# Mixed Nash equilibrium

I have game table posted below:

$$\begin{matrix} &\#2 \\ \#1 & \begin{array}{c|c|c|c} &D &E &F \\ \hline A &4,4 &6,6 &2,6 \\ \hline B &6,4 &2,2 &0,4 \\ \hline C &2,6 &8,4 &0,0 \end{array} \end{matrix}$$

Are there any Nash equilibrium where the agent 1 mixes between

(i) A & B

(ii) A and C

(iii) B & c?

What I did is that

For example I did first one

I calculate its expected utilities

$$u1(A, p2)= 4p_D+ 6 p_E+ 2+(1-p_D-p_E)$$

$$u1(B,p)=6p_D+2p_E$$

Since first payoff is higher than the second payoff, there is no such a mixed Nash equilibrium.

I think this approach to such a question seems wrong. Therefore I ask this question. What is your idea about the solution?

• How did you conclude that player 1 choosing $A$ has a higher payoff than choosing $B$? – Herr K. May 8 '18 at 13:58
• (-1) for cross-posting on MSE. – Herr K. May 8 '18 at 15:38
• Okay @HerrK. no worries about down voting. No body answer my question, so I published both websites in order to reach everyone. – mnm123 May 8 '18 at 20:19
• @HerrK. I only calculated expected utilities. $p_i$ means probability of i for i=D,E,F. But I know this is wrong. How should I do this type questions. – mnm123 May 8 '18 at 20:20
• If #1 chooses B, and #2 chooses D, then that is a nonstrict NE (#2 can do as good, but not better, by choosing F). That is a pure strategy, which can be considered to be a mixed strategy with one strategy having probability 1. – Acccumulation May 19 '18 at 16:35

(i) $A$ & $B$

If player 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$, where $0<p<1$, then player 2's expected payoff from playing

• $D$ is $4p+4(1-p) = 4$
• $E$ is $6p + 2(1-p) = 4p + 2$
• $F$ is $6p + 4(1-p) = 2p + 4$

Since payoff from playing $F$ is more than the payoff from playing any other strategy for player 2, he will always choose $F$ in response to any mixed strategy of player 1 where 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$. Therefore, $F$ is the best response of player 2 if 1 play a mixed strategy $A$ with probability $p$ and $B$ with probability $(1-p)$. However, given the strategy $F$ of player 2, player 1's best response is to play a pure strategy $A$. Therefore, there is no Nash equilibrium in which player 1 mixes between $A$ & $B$.

(ii) $A$ & $C$

If player 1 play $A$ with probability $p$ and $C$ with probability $(1-p)$, where $0<p<1$, then player 2's expected payoff from playing

• $D$ is $4p+6(1-p) = 6 - 2p$
• $E$ is $6p + 4(1-p) = 2p + 4$
• $F$ is $6p + 0(1-p) = 6p$

Since payoff from playing $E$ is more than the payoff from playing $F$, he will never choose $F$ in response to any mixed strategy of player 1 where 1 play $A$ with probability $p$ and $C$ with probability $(1-p)$. Therefore, $F$ is never a best response of player 2 if 1 play a mixed strategy $A$ with probability $p$ and $C$ with probability $(1-p)$. Now let us consider the following cases for $p$ :

• $0 < p < 0.5$ In this case player 2 has a unique best response and that is to play $D$, in response to which player 1 would like to play a pure strategy $B$, and not a mix of $A$ and $C$.
• $0.5 < p < 1$ In this case player 2 has a unique best response and that is to play $E$, in response to which player 1 would like to play a pure strategy $C$, , and not a mix of $A$ and $C$.
• $p = 0.5$ In this case player 2 is indifferent between playing $D$ and $E$, and if 2 mixes between them appropriately i.e. by choosing to play $D$ with probability $0.5$ and $E$ with probability $0.5$, it will leave player 1 with two best responses $A$ and $C$ and therefore, 1 can now choose to play $A$ and $C$ with equal probability. So, we've found a mixed strategy Nash Equilibrium in this case: Player 1 choose to play $A$ with probability 0.5 and $C$ with probability 0.5, and player 2 choose to play $D$ with probability 0.5 and $E$ with probability 0.5.

(iii) $B$ & $C$

If player 1 play $B$ with probability $p$ and $C$ with probability $(1-p)$, where $0<p<1$, then player 2's expected payoff from playing

• $D$ is $4p+6(1-p) = 6 - 2p$
• $E$ is $2p + 4(1-p) = 4 - 2p$
• $F$ is $4p + 0(1-p) = 4p$

Since payoff from playing $D$ is more than the payoff from playing any other strategy for player 2, he will always choose $D$ in response to any mixed strategy of player 1 where 1 play $B$ with probability $p$ and $C$ with probability $(1-p)$. Therefore, $D$ is the best response of player 2 if 1 play a mixed strategy $B$ with probability $p$ and $C$ with probability $(1-p)$. However, given the strategy $D$ of player 2, player 1's best response is to play a pure strategy $B$. Therefore, there is no Nash equilibrium in which player 1 mixes between $B$ & $C$.