
Operations Research  Game Theory
by
Elmer G. Wiens
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Game Theory  Introduction 
Battle of the Sexes  Prisoner's Dilemma  Free Rider  Game of Chicken 
Play a Game Online 
Metagame Example
Game Theory  Introduction
Max Euwe, 1935 Chess World Champion: Strategy requires thought; Tactics require observation.
introduction  twoperson zerosum  strategy  colonel blotto game  doc holliday's revenge  graphical game solution minmax matrix theorem  linear programming solution  play matrix game  properties of solutions  snowshapley theorem references
Game theory studies situations in which parties compete, and also possibly cooperate, to influence the outcome of the parties' interaction to each party's advantage. The situation involves conflict between the participants — called players — because some outcomes favour one player at the expense of the other players. What each player obtains from a particular outcome is called the player's payoff. Each player can choose among a number of actions to influence his payoff. However, each player's payoff depends on the other players' choices. Moreover, outcomes can also be influenced by "chance occurences." (see the Wikipedia's introduction to game theory.)
A twoperson game has two players. A game in which one player wins what the other player loses is called a zerosum game. The theory of twoperson zerosum games is the foundation of more complicated games, such as games with more than two players (nperson games), and games in which the players can benefit through cooperation, with or without collusion, side payments, or binding agreements. This introduction is primarily concerned with twoperson zerosum games.
The irreconcilable, conflicting interests between the two players in a zerosum game resemble parlour games and military encounters between enemy states. Players make moves and countermoves, until the rules of engagement declare the game is ended. The rules of engagement determine what each player can or must do at each stage (the available and/or required moves given the circumstances of the game at this stage) as the game unfolds. For example, in the game rock, paper, scissors both players simultaneously make one move, with rock beating scissors beating paper beating rock. While this game consists of only one move, games like chess require many moves to resolve the conflict.
A player's strategy is defined as a plan of action that determines the player's move at each stage of the game (depending on the circumstances of the game at each stage), from the player's first move all the way to the player's final move. Different ploys for play produce different strategies. A player's tactics specify a player's "line of attack/defense"
at a particular stage of the game. Different tactics yield different strategies. Suppose, for example, players agree to play "ten games" of rock, paper, scissors, a game of ten moves. One strategy is the tactic of selecting rock, paper, or scissors with probability 1/3 at each of the ten stages. In rapid play, however, choosing moves at random is difficult. Consequently, detectable patterns of play could emerge. Another strategy is the tactic of replicating the opposing player's previous move in the 3rd, 7th, and 10th stage (titfortat tactics), and "going with intuition" at the other stages. An alternative strategy is the exhaustive strategy whereby each player's move is determined, at a given stage, on the network of all possible moves both players made in the previous stages. Whether exhaustive strategies are actually available depends on such factors as the player's memory and the rate at which the game is played.
Thus, a player's strategy provides him with a move depending on what the other player has done, and/or chance events, as the game unfolds. Different moves and/or means to choose the move in a given situation are embedded in different strategies. A game with a finite number of stages and a finite number of moves at each stage is called a finite game. A finite game has a finite number of distinct strategies available to each player. The game of chess is a finite game, despite its complexity and vast number of strategies.
The choice of a particular strategy by each player determines the payoff to each player.
Colonel Blotto Attack and Defense Game
The following game of strategy illustrates the concepts introduced above.
Colonel Blotto must defend two cities with one indivisible regiment of soldiers. His enemy, Colonel Sotto, plans to attack one city with his indivisible regiment. City I has a value of 10 units, while City II has a value of 5 units. If Colonel Sotto attacks a defended city, Sotto loses the battle and obtains nothing. If Colonel Sotto attacks an undefended city, he obtains the value of the city, while Colonel Blotto loses the value of the city. Neither Colonel knows what the other plans to do.
Normal Form.
The following tableau summarizes this situation of conflict — a twoperson, zerosum game. This representation of the game is called the normal form. It lists Sotto's strategies vertically, and Blotto's strategies horizontally. The numbers represent the gains to Sotto and the losses to Blotto, for each choice of strategy.
Attack and Defense Tableau  Colonel Blotto 
Defend City I D1  Defend City II D2 
Colonel Sotto 
Attack City I A1  0  10 
Attack City II A2  5  0 
Colonel Sotto has two "pure strategies," Attack City I and Attack City II; Colonel Blotto has two "pure strategies," Defend City I and Defend City II. Each Colonel weighs each pure strategy against his enemy's pure strategies.
