# I Game Theory: Strategy for game with non-square payoff matrix

1. Apr 29, 2016

### Wminus

Hi, suppose two players are a playing a game with a non-square payoff matrix, like for example this one:
.....a.......b...
A: (1,3) (1,0
B: (0,0) (2,1)
C: (3,1) (0,3)

How would one go about finding an optimal mixed strategy for something like this? I mean, if this was a 3x3 matrix then one could find an optimal mixed strategy for each player by defining it as optimal when it gives equal expected payoff regardless of the opponent's choice, however this isn't possible when it's 3x2.

The only thing I can think of is by eliminating one of the rows (my intuition says that the 2nd row in the example above seems like a good candidate to remove) and making it 2x2, but I don't know how to logically go about this..

2. Apr 29, 2016

3. May 2, 2016

### akeleti8

Hi there. this is super cool. So this is how the pay for each player would be figured out. b1 b2 c1 c2
a1 1 3 d1 1 0
a2 0 0 d2 2 1
a3 3 1 d3 0 3
the pay for player a would be a1 = 1b1 +3b2
a2 = 0
a3 = 3b1 +b2
the pay is the same for all three if it is correct. so let's have them equal 1. b2 = 1 - b1
4b1+3= 0= 4b1 + 1 this is an equation that makes no sense.

There is a turn of a1 a2 a3 that the player will not play.

options a1 a2 a3 yield an even game. both get a pay of zero.

the pay for player d would be d1 = 1c1
d2 = 2c1 +c2
d3 = 3c2
the pay is the same for all three if it is correct. so let's have them equal 1. b2=1-b1
1c1 = 3 c1 + 1 = 3c1 + 3 this is an equation that makes no sense.

there is a turn of d1 d2 d3 that will not be played. the plays that will occur are either d1 d3 or d2 d3. d1 d2 would have you siding too much with c2.
they both get a pay of zero. the maximin entry of a is zero. so there will be a pay of zero or more. the minimax entry is 2 so player 2 will have a loss of 2 or more.

To help simplify this problem i get help from http://math.stackexchange.com/questions/964327/mixed-nash-equilibrium-for-non-square-matrix-game. You are a game person to do this problem. hope you are having a good day mate.