<p>(i) $A$ &amp; $B$</p>

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

<ul>
<li>$D$ is $4p+4(1-p) = 4$</li>
<li>$E$ is $6p + 2(1-p) = 4p + 2$</li>
<li>$F$ is $6p + 4(1-p) = 2p + 4$</li>
</ul>

<p>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 $A$ &amp; $B$.</p>

<p>(ii) $A$ &amp; $C$</p>

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

<ul>
<li>$D$ is $4p+6(1-p) = 6 - 2p$</li>
<li>$E$ is $6p + 4(1-p) = 2p + 4$</li>
<li>$F$ is $6p + 0(1-p) = 6p$</li>
</ul>

<p>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 $p$ :</p>

<ul>
<li>$0 &lt; p &lt; 0.5$ 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 $C$. </li>
<li>$0.5 &lt; p &lt; 1$ 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 $C$. </li>
<li>$p = 0.5$ 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.</li>
</ul>

<p>(iii) $B$ &amp; $C$</p>

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

<ul>
<li>$D$ is $4p+6(1-p) = 6 - 2p$</li>
<li>$E$ is $2p + 4(1-p) = 4 - 2p$</li>
<li>$F$ is $4p + 0(1-p) = 4p$</li>
</ul>

<p>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 $B$ &amp; $C$.</p>

CC BY-SA 4.0
2018-05-22T00:24:31.770Z
2018-05-22T06:43:53.660Z
2018-05-22T06:43:53.660Z
Amit
<p>I have game table posted below:</p>

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

<p>Are there any Nash equilibrium where the agent 1 mixes between </p>

<p>(i) A &amp; B</p>

<p>(ii) A and C</p>

<p>(iii) B &amp; c?</p>

<p>What I did is that</p>

<p>For example I did first one </p>

<p>I calculate its expected utilities </p>

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

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

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

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

CC BY-SA 4.0
2018-05-08T13:58:26.773Z
How did you conclude that player 1 choosing $A$ has a higher payoff than choosing $B$?
Herr K.
CC BY-SA 4.0
2018-05-08T15:38:54.180Z
(-1) for [cross-posting on MSE](https://math.stackexchange.com/q/2771497/70119).
Herr K.
CC BY-SA 4.0
2018-05-08T20:19:10.977Z
Okay @HerrK. no worries about down voting. No body answer my question, so I published both websites in order to reach everyone.
mnm123
CC BY-SA 4.0
2018-05-08T20:20:53.560Z
@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
CC BY-SA 4.0
2018-05-08T22:28:16.603Z
You *wrote down* the expected utilities, but you didn't *calculate* them, i.e. finding values for the $p_i$'s. For example, when $p_D=1$ and $p_E=0$, the payoff from $B$ is higher than the payoff from $A$. There exist values for $p_i$'s that can equate the two payoffs.
Herr K.
CC BY-SA 4.0
2018-05-09T03:10:36.867Z
Yes, I could not calculate perfectly. How can I do it? @HerrK.
mnm123
CC BY-SA 4.0
2018-05-18T20:31:25.200Z
B,D is a nonstrict NE, so if you consider a pure strategy to be a subset of mixed strategies, then that is a mixed strategy that satisfies (i) and (iii).
Acccumulation
CC BY-SA 4.0
2018-05-19T07:57:16.930Z
@Acccumulation I can imagine what you said. But I cannot imply I am a bit confuse. Please can you show more precisely?
mnm123
CC BY-SA 4.0
2018-05-19T16:35:24.670Z
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
<microeconomics><game-theory><nash-equilibrium><self-study>
Mixed Nash equilibrium
mnm123
# Mixed Nash equilibrium

(Asked by: username_0 on 2018-05-08)

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?

**Tags:** microeconomics, game-theory, nash-equilibrium, self-study

> **Comments:**
> **username_1**: How did you conclude that player 1 choosing $A$ has a higher payoff than choosing $B$?  
> **username_1**: (-1) for [cross-posting on MSE](https://math.stackexchange.com/q/2771497/70119).  
> **username_0**: Okay @HerrK. no worries about down voting. No body answer my question, so I published both websites in order to reach everyone.  
> **username_0**: @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.  
> **username_1**: You *wrote down* the expected utilities, but you didn't *calculate* them, i.e. finding values for the $p_i$'s. For example, when $p_D=1$ and $p_E=0$, the payoff from $B$ is higher than the payoff from $A$. There exist values for $p_i$'s that can equate the two payoffs.  
> **username_0**: Yes, I could not calculate perfectly. How can I do it? @HerrK.  
> **username_2**: B,D is a nonstrict NE, so if you consider a pure strategy to be a subset of mixed strategies, then that is a mixed strategy that satisfies (i) and (iii).  
> **username_0**: @username_2 I can imagine what you said. But I cannot imply I am a bit confuse. Please can you show more precisely?  
> **username_2**: 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.  

---

## 1 answers

### Answer by username_3 on 2018-05-22. Score: 5 | Accepted answer

(i) $A$ & $B$

If player 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$, where $0

* $D$ 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

* $D$ 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

* $D$ 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 $B$ & $C$.


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