1. Example 31: Solve the following game by equal gains method: Y I II I II I X II II 4 1 4 1 2 3 2 3

2. Solution. Let p1 be the probability that X selects strategy I. Then (1-p1) be the probability that X selects strategy II. Y I II I II I p1 I p1 X II 1-p1 II 1-p1 If player Y selects strategy I the pay off of X will be: 4p1 + 2(1-p1) 4 1 4 1 2 3 2 3

3. Similarly, If player Y selects strategy II the pay off over X will be : 1p1 + 3(1-p1)  Since the pay off under both the situation is bound to be equal. 4p1+ 2(1-p1) = 1p1 +3(1-p1) 4p1+ 2(1-p1) = 1p1 +3(1-p1) 4p1 + 2 – 2p1 = p1 + 3 – 3p1 4p1 + 2 – 2p1 = p1 + 3 – 3p1 4p1 – 2p1 – p1 + 3p1 = 3 – 2 4p1 – 2p1 – p1 + 3p1 = 3 – 2 So, p1 = ¼, (1-p1) = ¾

4. Let q1 be the probability that Y selects strategy I. Then (1-q1) be the probability that Y selects strategy II. Y I II I II I X II II q1 (1-q1) q1 (1-q1) If player X selects strategy I the pay off over Y will be : 4q1 + 1(1-q1) 4q1 + 1(1-q1) 4 3 4 3 2 3 2 3

5. If player X selects strategy II the pay off over Y will be : 2q1 + 3(1-q1) Similarly, we can determine the pay off to Mr. Y: 4q1 + 1(1-q1) = 2q1 + 3(1-q1) 4q1 + 1(1-q1) = 2q1 + 3(1-q1) 4q1 + 1 – q1 = 2q1 + 3 – 3q1 4q1 + 1 – q1 = 2q1 + 3 – 3q1 4q1 + 3q1 – q1 – 2q1 = 3 – 1 4q1 + 3q1 – q1 – 2q1 = 3 – 1So, q1= ½, (1-q1) = ½ q1= ½, (1-q1) = ½

6. Value of Game = {(Expected pay off of player X when player Y uses strategy I) x (Probability of player Y using strategy I)} or {(Expected pay off of player X when player Y uses strategy II) x (Probability of player Y using strategy II)} {(Expected pay off of player X when player Y uses strategy II) x (Probability of player Y using strategy II)} Therefore, {(4p1 + 2(1-p1)q1} + {[1p1 + 3(1-p1)](1-q1)} {(4p1 + 2(1-p1)q1} + {[1p1 + 3(1-p1)](1-q1)}

7. {(4p1 + 2 – 2p1)q1} + {[1p1+ 3 – 3p1}](1-q1) {[2p1 + 2]q1} + [-2p1 + 3](1-q1)} 2p1q1 + 2q1 + {-2p1(1-q1) + 3(1-q1)} 2p1q1 + 2q1 – 2p1 + 2p1q1 + 3 – 3q1 4p1q1 – 1q1 – 2p1 + 3 [Equation 1] Now put the values of p1= ¼, (1-p1) = ¾, & q1= ½, (1-q1)= ½ in equation 1.

8. [ 4 x ¼ x ½ ] – [ 1 x ½ ] – [ 2 x ¼ ] + 3 [ 4 x ¼ x ½ ] – [ 1 x ½ ] – [ 2 x ¼ ] + 3 ½ - ½ - ½ + 3 = ( -1+ 6 )/2 ½ - ½ - ½ + 3 = ( -1+ 6 )/2 Value of Game = 5/2 Ans. Value of Game = 5/2 Ans.

9. Probability of Mr. X to select strategy : I = 1/4 I = 1/4 II = 3/4 II = 3/4 Probability of Mr. Y to select strategy : I = 1/2 I = 1/2 II = 1/2 II = 1/2