Examining his options, Colonel Sotto thinks, "Blotto might defend City I, since City I is worth more than City II. Why don't I attack City II"? Conversely, Colonel Blotto thinks, "Sotto might attack City II, since he thinks I will defend City I. Why don't I defend City II"? Thinking further into the game, Sotto thinks, "If Blotto is trying to read my mind, he might defend City II. Maybe I should attack City I after all."
Each Colonel's attempt to read his enemy's mind degenerates into an infinite regression.
However, a solution concept lurks in the Colonels' "gedank" experiments. Sotto attacks each city with positive probability, while Blotto defends each city with positive probability. Furthermore, since City I is worth twice as much as City II, the probability that Blotto defends City I should be twice as great as the probability that he defends City II. Conversely, since Sotto loses in a headtohead confrontation with Blotto, the probability that Sotto attacks City II should be twice as great as the probability that he attacks City I. Both probabilities for each Colonel must addup to one. Therefore, the optimal strategy for Colonel Blotto is to defend City I with probability 2/3, and to defend City II with probability 1/3. In the meantime, the optimal strategy for Colonel Sotto is to attack City I with probability 1/3, and to attack City II with probability 2/3.
Strategies consisting of probability weighted pure strategies are called mixed strategies.
If Colonel Sotto plays the mixed strategy (Attack City I, Attack City II) = (1/3, 2/3), he expects to gain:
0 * 1/3 + 5 * 2/3 = 3 1/3 units if Blotto plays Defend City I,
10 * 1/3 + 0 * 2/3 = 3 1/3 units if Blotto plays Defend City II.
If Colonel Blotto plays the mixed strategy (Defend City I, Defend City II) = (2/3, 1/3), he expects to lose:
0 * 2/3 + 10 * 1/3 = 3 1/3 units if Sotto plays Attack City I,
5 * 1/3 + 0 * 2/3 = 3 1/3 units if Sotto plays Attack City II.
Colonel Sotto's strategy assures him of an expected gain of at least 3 1/3 units; Colonel Blotto's strategy assures him of an expected loss of not more than 3 1/3 units. The value of the game is 3 1/3 units.
Extensive Form.
The next diagram shows the extensive form, or game tree, for the Colonel Blotto Game. The blue circle represents Colonel Sotto's decision node, while the red circles represent Colonel Blotto's decision nodes. Colonel Blotto's decision nodes are contained in the yellow ellipse, because Blotto does not know which node he actually is in, since Blotto and Sotto choose their strategies simultaneously. The numbers at the green nodes are the payoffs to (Sotto, Blotto) with the strategies chosen.
The yellow ellipse is called the information set for Colonel Blotto, since Blotto just knows that he is at one or the other node in the ellipse.
The game is symmetrical in the sense that I can start the game tree with Colonel Blotto's decision node instead.
"Doc" Holliday's Revenge
On October 26, 1881, the bad blood between the Earps, Clantons, and McLaurys came to a head at the OK Corral. Billy Clanton, Frank McLaury, and Tom McLaury were killed. Doc Holliday, Virgil and Morgan Earp were injured. Miraculously, Wyatt Earp was unharmed, while the unarmed Isaac "Ike" Clanton survived by running away. Many people believed that Doc Holliday shotgunned an unarmed Tom McLaury in the back as he was attempting to flee the scene.
Ike Clanton and his friends and associates, known as the "cowboys," swore to get their revenge on the Earps and Holliday. In the ensuing months, Morgan Earp was murdered and Virgil Earp seriously wounded in an ambush. A few days later, Wyatt Earp apparently shot and killed Frank Stillwell, a Clanton associate, and another man believed involved in the ambush. Over the next few years, many more of the "cowboys" were killed.
Although close to death from tuberculosis, in 1887 Doc Holliday decides to look up Ike Clanton and to settle their differences once and for all. On June 1, 1887, Doc Holliday and J.V. Brighton corner Ike Clanton near Springerville, Arizona. Doc tells J.V. to stay out of it for the time being.
Ike and Doc have each a pistol and shotgun. Ike and Doc, spurnning their pistols in favour of their shotguns, press forward towards each other. At long range, Ike — with his cowboy background — is a better shot than Doc. At middle range, Doc — the seasoned gunfighter — outguns Ike. Both desperadoes are deadly at close range. The probabilities that either rogue will kill the other with a blast from his singleshot shotgun appear in the following table:
 Kill Probability 
Long Range  Middle Range  Close Range 
Ike Clanton  0.5  0.6  1.0 
Doc Holliday  0.3  0.8  1.0 
The payoff to Doc is 10 units if Doc survives and Ike is killed, 10 units if Doc is killed and Ike survives. Doc and Ike's strategies are:
L  blast at long range 
M  if enemy has not shot, blast at middle range; otherwise, blast at close range 
C  blast enemy at close range 
Normal Form.
