Game Theory Evolving: A Problem-Centered Introduction to Modeling Strategic Interaction - Second Edition / Edition 2

Game Theory Evolving: A Problem-Centered Introduction to Modeling Strategic Interaction - Second Edition / Edition 2

by Herbert Gintis
ISBN-10:
0691140510
ISBN-13:
9780691140513
Pub. Date:
02/15/2009
Publisher:
Princeton University Press
ISBN-10:
0691140510
ISBN-13:
9780691140513
Pub. Date:
02/15/2009
Publisher:
Princeton University Press
Game Theory Evolving: A Problem-Centered Introduction to Modeling Strategic Interaction - Second Edition / Edition 2

Game Theory Evolving: A Problem-Centered Introduction to Modeling Strategic Interaction - Second Edition / Edition 2

by Herbert Gintis
$63.0 Current price is , Original price is $63.0. You
$63.00 
  • SHIP THIS ITEM
    Qualifies for Free Shipping
  • PICK UP IN STORE
    Check Availability at Nearby Stores
$63.00 
  • SHIP THIS ITEM

    Temporarily Out of Stock Online

    Please check back later for updated availability.

    • Condition: Good
    Note: Access code and/or supplemental material are not guaranteed to be included with used textbook.

Overview

Since its original publication in 2000, Game Theory Evolving has been considered the best textbook on evolutionary game theory. This completely revised and updated second edition of Game Theory Evolving contains new material and shows students how to apply game theory to model human behavior in ways that reflect the special nature of sociality and individuality. The textbook continues its in-depth look at cooperation in teams, agent-based simulations, experimental economics, the evolution and diffusion of preferences, and the connection between biology and economics.


Recognizing that students learn by doing, the textbook introduces principles through practice. Herbert Gintis exposes students to the techniques and applications of game theory through a wealth of sophisticated and surprisingly fun-to-solve problems involving human and animal behavior. The second edition includes solutions to the problems presented and information related to agent-based modeling. In addition, the textbook incorporates instruction in using mathematical software to solve complex problems. Game Theory Evolving is perfect for graduate and upper-level undergraduate economics students, and is a terrific introduction for ambitious do-it-yourselfers throughout the behavioral sciences.

  • Revised and updated edition relevant for courses across disciplines
  • Perfect for graduate and upper-level undergraduate economics courses
  • Solutions to problems presented throughout
  • Incorporates instruction in using computational software for complex problem solving
  • Includes in-depth discussions of agent-based modeling


Product Details

ISBN-13: 9780691140513
Publisher: Princeton University Press
Publication date: 02/15/2009
Edition description: Second
Pages: 408
Product dimensions: 6.90(w) x 9.90(h) x 1.20(d)

About the Author

Herbert Gintis holds faculty positions at the Santa Fe Institute, Central European University, and University of Siena. He has coedited numerous books, including Moral Sentiments and Material Interests, Unequal Chances (Princeton), and Foundations of Human Sociality.

Read an Excerpt

Game Theory Evolving

A Problem-Centered Introduction to Modeling Strategic Interaction
By Herbert Gintis

Princeton University Press

Copyright © 2009 Princeton University Press
All right reserved.

ISBN: 978-0-691-14051-3


Chapter One

Probability Theory

Doubt is disagreeable, but certainty is ridiculous.

Voltaire

1.1 Basic Set Theory and Mathematical Notation

A set is a collection of objects. We can represent a set by enumerating its objects. Thus,

A = {1, 3, 5, 7, 9, 34}

is the set of single digit odd numbers plus the number 34. We can also represent the same set by a formula. For instance,

A = {x|x [member of] N [conjunction] (x < 10 [conjunction] x is odd) [disjunction] (x = 34)}.

In interpreting this formula, N is the set of natural numbers (positive integers), "|" means "such that," "[member of]" means "is a element of," [conjunction] is the logical symbol for "and," and [disjunction] is the logical symbol for "or." See the table of symbols in chapter 14 if you forget the meaning of a mathematical symbol.

The subset of objects in set X that satisfy property p can be written as

{x [member of] X|p(x)}.

The union of two sets A, B [subset] X is the subset of X consisting of elements of X that are in either A or B:

A [union] B = {x|x [member of] A [disjunction] x [member of] B}.

