix} \qquad B = \begin{pmatrix} 3 & 5\\ 0 & 1 \end{pmatrix} $
This is not a Prisoner's Dilemma $A \ne B ^ T$
$ A = \begin{pmatrix} 6 & 0\\ 12 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 6 & 12\\ 0 & 0 \end{pmatrix} $
This is not a Prisoner's Dilemma $A \ne B ^ T$
Obtain the Markov chain representation for a match between reactive strategies with the following vectors:
Bookwork: this is a substitution exercise using https://vknight.org/gt/chapters/09/#Markov-chain-representation-of-a-Match-between-two-reactive-strategies
$p=(1/2, 1/2)\qquad q=(1/2, 1/2)$
$$M = \begin{pmatrix} 1/4&1/4&1/4&1/4\\ 1/4&1/4&1/4&1/4\\ 1/4&1/4&1/4&1/4\\ 1/4&1/4&1/4&1/4 \end{pmatrix}$$
$p=(1/4, 1/2)\qquad q=(1/2, 1/4)$
$$M = \begin{pmatrix} 1/8&1/8&3/8&3/8\\ 1/4&1/4&1/4&1/4\\ 1/16&3/16&3/16&9/16\\ 1/8&3/8&1/8&3/8 \end{pmatrix}$$
$p=(1/3, 1/3)\qquad q=(2/3, 1/4)$
$$M = \begin{pmatrix} 2/9 & 1/9 & 4/9 & 2/9\\ 2/9 & 1/9 & 4/9 & 2/9\\ 1/12 & 1/4 & 1/6 & 1/2\\ 1/12 & 1/4 & 1/6 & 1/2 \end{pmatrix}$$
Obtain the utilities for both players for the vectors of question 3.
Bookwork: this is a substitution exercise using https://vknight.org/gt/chapters/09/#Theorem:-steady-state-probabilities-for-match-between-reactive-players
Here is some python code that also carries out these calculations:
In [1]:
import numpy as np
import itertools
def make_matrix(p, q):
"""
Code to obtain Markov chain representation of match between two reactive players.
"""
M = [[ele[0] * ele[1] for ele in itertools.product([player, 1 - player],
[opponent, 1 - opponent])]
for opponent in q for player in p]
return np.array(M)
def theoretic_steady_state(p, q):
r_1 = p[0] - p[1]
r_2 = q[0] - q[1]
s_1 = (q[1] * r_1 + p[1]) / (1 - r_1 * r_2)
s_2 = (p[1] * r_2 + q[1]) / (1 - r_1 * r_2)
return np.array([s_1 * s_2, s_1 * (1 - s_2), (1 - s_1) * s_2, (1 - s_1) * (1 - s_2)])
def theoretic_utility(p, q, rstp=np.array([3, 0, 5, 1])):
pi = theoretic_steady_state(p, q)
return np.dot(pi, rstp)
In [2]:
import sympy as sym
for p, q in [([sym.S(1) / 2, sym.S(1) / 2], [sym.S(1) / 2, sym.S(1) / 2]),
([sym.S(1) / 4, sym.S(1) / 2], [sym.S(1) / 2, sym.S(1) / 4]),
([sym.S(1) / 3, sym.S(1) / 3], [sym.S(2) / 3, sym.S(1) / 4])]:
print("=====")
print(p, q)
print("gives:")
print(make_matrix(p, q))
print("With utility:", theoretic_utility(p, q), theoretic_utility(q, p))
5. Assuming $p=(x, 1/2)$, find the optimal $x$ against the following players:
A part of this question involves bookwork: this is a substitution exercise using https://vknight.org/gt/chapters/09/#Theorem:-steady-state-probabilities-for-match-between-reactive-players. This is a substitution exercise to obtain a formula for the utility of $p$ as a function of $x$.
$q=(1, 0)$
$$u(x)=\frac{(-10x + 4(x - 1)^2 + 13)}{(2x - 3)^2}$$
The derivative of this function is given by:
$$ \frac{2(6x - 7)}{(2x - 3)^3} $$
This derivative has zero for $x=7/6$ which is $>1$. Thus the utility is monotic increasing over the interval $[0, 1]$. We have (by substitution):
$$u(0)=17/9\qquad u(1)=3$$
Thus $u(x)$ is an increasing function so the optimal value of $x$ is $1$.
Against a player that is unforgiving (reacts to defection with defection), given that our player will play randomly against a defection it is better to always cooperate.
$q=(1/2, 1/2)$
$$u(x)=-3x/4+21/8$$
This is a decreasing function so the optimal value of $x$ is $0$.
Against a random player (who takes no notice of what we do) it is better to defect.
Below is some code to verify this calculations.
In [3]:
x = sym.Symbol("x")
for q in [(sym.S(1), sym.S(0)), (sym.S(1) / 2, sym.S(1) / 2)]:
utility = theoretic_utility((x, sym.S(1) / 2), q)
print(utility.simplify(), utility.subs({x: 0}), utility.subs({x: 1}), sym.diff(utility, x).simplify(), sym.solveset(sym.diff(utility, x), x))