GAME THEORY : SOLVING 2Xn OR mX2 GAMES

When is this used

• When either player A or player B has only 2 strategies and the other player has more than 2 strategies

• No saddle point is found• 2 X n game means: player A has only 2

strategies and Player B has n strategies• m X 2 game means: player A has m strategies

and Player B has only 2 strategies

Algorithm for 2 X n game

• Step 1 : Reduce the payoff matrix by applying dominance property, if it exists

• Step 2: Let, x be the probability for Player A taking alternative 1, then (1-x) becomes the probability for alternative 2. Derive the expected gain function of player A

• Step 3 :For each gain function find values when x = 0 & x=1

• Step 4 : plot gain. Keep x value on X axis and gain on Y axis

• Step 5: Since player A is a maximin player, find the highest intersection point in the lower boundary of the graph. Let it be the maximum point

Algorithm for 2 X n game (contd..)

• Step 6 : If the no. of lines passing through the maximum point is only two, from a 2 x 2 payoff matrix from the original problem by retaining only the columns corresponding to those two lines and go to step 8, otherwise go to step 7

• Step 7: If more than two lines are passing through the maximum point, identify any two lines with opposite slopes passing through that point. Then form 2x2 payoff matrix from the original problem by retaining only the columns corresponding to those two lines which are having opposite slopes

• Step 8 :Solve the 2x2 game with oddments and find strategies for player A & player B and also the value of the game

Solve the below 2 x 5 game graphically

3 0 6 -1 7

-1 5 -2 2 1

Player B 1 2 3 4 5

1Player A 2

Solving

3 0 6 -1 7

-1 5 -2 2 1

Player B 1 2 3 4 5

1Player A 2

Row minimum-1 (Maximin)-2

Column maximum 3 5 6 2 7 (Minimax)

Maximin = Minimax. No saddle point

Checking for dominancy in column (as only 2 rows are present)

3 0 6 -1 7

-1 5 -2 2 1

Player B 1 2 3 4 5

1Player A 2

Column dominancy:Column 4 cell values are < column 2 values. So column 4 is dominant. Delete column 2

Column dominancy:Column 43 cell values are < column 5 values. So column 3 is dominant. Delete column 5

3 6 -1

-1 -2 2

After deleting column 2 & 5

Player B 1 3 4

1Player A 2

Now derive the gain function

3 6 -1

-1 -2 2

Player B 1 3 4

1Player A 2

B’s alternative A’s expected payoff function

1 3x + (-1 * (1-x) ) = 3x-1+x = 4x-1

3 6x + (-2 * (1-x) ) = 6x -2 +2x = 8x-2

4 (-1 * x ) + (2* (1-x) ) = -1x +2 – 2x = -3x +2

Now compute gain for x = 0 & x=1

B’s alternative

A’s expected payoff function

A’s expected gain

X = 0 X = 1

1 4x-1 -1 3

3 8x-2 -2 6

4 -3x +2 2 -1

Now plot the graph

Player A maximin player, shade the lowest boundary of the graph. Intersection points are a,b,c & d. Highest point is c. Optimal solution exists at c. Check the no. of lines passing thru c. 2 lines. So consider the corresponding columns (B1 & B4) to make 2X2 matrix

Now we have got 2X2 matrix and can solve through oddments

3 -1

-1 2

Player B 1 4

1Player A 2

Oddments 3 4

Oddments 3 4

Calculating probability P1 = 3/7, P2 = 4/7, q1 = 3/7 & q2 = 4/7

Calculating value V = (3*3) + (-1*4) / (3+4) = (9-4) /7 = 5/7