The intersection of two sets A, B [subset] X is the subset of X consisting of elements of X that are in both A or B:

A [intersection] B = {x|x [member of] A [conjunction] x [member of] B}.

If a [member of] A and b [member of] B, the ordered pair (a,b) is an entity such that if (a, b) = (c, d), then a = c and b = d. The set {(a, b)|a [member of] A [conjunction] b [member of] B} is called the product of A and B and is written A x B. For instance, if A = B = R, where R is the set of real numbers, then A x B is the real plane, or the real two-dimensional vector space. We also write

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

A function f can be thought of as a set of ordered pairs (x, f(x)). For instance, the function f(x) = [chi square] is the set

{(x, y)|(x, y [member of] R) [conjunction] (y = [chi square])}

The set of arguments for which f is defined is called the domain of f and is written dom (f). The set of values that f takes is called the range of f and is written range (f). The function f is thus a subset of dom(f) x range(f). If f is a function defined on set A with values in set B, we write f : A -> B.

1.2 Probability Spaces

We assume a finite universe or sample space [OMEGA] and a set X of subsets A, B, C, ... of [OMEGA], called events. We assume X is closed under finite unions (if [A.sub.1], [A.sub.2], ... [A.sub.n] are events, so is [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]), finite intersections (if [A.sub.1], ..., [A.sub.n] are events, so is [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]), and complementation (if A is an event so is the set of elements of [OMEGA] that are not in A, which we write [A.sup.c]). If A and B are events, we interpret A [intersection] B = AB as the event "A and B both occur," A [intersection] B as the event "A or B occurs," and [A.sup.c] as the event "A does not occur."

For instance, suppose we flip a coin twice, the outcome being HH (heads on both), HT (heads on first and tails on second), TH (tails on first and heads on second), and TT (tails on both). The sample space is then [OMEGA] = {HH, TH, HT, TT}. Some events are {HH, HT} (the coin comes up heads on the first toss), {TT} (the coin comes up tails twice), and {HH, HT, TH} (the coin comes up heads at least once).

The probability of an event A [member of] X is a real number P[A] such that 0 [less than or equal to] P[A] [less than or equal to] 1. We assume that P[[OMEGA]] = 1, which says that with probability 1 some outcome occurs, and we also assume that if [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], where [A.sub.i] [member of] X and the {[A.sub.i]} are disjoint (that is, [A.sub.i] [intersection] [A.sub.j] = 0 for all i [not equal to] j), then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], which says that probabilities are additive over finite disjoint unions.

1.3 De Morgan's Laws

Show that for any two events A and B, we have

[(A [intersection] B).sup.c] = [A.sup.c] [intersection] [B.sup.c]

and

[(A [intersection] B).sup.c] = [A.sup.c] [intersection] [B.sup.c].

These are called De Morgan's laws. Express the meaning of these formulas in words.

Show that if we write p for proposition "event A occurs" and q for "event B occurs," then

not (p or q) [??] (not p and not q);

not (p and q) [??] (not p or not q).

The formulas are also De Morgan's laws. Give examples of both rules.

1.4 Interocitors

An interocitor consists of two kramels and three trums. Let [A.sub.k] be the event "the kth kramel is in working condition," and [B.sub.j] is the event "the jth trum is in working condition." An interocitor is in working condition if at least one of its kramels and two of its trums are in working condition. Let ITLITL be the event "the interocitor is in working condition." Write ITLITL in terms of the [A.sub.k] and the [B.sub.j].

1.5 The Direct Evaluation of Probabilities

Theorem 1.1 Given [a.sub.1], ..., [a.sub.n] and [b.sub.1], ..., [b.sub.m], all distinct, there are n x m distinct ways of choosing one of the [a.sub.i] and one of the [b.sub.j]. If we also have [c.sub.1], ..., [c.sub.r], distinct from each other, the [a.sub.i] and the [b.sub.j], then there are n x m x r distinct ways of choosing one of the [a.sub.i], one of the [b.sub.j], and one of the [c.sub.k].

Apply this theorem to determine how many different elements there are in the sample space of

a. the double coin flip

b. the triple coin flip

c. rolling a pair of dice

Generalize the theorem.