Always the consummate gambler, Doc Holliday correctly computes his payoff matrix as:
Shootout Payoff  Ike's Strategies 
 I_{L}  I_{M}  I_{C} 
Doc's Strategies  D_{L}  2  7  7 
D_{M}  0  2  6 
D_{C}  0  2  0 
Furthermore, Doc realizes that strategy D_{M} dominates strategies D_{L} and D_{C}, in that his expected payoff is higher with D_{M} no matter what Ike does. If Ike is rational enough — if Ike correctly computes his payoff matrix as the negative of Doc's payoff matrix — Ike will choose strategy I_{L}, because I_{L} dominates strategies I_{M} and I_{C} provided Doc chooses strategy D_{M}.
The pair of pure strategies, the dominant strategies (D_{M}, I_{L}), is the "solution" to the game. The value of the game is 0, the payoff when Doc chooses strategy D_{M} and Ike chooses I_{L}.
The pair of strategies, (D_{M}, I_{L}), determines a saddle point of the payoff matrix. Notice that the payoff matrix entry, "0," is the largest entry in the I_{L} column and the smallest entry in the D_{M} row. A strategy pair whose payoff matrix entry is a saddle point is called an equilibrium strategy pair.
Extensive Form.
The next diagram shows the extensive form, or game tree, for Doc Holliday's Game. The blue circle represents Doc Holliday's decision node, while the red circles represent Ike Clanton's decision nodes. Ike Clanton's decision nodes are contained in the yellow ellipse, because Ike does not know which node he actually is in, since Ike and Doc choose their strategies simultaneously. The numbers at the green nodes are the payoffs to (Doc, Ike) with the strategies chosen.
The yellow ellipse is called the information set for Ike Clanton, since Ike just knows that he is at one of the three nodes in the ellipse.
The game is symmetrical in the sense that I can start the game tree with Ike Clanton's decision node instead.
Doc Holliday survived the showdown, apparently dying from tuberculosis on November 8, 1887 in Colorado. Not so funny, after all.
Graphical Solution of a 2 x n or m x 2 Game
A twoperson zerosum game with a payoff matrix with dimensions either 2 x n, or m x 2 can be solved graphically. For example, consider the game with the 2 by 4 payoff matrix:
PayOff Matrix  Player II's Strategies 
II_{1}  II_{2}  II_{3}  II_{3} 
Player I's Strategies 
I_{1}  4  3  1  3 
I_{2}  0  5  2  1 
Since this matrix does not have a saddle point, one must resort to mixed strategies for a solution. If Player I plays strategy I_{1} with probability ß, and strategy I_{2} with probability (1  ß) his expected returns, R, against each of Player II's strategies are:
II's strategy  Player I's Return 
II_{1}  4 * ß + 0 * (1  ß) = 0 + 4 * ß = R_{1} 
II_{2}  3 * ß + 5 * (1  ß) = 5  2 * ß = R_{2} 
II_{3}  1 * ß + 2 * (1  ß) = 2  1 * ß = R_{3} 
II_{4}  3 * ß + 1 * (1  ß) = 1 + 2 * ß = R_{4} 
Player I's expected return against a particular strategy of Player II is a function of ß, a straight line. The next diagram graphs these straightline functions as ß varies from 0 to 1. The probability ß is graphed along the xaxis, while Player I's return R_{i} against Player II's ith strategy is graphed along the yaxis.
Player I's "Optimal" Strategy.
If Player I sets ß = .4, Player I can expect at least a return R = 1.6 against any strategy or combination of strategies that Player II chooses to play. The (ß, R) = (0.4, 1.6) point occurs at the intersection of the R_{1} and R_{3} lines.
Player II's "Optimal" Strategy.
Since Player's I's mixed strategy (0.4, 0.6) assures him of an expected gain of 1.6 units, Player II wants a strategy that will limit his expected loss to 1.6 units. Moreover, Player II can restrict his choice of strategies to II_{1} and II_{3}, because the diagram shows that strategies II_{2} and II_{4} expose him to losses greater than 1.6 units.
