Question

Solve the follwing game theory matrix,

Find the optimal strategies for both players

	a	b	c
A	-4	2	3
B	-1	0	4
C	2	-1	-3

Answer 1

Let the probabilities of the strategies of both players be (a,b,c) and (A,B,C) and v be the value of the game.

Then by condition we have,

-4a+2b+3c = v.................(i)

-a+4c = v....................(ii)

2a-b-3c = v...................(iii)

-4A-B+2C = v...........(iv)

2A-C = v..................(v)

3A+4B-3C = v..............(vi)

a+b+c = 1...............(vii)

A+B+C = 1.........(viii)

a,b,c,A,B,C 0

Multiplying (ii) by 2 and adding with (i) we get, -3c = 3v

i.e., c = -v

From (ii) we have, a = 4c-v

i.e., a = 4(-v)-v

i.e., a = -5v

Putting this in (i) we get, -4(-5v)+2b+3c = v

i.e., 20v+2b-3v = v

i.e., 17v+2b = v

i.e., 2b = -16v

i.e., b = -8v

Now putting this values in(vii) we get,

-5v-8v-v = 1

i.e., -14v = 1

i.e., v = -1/14

Therefore, a = 5/14, b = 8/14 = 4/7, c = 1/14.

Multiplying (v) by 2 and adding with (iv) we get, -B = 3v

i.e., -B = -3/14

i.e., B = 3/14

Multiplying (v) by 3 and subtracting from (vi) we get,

-3A+4B = -2v

i.e., -3A+12/14 = 2/14

i.e., -3A = -10/14

i.e., A = 10/42

i.e., A = 5/21

Putting this in (v) we get,

2*(5/21)-C = -1/14

i.e., 10/21-C = -1/14

i.e., C = (10/21)+(1/14)

i.e., C = 23/42

Therefore, the probabilities of the strategies of both players be (5/14,4/7,1/14) and (5/21,3/14,23/42).

The value of the game is : v = -1/14.