1.6 Probability as Frequency

Suppose the sample space [OMEGA] consists of a finite number n of equally probable elements. Suppose the event A contains m of these elements. Then the probability of the event A is m/n.

A second definition: Suppose an experiment has n distinct outcomes, all of which are equally likely. Let A be a subset of the outcomes, and n(A) the number of elements of A. We define the probability of A as P[A] = n(A)/n.

For example, in throwing a pair of dice, there are 6 x 6 = 36 mutually exclusive, equally likely events, each represented by an ordered pair (a, b), where a is the number of spots showing on the first die and b the number on the second. Let A be the event that both dice show the same number of spots. Then n(A) = 6 and P[A] = 6/36 = 1/6.

A third definition: Suppose an experiment can be repeated any number of times, each outcome being independent of the ones before and after it. Let A be an event that either does or does not occur for each outcome. Let [n.sub.t](A) be the number of times A occurred on all the tries up to and including the [t.sup.th] try. We define the relative frequency of A as [n.sub.t](A)/t, and we define the probability of A as

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

We say two events A and B are independent if P[A] does not depend on whether B occurs or not and, conversely, P[B] does not depend on whether A occurs or not. If events A and B are independent, the probability that both occur is the product of the probabilities that either occurs: that is,

P[A and B] = P[A] x P[B].

For example, in flipping coins, let A be the event "the first ten flips are heads." Let B be the event "the eleventh flip is heads." Then the two events are independent.

For another example, suppose there are two urns, one containing 100 white balls and 1 red ball, and the other containing 100 red balls and 1 white ball. You do not know which is which. You choose 2 balls from the first urn. Let A be the event "The first ball is white," and let B be the event "The second ball is white." These events are not independent, because if you draw a white ball the first time, you are more likely to be drawing from the urn with 100 white balls than the urn with 1 white ball.

Determine the following probabilities. Assume all coins and dice are "fair" in the sense that H and T are equiprobable for a coin, and 1, ..., 6 are equiprobable for a die.

a. At least one head occurs in a double coin toss.

b. Exactly two tails occur in a triple coin toss.

c. The sum of the two dice equals 7 or 11 in rolling a pair of dice.

d. All six dice show the same number when six dice are thrown.

e. A coin is tossed seven times. The string of outcomes is HHHHHHH.

f. A coin is tossed seven times. The string of outcomes is HTHHTTH.

1.7 Craps

A roller plays against the casino. The roller throws the dice and wins if the sum is 7 or 11, but loses if the sum is 2, 3, or 12. If the sum is any other number (4, 5, 6, 8, 9, or 10), the roller throws the dice repeatedly until either winning by matching the first number rolled or losing if the sum is 2, 7, or 12 ("crapping out"). What is the probability of winning?

1.8 A Marksman Contest

In a head-to-head contest Alice can beat Bonnie with probability p and can beat Carole with probability q. Carole is a better marksman than Bonnie, so p > q. To win the contest Alice must win at least two in a row out of three head-to-heads with Bonnie and Carole and cannot play the same person twice in a row (that is, she can play Bonnie-Carole-Bonnie or Carole-Bonnie-Carole). Show that Alice maximizes her probability of winning the contest playing the better marksman, Carole, twice.

1.9 Sampling

The mutually exclusive outcomes of a random action are called sample points. The set of sample points is called the sample space. An event A is a subset of a sample space [OMEGA]. The event A is certain if A = [OMEGA] and impossible if A = 0 (that is, A has no elements). The probability of an event A is P[A] = n(A)/n([OMEGA]), if we assume [OMEGA] is finite and all [omega] [member of] [OMEGA] are equally likely.

a. Suppose six dice are thrown. What is the probability all six die show the same number?

b. Suppose we choose r object in succession from a set of n distinct objects [a.sub.1], ..., [a.sub.n], each time recording the choice and returning the object to the set before making the next choice. This gives an ordered sample of the form ([b.sub.1], ..., [b.sub.r]), where each [b.sub.j] is some [a.sub.i]. We call this sampling with replacement. Show that, in sampling r times with replacement from a set of n objects, there are [n.sub.r] distinct ordered samples.

c. Suppose we choose r objects in succession from a set of n distinct objects [a.sub.1], ..., [a.sub.n], without returning the object to the set. This gives an ordered sample of the form ([b.sub.1], ..., [b.sub.r]), where each [b.sub.j] is some unique [a.sub.i]. We call this sampling without replacement. Show that in sampling r times without replacement from a set of n objects, there are

n(n - 1) ... (n - r + 1) = n!/(n - r)!

distinct ordered samples, where n! = n x (n - 1) x ... x 2 x 1.

