Game Theory problem using Matrix method Method & Example-1

1. Method & Example-1

We use cookies to improve your experience on our site and to show you relevant advertising. By browsing this website, you agree to our use of cookies. Learn more

Support us

Home > Operation Research calculators > Game Theory >> Saddle Point example

7. Matrix method example ( Enter your problem )

( Enter your problem )

Method & Example-1
Example-2

Other related methods

6. Arithmetic method
(Previous method)

2. Example-2
(Next example)

Player `B`

Player `A`

Player `A`

Method

Step-1:	Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[p_1,p_2]]`
Step-2:	Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[q_1,q_2]]`
Step-3:	Value of the game = (Player A's optimal strategies) `xx` (Payoff matrix P) `xx` (Player B's optimal strategies) `V = [[p_1,p_2]] xx P xx [[q_1],[q_2]]`
	Here `P_(Cof)` = Cofactor matrix of P and `P_(Adj)` = Adjoint matrix of P

Example-1
Find Solution of game theory problem using matrix method
_{Player A}\^{Player B} B1 B2 B3
A1 1 7 2
A2 6 2 7
A3 5 1 6

Solution:
1. Saddle point testing
Players

		Player `B`
		`B_1`	`B_2`	`B_3`
Player `A`	`A_1`	1	7	2
	`A_2`	6	2	7
	`A_3`	5	1	6

We apply the maximin (minimax) principle to analyze the game.

		Player `B`
		`B_1`	`B_2`	`B_3`	Row Minimum
Player `A`	`A_1`	1	7	2	`1`
	`A_2`	(6)	[2]	7	`[2]`
	`A_3`	5	1	6	`1`
	Column Maximum	`(6)`	`7`	`7`

Select minimum from the maximum of columns
Column MiniMax = (6)

Select maximum from the minimum of rows
Row MaxiMin = [2]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.

2. Dominance rule to reduce the size of the payoff matrix
Using dominance property

		Player `B`
		`B_1`	`B_2`	`B_3`
Player `A`	`A_1`	1	7	2
	`A_2`	6	2	7
	`A_3`	5	1	6

row-3 `<=` row-2, so remove row-3

		Player `B`
		`B_1`	`B_2`	`B_3`
Player `A`	`A_1`	1	7	2
Player `A`	`A_2`	6	2	7

column-3 `>=` column-1, so remove column-3

reduced matrix

For this reduced matrix, calculate `P_(Adj)` and `P_(Cof)`

`P_(Adj) = [[2,-7],[-6,1]]`

and `P_(Cof) = [[2,-6],[-7,1]]`

Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]])`

`=([[1,1]][[2,-7],[-6,1]])/([[1,1]][[2,-7],[-6,1]][[1],[1]])`

`=([[-4,-6]])/(-10)`

`=[[2/5,3/5]]`

`p_1=2/5` and `p_2=3/5`, where `p_1` and `p_2` represent the probabilities of player A's, using his strategies `A_1` and `A_2` respectively.

Similarly,
Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]])`

`=([[1,1]][[2,-6],[-7,1]])/([[1,1]][[2,-7],[-6,1]][[1],[1]])`

`=([[-5,-5]])/(-10)`

`=[[1/2,1/2]]`

`q_1=1/2` and `q_2=1/2`, where `q_1` and `q_2` represent the probabilities of player B's, using his strategies `B_1` and `B_2` respectively.

Hence, Value of the game `V` = (Player A's optimal strategies) `xx` (Payoff matrix `P_(ij)`) `xx` (Player B's optimal strategies)

`V=[[2/5,3/5]][[1,7],[6,2]][[1/2],[1/2]]=4`

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here