If Player II plays strategy II_{1} with probability µ, and strategy II_{3} with probability (1  µ) his expected losses L against Player I's mixed strategies are:
L = R_{1} * µ + R_{3} * (1  µ),
L = (4 * ß) * µ + (2  ß) * (1  µ), or
L = (5 * µ  1) * ß + (2  2 * µ)

If Player II sets µ = 1 / 5 = .2 , his expected losses are:
L = (5 * (1/5)  1) * ß + (2  2 * (1/5) ) = 8 / 5 = 1.6

Since Player II can limit his losses independently of Player I's choice of mixed strategy, Player II's mixed strategy (0.2, 0, 0.8, 0) guarantees him of an expected loss of no greater than L = 1.6.
Game Solution.
Player I plays a mixed stragey of (0.4, 0.6) and Player II plays a mixed strategy of (0.2, 0, 0.8, 0). The value of the game is 1.6 units.
The MinMax Theorem for Matrix Games
A matrix game can be described by its payoff matrix, an m by n matrix A = [a_{i, j}], the set X of all probability distributions x on the integers {1, 2, . . . , m}, and the set Y of all probability distributions y on the intergers {1, 2, . . . , n}. Thus
x^{T} = (x_{1}, x_{2}, ... , x_{m}), with x_{1} + x_{2} + ... + x_{m} = 1
y^{T} = (y_{1}, y_{2}, ... , y_{n}), with y_{1} + y_{2} + ... + y_{n} = 1

A = 
a_{1,1} a_{1,2} . . . . . a_{1,n1} a_{1,n}
a_{2,1} a_{2,2} . . . . . a_{2,n1} a_{2,n}
. . . . . . . . . . .
a_{m,1} a_{m,2} . . . a_{m,n1} a_{m,n}

If Player I selects a probability distribution x and Player II selects a probability distribution y, the payoff to Player I is P(x, y) while the payoff to Player II is P(x, y), where:
ie, the expected payoff equals the product of the row vector x^{T} and the matrix A, (another row vector), and the column vector y^{}
Lower Value: MaxMin Game.
If Player I announes that he will use a particular strategy x in X, Player II should choose the strategy y_{0} in Y that minimizes the expected payoff to Player I. That is, Player II should search for that y_{0} where:
P(x, y_{0}) = min_{y in Y} P(x, y)
Therefore, Player I should choose x_{0} in X so that:
v_{I} = min_{y in Y} P(x_{0}, y) = max_{x in X} [ min_{y in Y} P(x, y) ]
The number v_{I} is called the lower value, or the maxmin value, of the game.
Upper Value: MinMax Game.
If Player II announes that he will use a particular strategy y in Y, Player I should choose the strategy x_{0} in X that maximizes the expected payoff to Player I. That is, Player I should search for that x_{0} where:
P(x_{0}, y) = max_{x in X} P(x, y)
Therefore, Player II should choose y_{0} in Y so that:
v_{II} = max_{x in X} P(x, y_{0}) = min_{y in Y} [ max_{x in X} P(x, y) ]
The number v_{II} is called the upper value, or the minmax value, of the game.
MinMax Theorem for Matrix Games.
If x_{0} and y_{0} are the strategies chosen as indicated above, then:
v_{I} = v_{II} = v = the value of the game = P(x_{0}, y_{0})
If x is any mixed strategy in X, and y is any mixed strategy in Y, then:
P(x, y_{0}) <= P(x_{0}, y_{0}) <= P(x_{0}, y)
Thus the strategy pair (x_{0}, y_{0}) is the solution to the game. The two strategies provide a saddle point in mixed strategies in the Cartesian product domain of X x Y.
Linear Programming Solution for Matrix Games
One can conveniently illlustrate the method used to turn a matrix game into a linear programming problem with an example (Hadley, 427).
Suppose that player I has 3 strategies, player II has 5 strategies, and the payoff matrix A is given by:
P l a y e r
I  P l a y e r II 
 = A

If player I plays strategy 2, his payoff depends on which strategy player II has chosen to play. If player II knows player I is playing strategy 2, she should play her 4th strategy, because then she only pays him 1 unit. But if player I plays strategy 1 instead, she must pay him 9 units. I assume that players know the payoff matrix A, but that players 'announce' their strategies simultaneously. Player I wants to maximize the payoff; player II wants to minimize the payoff.
The MinMax Theorem suggests each player determine the choice of strategies on the basis of a probability distribution over the player's list of strategies. That is, player I chooses row 1, 2, or 3 with probabilities x_{1}, x_{2},and x_{3}. Player II chooses the columns with probabilities y_{1}, y_{2}, y_{3}, y_{4}, and y_{5}. Note that the x's and y's must be nonnegative and each set must add to one.
