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I am studying the following example but don't understand how the solution works:

  • Consider the following bimatrix game:

    $$ (A, B) = \begin{pmatrix} 4 & 2 & 0 & 0 \\ 0 & 0 & 1 & 3 \end{pmatrix} $$

  • There are two obvious pure Nash equilibria: ((p, q) = ((1, 0), (1, 0))) and (((0, 1), (0, 1))).

  • The game also has a mixed Nash equilibrium. Suppose the row player uses strategy (x = (x, 1 - x)) with (0 \leq x \leq 1). Then: $$ xB = (2x, 3 - 3x) $$ Setting (2x = 3 - 3x), we find (x = 0.6), and the corresponding mixed Nash equilibrium is ((0.6, 0.4)).

Why is the solution of the example set equation for (xB) and (Ay^T), rather than (xA) and (By^T)? Can anyone explain?

Here are the definitions my book used so far:

In a bimatrix game, two players (player I and player II) choose their strategies simultaneously. The payoffs to the players depend on the strategies used by both players. Unlike zero-sum games, there is no assumption on the sum of payoffs to the players. A bimatrix game can be represented by two matrices, denoted as (A) and (B).

  1. Definition of Bimatrix Game:

    • The normal form of a bimatrix game is given by a pair of (m \times n) matrices ((A, B)).
    • Matrix (A) represents the payoffs for the row player (player I), and matrix (B) represents the payoffs for the column player (player II).
    • Suppose the row player uses strategy (x \in P_m) and the column player uses strategy (y \in P_n). The payoff to the row player and column player are given by the following functions:
      • Row player payoff: (\pi(x, y) = xAy^T)
      • Column player payoff: (\rho(x, y) = xBy^T)
  2. Safety Levels:

    • The safety level (or security level) of the row player is defined as: [ \mu = \max_{x \in P_m} \min_{y \in P_n} xAy^T = \nu(A) ] where (\nu(A)) denotes the value of matrix (A) when considered as the game matrix of a two-person zero-sum game.
    • The safety level of the column player is defined as: [ \nu = \max_{y \in P_n} \min_{x \in P_m} xBy^T = \nu(B^T) ] where (\nu(B^T)) is the value of the transpose of matrix (B).
  3. Nash Equilibrium:

    • A pair of strategies ((p, q)) is considered a Nash equilibrium (or mixed Nash equilibrium) for the bimatrix game ((A, B)) if the following conditions hold:
      • (xAq^T \leq pAq^T) for any (x \in P_m)
      • (pBy^T \leq pBq^T) for any (y \in P_n)
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  • $\begingroup$ I do not understand the notation. If the payoff to the row player $A = \begin{pmatrix} 4 & 2 \\ 0 & 0 \end{pmatrix}$, how can the row player choosing $1$ (the bottom row) rather than $0$ ever be part of a Nash equilibrium. $\endgroup$
    – Henry
    Commented May 19 at 21:11

2 Answers 2

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The row player's payoff is given by $$\pi(x, y) = xAy^T$$, where $x$ represents the row player's mixed strategy and $y$ represents the column player's strategy.

Similarly, the column player's payoff is given by $$\rho(x, y) = xBy^T$$.

By convention, we express the row player's strategy first (i.e., $x$) and then multiply it with the matrix representing the opponent's strategy (i.e., $B$ or $A$).

So, we use $xB$ for the row player's payoff and $Ay^T$ for the column player's payoff.

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The row player aims to maximize their expected payoff $\pi(x, y)$ by choosing an optimal mixed strategy $x$. Similarly, the column player aims to maximize their expected payoff $\rho(x, y)$ by choosing an optimal strategy $y$. The Nash equilibrium occurs when both players have no incentive to unilaterally change their strategies, given the opponent’s strategy.

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  • $\begingroup$ Sorry I am not familiar with how the $xB$ and $Ay^T$ comes out, how you relate to it with your answer? $\endgroup$
    – user1101956
    Commented May 19 at 3:55
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    $\begingroup$ Sorry but I think you are not answering the question, instead you are just restating the definition. $\endgroup$
    – user1055322
    Commented May 19 at 4:00

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