Example 31:
Solve the following game by equal gains method:
Y
I
II
I
4
1
X
II
2
3
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
p1
4
1
X
II
1-p1
2
3
If player Y selects strategy I the pay off of X will be:
4p1 + 2(1-p1)
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 – 2p1 = p1 + 3 – 3p1
4p1 – 2p1 – p1 + 3p1 = 3 – 2
So,
p1 = ¼ , (1-p1) = ¾
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
4
3
X
3
II 2
q1
(1-q1)
If player X selects strategy I the pay off over Y will be :
4q1 + 1(1-q1)
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 – q1 = 2q1 + 3 – 3q1
4q1 + 3q1 – q1 – 2q1 = 3 – 1
So,
q1= ½ ,
(1-q1) = ½
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)}
Therefore ,
{(4p1 + 2(1-p1)q1} + {[1p1 + 3(1-p1)](1-q1)}
{(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 .
[4x¼x½]–[1x½]–[2x¼]+3
½ - ½ - ½ + 3 = ( -1+ 6 )/2
Value of Game = 5/2 Ans.
Probability of Mr. X to select strategy :
I = 1/4
II = 3/4
Probability of Mr. Y to select strategy :
I = 1/2
II = 1/2