To turn this into a linear programming problem write:
Primal Linear Program
Maximize P = x_{4}, with x_{4} nonnegative 
3*x_{1}  + 2*x_{2}  + 1*x_{3} 
 1*x_{4}  >= 0 
5*x_{1}  + 6*x_{2}  + 7*x_{3}   1*x_{4} 
>= 0 
0*x_{1}  + 8*x_{2}  + 4*x_{3}   1*x_{4} 
>= 0 
9*x_{1}  + 1*x_{2}  + 9*x_{3}   1*x_{4} 
>= 0 
6*x_{1}  + 2*x_{2}  + 3*x_{3}   1*x_{4} 
>= 0 
1*x_{1}  + 1*x_{2}  + 1*x_{3}  + 0*x_{4} 
= 1 

where the extra variable, x_{4}, is the expected payoff (the value) of the game.
The equation,
Maximize P = x_{4}, with x_{4} nonnegative,
is the objective function. Player I wants to maximize the objective function, the expected payoff.
Then there is an equation for each column of A. Why? If player II plays her 1st strategy, the expected payoff to player I is:
3*x_{1} + 2*x_{2} + 1*x_{3}.
If player I choses his probability distribution optimally, his expected payoff should be greater than or equal x_{4}, the expected value of the game. This argument applies to each of A's columns.
The dual linear program to the primal linear program is:
Dual Linear Program
Minimize D = y_{6}, with y_{6} nonnegative 
3*y_{1}  + 5*y_{2}  + 0*y_{3} 
+ 9*y_{4}  + 6*y_{5}   1*y_{6}  <= 0 
2*y_{1}  + 6*y_{2}  + 8*y_{3} 
+ 1*y_{4}  + 2*y_{5}   1*y_{6}  <= 0 
1*y_{1}  + 7*y_{2}  + 4*y_{3} 
+ 9*y_{4}  + 3*y_{5}   1*y_{6}  <= 0 
1*y_{1}  + 1*y_{2}  + 1*y_{3} 
+ 1*y_{4}  + 1*y_{5}  + 0*y_{6}  = 1 

The equation,
Minimize D = y_{6}, with y_{6} nonnegative,
is the objective function. Player II wants to minimize the objective function, the expected payoff.
Then there is an equation for each row of A. Why? If player I plays his 1st strategy, the expected payoff to player II is:
3*y_{1} + 5*y_{2} + 0*y_{3} + 9*y_{4} + 6*y_{5}.
If player II choses her probability distribution optimally, her expected payoff should be less than or equal y_{6}, the expected value of the game. This argument applies to each of A's rows.
If one solves this linear programming problem using one of the available methods, the optimal progams:
x^{T} = (x_{1}, x_{2}, x_{3}) = (0.667, 0.333, 0) and
y^{T} = (y_{1}, y_{2}, y_{3}, y_{4}, y_{5}) = (0.889, 0, 0.111, 0, 0),
are the MaxMin and MinMax solutions, while the value, x_{4} = y_{6} = 2.667, is the value of the matrix game.
Value of the Game  Nonnegative.
To ensure that the value of an arbitrary matrix game is nonnegative when using linear programming to find its solution, add the absolute value of the most negative entry of the payoff matrix to each entry of the payoff matrix. After solving the modified game, subtract this constant from its value to get the value of the original matrix game.
You can play this matrix game using the form below. You are the row player, Player I.
Use the bonus random number to select the row you want to play. Can you can beat the value of the game over a sequence of games?
Properties of Optimal Mixed Strategies.
Suppose Player I plays his optimal strategy , x^{T} = (x_{1}, x_{2}, x_{3}) = (0.667, 0.333, 0), and Player II plays one of her pure strategies. The vector x^{T} * A shows the amount Player II can expect to pay to Player I for each pure strategy she plays.
x^{T} * A =  (2/3, 1/3, 0) * 
 = (2.6667 5.3333 2.6667 6.3333 4.6667) 
Notice that only pure strategies 1 and 3 ensure Player II of paying the minimum of 2.6667 units, when Player I plays his optimal mixed strategy.
Likewise, the vector A * y shows the amount Player I can expect to receive from Player II for each pure strategy he plays against her optimal strategy.
