# Homework Help: Game theory

1. Feb 20, 2005

### msmith12

So I know this isn't exactly a math/physics question, but game theory is close enough so that hopefully someone can help me out.

I am supposed to find ALL Nash Equilibrium for a given three player game such that player 1 has the choices U,D... player 2 has the choices L,R... and player three can choose A,B.

the payoffs are represented by two matrices. The first step is really easy in that player two's dominate strategy is R... this leaves a matrix with the values

09,07|06,15
14,12|05,10

where the first number is the utility to player 1, and the second is the utility to player 3.

This leaves two pure strategy nash equilibrium of 06,15 and 14,12. I need to find the mixed strategy nash equilibrium...

So, setting the probability that player one chooses U is x, and D is 1-x, and the probability that player three chooses A is y, and B is 1-y.

From here I get the two equations
$$y(9x+6(1-x))=(1-y)(14x-5(1-x)).... y(3x+6)+y(9x+5)=9x+5.... y(12x+11)=9x+5.... y=\frac{9x+5}{12x+11}$$
and
$$x(-5y+12)=(1-x)(5y+10).... 22x=5y+10.... x=\frac{5y+10}{22}$$
from here, I thought that I could just plug the second equation into the first, and get a value for y, which would give a value for x, and then I would have my probabilities, but this isnt working for me... Am I doing the set-up incorrectly? Am I missing something? Thanks for any help

~Confused