Solving a Three-Player Nash Equilibrium

[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
$$<br /> y(9x+6(1-x))=(1-y)(14x-5(1-x))...<br /> <br /> y(3x+6)+y(9x+5)=9x+5...<br /> <br /> y(12x+11)=9x+5...<br /> <br /> y=\frac{9x+5}{12x+11}$$
and
$$x(-5y+12)=(1-x)(5y+10)...<br /> <br /> 22x=5y+10...<br /> <br /> x=\frac{5y+10}{22}$$
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 isn't working for me... Am I doing the set-up incorrectly? Am I missing something? Thanks for any help

~Confused

(sorry about the TEX formatting, i couldn't get carriage return to work)

 

[wais]

First of all, great job on finding the pure strategy Nash equilibria! Now, for finding the mixed strategy Nash equilibrium, your set-up seems correct. However, your method of solving for y might be causing some confusion. Instead of plugging in the second equation into the first, try solving for x in the second equation and then substituting that value into the first equation to solve for y. This should give you the probabilities for player 1 and player 3 to choose U and A, respectively.

Another method you could use is to set up a system of equations with x and y as the variables and solve for them simultaneously. For example, using the second equation, you can solve for x in terms of y and then substitute that into the first equation to solve for y. This might be a bit more straightforward and less confusing.

One more thing to keep in mind is that there may be multiple solutions for x and y that satisfy the equations. In this case, you can check each solution by plugging them into the original equations and seeing if they hold true. If they do, then you have found another mixed strategy Nash equilibrium.

I hope this helps and good luck with finding the mixed strategy Nash equilibrium!