1.10 Aces Up

A deck of 52 cards has 4 aces. A player draws 2 cards randomly from the deck. What is the probability that both are aces?

1.11 Permutations

A linear ordering of a set of n distinct objects is called a permutation of the objects. It is easy to see that the number of distinct permutations of n > 0 distinct objects is n! = n x (n - 1) x ... x 2 x 1. Suppose we have a deck of cards numbered from 1 to n > 1. Shuffle the cards so their new order is a random permutation of the cards. What is the average number of cards that appear in the "correct" order (that is, the kth card is in the kth position) in the shuffled deck?

1.12 Combinations and Sampling

The number of combinations of n distinct objects taken r at a time is the number of subsets of size r, taken from the n things without replacement. We write this as [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. In this case, we do not care about the order of the choices. For instance, consider the set of numbers (1,2,3,4}. The number of samples of size two without replacement = 4!/2! = 12. These are precisely {12,13,14,21,23,24,31,32,34,41,42,43}. The combinations of the four numbers of size two (that is, taken two at a time) are {12,13,14,23,24,34}, or six in number. Note that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. A set of n elements has n!/r!(n - r)! distinct subsets of size r. Thus, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

1.13 Mechanical Defects

A shipment of seven machines has two defective machines. An inspector checks two machines randomly drawn from the shipment, and accepts the shipment if neither is defective. What is the probability the shipment is accepted?

1.14 Mass Defection

A batch of 100 manufactured items is checked by an inspector, who examines 10 items at random. If none is defective, she accepts the whole batch. What is the probability that a batch containing 10 defective items will be accepted?

1.15 House Rules

Suppose you are playing the following game against the house in Las Vegas. You pick a number between one and six. The house rolls three dice, and pays you $1,000 if your number comes up on one die, $2,000 if your number comes up on two dice, and $3,000 if your number comes up on all three dice. If your number does not show up at all, you pay the house $1,000. At first glance, this looks like a fair game (that is, a game in which the expected payoff is zero), but in fact it is not. How much can you expect to win (or lose)?

1.16 The Addition Rule for Probabilities

Let A and B be two events. Then 0 [less than or equal to] P[A] [less than or equal to] 1 and

P[A [union] B] = P[A] + P[B] - P[AB].

If A and B are disjoint (that is, the events are mutually exclusive), then

P[A [union] B] = P[A] + P[B].

(Continues...)



Excerpted from Game Theory Evolving by Herbert Gintis Copyright © 2009 by Princeton University Press. Excerpted by permission.
All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.
Excerpts are provided by Dial-A-Book Inc. solely for the personal use of visitors to this web site.

Table of Contents

Contents

Preface....................xv
1 Probability Theory....................1
2 Bayesian Decision Theory....................18
3 Game Theory: Basic Concepts....................32
4 Eliminating Dominated Strategies....................52
5 Pure-Strategy Nash Equilibria....................80
6 Mixed-Strategy Nash Equilibria....................116
7 Principal-Agent Models....................162
8 Signaling Games....................179
9 Repeated Games....................201
10 Evolutionarily Stable Strategies....................229
11 Dynamical Systems....................247
12 Evolutionary Dynamics....................270
13 Markov Economies and Stochastic Dynamical Systems....................297
14 Table of Symbols....................319
15 Answers....................321
Sources for Problems....................373
References....................375
Index....................385

What People are Saying About This

Michael Suk-Young Chwe

There is a great need for a problem book in game theory, and Gintis is correct to emphasize that learning game theory requires the solving of problems.

From the Publisher

"Mathematically rigorous, computationally adroit, rich in illuminating problems, and engagingly written, Game Theory Evolving will be invaluable to students and researchers across the social sciences."—Joshua M. Epstein, Brookings Institution and Santa Fe Institute

"An inspiring introduction to the potential and various applications of game theory."—Jan Ekman, Ecological Economics

From the B&N Reads Blog

Customer Reviews