A * y = 
 * (0.88888889, 0, 0.11111111, 0, 0)^{T} = 
= (2.6667, 2.6667, 1.3333)^{T} 
Notice that only pure strategies 1 and 2 ensure Player I of receiving the maximum of 2.6667 units, when Player II plays her optimal mixed strategy.
Worthwhile Strategies.
Pure strategies that appear in an optimal strategy with positive probability are called worthwhile strategies (Thomas, 37).
Set of Optimal Mixed Strategies.
For each player in a twoperson zerosum matrix game, the set of optimal mixed strategies is a closed, convex set (Karlin, 36).
For each player, the optimal strategy calculated from the equivalent linear programming problem is an extreme point of the player's set of optimal strategies. This follows because in a linear programming problem, an optimal program obtains at a vertex of the feasible set, and the feasible set must contain the set of optimal strategies.
Pure Strategy as a Mixed Strategy.
A pure strategy is a mixed strategy with the component corresponding to the probability of playing the pure strategy equal to 1.
The SnowShapley Theorem for Matrix Games.
Consider again the twoperson zerosum game with payoff matrix:
This game has optimal mixed strategies:
x^{T} = (x_{1}, x_{2}, x_{3}) = (0.667, 0.333, 0) and
y^{T} = (y_{1}, y_{2}, y_{3}, y_{4}, y_{5}) = (0.889, 0, 0.111, 0, 0),
If one strikes out the rows and columns of A corresponding to strategies played with zero probability, one gets the reduced matrix M corresponding to the reduced matrix game with payoff matrix M:
This procedure of striking out "worthless strategies" produces a matrix M whose inverse M^{(1)} exists:
Furthermore, letting e^{T} = (1, 1), ie the vector whose components are all one, the value of the game, v, is given by:
v = e^{T} * M^{(1)} * e = 2.6667.
Moreover, the optimal strategies for Player I and Player II in the reduced game are:
x^{0} = v * e^{T} * M^{(1)} = (0.6667, 0.3333), and
y^{0} = v * M^{(1)} * e = (0.8889, 0.1111).
Finally, the value of the original game is also v, and the optimal strategies for Player I and Player II in the original game are obtained by setting to zero the probability of playing a worthless strategy.
SnowShapley Theorem.
Consider any twoperson zerosum matrix game with payoff matrix A. If x is an extreme point of X^{0}, Player I's set of optimal strategies, and if y is an extreme point of Y^{0}, Player II's set of optimal strategies, there exists a square, nonsingular submatrix M of A such that:
v = e^{T} * M^{(1)} * e,
x^{0} = v * e^{T} * M^{(1)},
y^{0} = v * M^{(1)} * e,
where e is the vector of the same dimension as M all of whose components equal 1, v is the value of the game, and x^{0} and y^{0} correspond to the nonzero components of x and y, respectively.
References
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 Brahms, Steven J. Biblical Games: A Strategic Analysis of Stories in the Old Testament. Cambridge, Mass: MIT Press, 1980.
 Brahms, Steven J. Theory of Moves. Cambridge: Cambridge UP, 1994.
 Dresher, Melvin. Games of Strategy: Theory and Applications. Englewood Cliffs: PrenticeHall, 1961.

Hadley G., Linear Programming. Reading, Mass.: AddisonWesley, 1962;
 Howard, Nigel. Paradoxes of Rationality: Theory of Metagames and Political Behavior. Cambridge: MIT, 1971.
 Intriligator, Michael D. Mathematical Optimization and Economic Theory. Englewood Cliffs: PrenticeHall, 1971.
 Karlin, Samuel. Mathematical Methods and Theory of Games, Programming, and Economics. Reading, Mass: AddisonWesley, 1959.
 Nash, John F. "Noncooperative Games." Annals of Mathematics. 54 (1951): 286295.
 Oxford Dictionary: The Concise Oxford Dictionary of Current English. 5th ed. Ed. H.W. Fowler and F.G. Fowler. Oxford: Oxford UP, 1964
 Owen, Guillermo. Game Theory, 2nd Edition. New York: Academic, 1982.
 Rapoport, Anatol. TwoPerson Game Theory: The Essential Ideas. Ann Arbor: U Michigan, 1966.

Restrepo, Rodrigo A. Theory of Games and Programming: Mathematics Lecture Notes. Vancouver: University of B.C., 1967.
 Thomas, L. C. Games, Theory and Applications. New York: John Wiley, 1984.
 Wiens, Elmer G. Reduction of Games Using Dominant Strategies. Vancouver: UBC M.Sc. Thesis, 1969.

