Find Solution of game theory problem using Bimatrix method
| Player A\Player B | L | R |
| U | 2,-3 | 1,2 |
| D | 1,1 | 4,-1 |
Solution:
| | | Player `B` | | |
| | | `L` | `R` | | |
| Player `A` | `U` | | 2, -3 | 1, 2 | |
| `D` | | 1, 1 | 4, -1 | |
| | | Player `B` | | |
| | | `color{green}{L}` | `color{green}{R}` | | |
| Player `A` | `color{red}{U}` | | 2 , -3 | 1 , 2 | |
| `color{red}{D}` | | 1 , 1 | 4 , -1 | |
The cells with both entries underlined represents pure strategy nash equilibrium.
The game has no pure strategy nash equilibrium
Let p be the probability of Player-1 and q be probability of Player-2.
For Player-1
`E(U)=2*q+1*(1-q)`
`E(D)=1*q+4*(1-q)`
`E(U)=E(D)`
`=>2q+(1-q)=q+4(1-q)`
`=>4q=3`
`=>q=3/4`
`=>(1-q)=1-3/4=1/4`
For Player-2
`E(L)=-3*p+1*(1-p)`
`E(R)=2*p-1*(1-p)`
`E(L)=E(R)`
`=>-3p+(1-p)=2p-(1-p)`
`=>-7p=-2`
`=>p=2/7`
`=>(1-p)=1-2/7=5/7`
The payoff for player-1 is
`(2xx3/4)+(1xx1/4)=(1xx3/4)+(4xx1/4)=7/4`
The payoff for player-2 is
`(-3xx2/7)+(1xx5/7)=(2xx2/7)+(-1xx5/7)=-1/7`
This material is intended as a summary. Use your textbook for detail explanation.
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