# How Do You Solve 3×3 Game Theory?

//

Jane Flores

Game theory is a fascinating subject that has practical applications in various fields such as economics, politics, and biology. In this article, we will explore the game theory problem of solving a 3×3 matrix game.

Firstly, let us understand what a 3×3 matrix game is. In game theory, a matrix game is a two-player game where each player has a finite set of strategies to choose from.

A 3×3 matrix game is one where each player has three possible strategies to choose from. The outcome of the game depends on the strategies chosen by both players.

Now, let’s move on to solving the 3×3 matrix game. The first step in solving a matrix game is to construct the payoff matrix. The payoff matrix shows the payoffs for each player for every combination of strategies they can choose.

Payoff Matrix:

Player A\Player B | Strategy 1 | Strategy 2 | Strategy 3 |
— | — | — | — |
Strategy 1 | (a11, b11) | (a12, b12) | (a13, b13) |
Strategy 2 | (a21, b21) | (a22, b22) | (a23, b23) |
Strategy 3 | (a\_31\_\>, b\_31\_\>)| (a\_32\_\>, b\_32\_\>)| (a\_33\_\>, b\_33\_\>)|

In the above payoff matrix, aij represents the payoff for Player A and bij represents the payoff for Player B when Player A chooses Strategy i and Player B chooses Strategy j.

Once we have constructed the payoff matrix, we can solve the game by using different methods such as graphical method or algebraic method. In this article, we will use the algebraic method to solve the game.

The first step in solving the game using algebraic method is to find the expected payoff for each player for every strategy. The expected payoff is calculated by taking a weighted average of all possible outcomes of the game where each outcome is weighted by its probability.

Expected Payoff:

Player A\Player B | Strategy 1 | Strategy 2 | Strategy 3 |
— | — | — | — |
Strategy 1 | (a11+a21+a\_31\_\>)/3 | (a12+a22+a\_32\_\>)/3 | (a13+a23+a\_33\_\>)/3 |
Strategy 2 | (a11+a21+a\_31\_\>)/3 | (a12+a22+a\_32\_\>)/3 | (a13+ a < sub > 23 < / sub > + a < sup > _33 )/3 |
Strategy 3 | (a11+a\_31\_\>+a21)/3| (a12+a\_32\_\>+a22)/3| (a13+a\_33\_\>+a23)/3|

Similarly, we can calculate the expected payoff for Player B for every strategy by taking a weighted average of all possible outcomes of the game where each outcome is weighted by its probability.

Player A\Player B | Strategy 1 | Strategy 2 | Strategy 3 |
— | — | — | — |
Strategy 1 | (b11+ b < sub > _12 < / sub > + b < sub > _13 < / sub > )/3 | (b < sub > _21 + b < sub > _22 + b < sub > _23 )/3 | (b < sup > _31 + b < sup > _32 + b < sup > _33 )/3 |
Strategy 2 | (b < sub > _11 + b < sub > _12 + b < sup > _13 )/3| (b < sub>_21
+b_22+b_23)/3| (b_31+b_32+b_33)/3|
Strategy 3| (b_11+b__13+b_12)/3| (b_21+b__23+b_22)/3| (b_31+b__33+b_23)/3|

After calculating the expected payoffs, we can determine the optimal strategy for each player. The optimal strategy is the one that maximizes the expected payoff for each player.

Optimal Strategy:

Player A:
– If (a11+a21+a\_31\_\>)/3 > (a\_31\_\>+a21+a\_31\_\>)/3 and (a11+a21+a\_31\_\>/3) > (a\_32\_\>+ a < sub > 23 < / sub > + a < sup > _33 )/3, then Player A should choose Strategy 1.
– If (a _ 32 + a < su p > _ 33 + a < sub > 23 )/3>( a < sub >­11 ­­+ a < sub >­12 ­­+ a < sup >­13 ­­ )/3 and ( a < su b >­32 ­­+ a < su p >­33 ­­+ a < sub >­23 ­­ )/3 > ( a < su p >_31 + a < sub >21 + a < su p >_32)/3, then Player A should choose Strategy 2.
– If (a\_31\_\>+a21+a\_31\_\>/3)>(a­32­­+a­33&sh y;­+a23)/3 and (a\_31\_\>+a21+a\_31\_\>/3)>(a_13+b_22+b__33)/3, then Player A should choose Strategy 3.

Player B:
– If (b_< sub >11 + b < sup > _21 + b < sup > _31 /3)>(b_< sup >_13+b_< sub >22+b_< sub >23/3) and (b_< sub >11 +b_< sup>_12 +b_< sup>_13 ) / 3 >(b _ < sub>31 +b _ < su p>_32 + b _ < su p>_33 ) / 3, then Player B should choose Strategy 1.
– If (b_< sub >21 + b < sup > _22 + b < sup > _23 /3)>(b_< su p>_13+b_< sub >22+b__33/3 ) and (b_21+b_22