Introduction to Game Theory

In this document we provide a couple of exercises that help understand the basic concepts of game theory, including pure and mixed strategies, zero sum and non-zero sum games, and saddle-point and Nash solutions. The definition for concepts and the ideas for these exercises were taken in part from [1, 2]. Before we start, let us introduce some notation. We are going to assume that we have only two “players” that we call “decision-makers.” These players are denoted by P1 and P2, and the players have cost functions J1 and J2, respectively. The cost functions represent the gains or losses of each player according to the decision that they make. In this case P1 (P2) wants to minimize J1 (J2), so we think of J1 (J2) as quantifying losses (cost). We assume that both players make their decisions at the same time and with full information. We let θ and θ denote the “decision variables” of P1 and P2 respectively, and we suppose that θ has a finite number of Dk decisions (i.e., θ ∈ {1, 2, . . . , Dk}). For convenience, we use θ = i and θ = j for P1 and P2, respectively. Each decision will give a cost of J1(ij) and J2(ij). In the two player case it is convenient to represent each of these matrices as D1 ×D2 matrices. We are going to denote these “cost matrices” as J ij 1 and J ij 2 , respectively. If J ij 1 ≤ 0 this represents that P2 pays the associated cost to player 1, if P1 plays strategy i and P2 plays strategy j. A “bimatrix” game is a game constituted by two (D1 ×D2) matrices J1 and J2 with each pair of entries (J ij 1 , J ij 2 ) denoting the outcome of the game corresponding to the decision made by the two players. A special type of game is the well known “zero sum” game, in which the cost matrices are defined so that J ij 1 + J ij 2 = 0. If the sum of that game were equal to a constant c, we can redefine the cost functions such that the constant c is incorporated in the cost, and we will obtain J ij 1 + J ij 2 = 0. It is clear then that in a zero sum game player 2 gets all the cost of P1 and vice versa. Hence, while P1 wants to minimize J1, P2 tries to maximize J1.