<p>Let give a counterexample showing that <em>player i who will be strictly better off by deviating to some pure strategy</em> is in general <strong>false</strong>
We are going to introduce a game for which</p>
<ol>
<li>The chosen mixed strategy profile <em>is not a Nash equilibrium</em>;</li>
<li><em>No player can improve his payoff by deviating to a pure strategy</em>.</li>
</ol>
<p>Let's define the two-player game A and B, with the following two strategies (options): &quot;1&quot; and &quot;2&quot;.
The payoff matrix is:
Let give a counterexample showing that <em>player i-th who will be strictly better off by deviating to some pure strategy</em> is in general <strong>false</strong>.</p>
<p>Let's define the two-player game <strong>A</strong> and <strong>B</strong> having the following two strategies (options): <em>1</em> and <em>2</em>.
The payoff matrix is</p>
<div class="s-table-container"><table class="s-table">
<thead>
<tr>
<th></th>
<th>(B: 1)</th>
<th>(B: 2)</th>
</tr>
</thead>
<tbody>
<tr>
<td><strong>(A: 1)</strong></td>
<td>(2, 2)</td>
<td>(1, 3)</td>
</tr>
<tr>
<td><strong>(A: 2)</strong></td>
<td>(3, 1)</td>
<td>(0, 0)</td>
</tr>
</tbody>
</table></div>
<p>If both players play the following mixed strategies:</p>
<ul>
<li><strong>A</strong> plays <em>1</em> with probability p = 0.5 and <em>2</em> with probability 1-p = 0.5</li>
</ul>
<p>Payoff for <strong>A</strong> <span class="math-container">$ = p \cdot \left(q \cdot 2 + (1-q) \cdot 1\right) + (1-p) \cdot \left(q \cdot 3 + (1-q) \cdot 0\right) = 0.5 \cdot \left(0.5 \cdot 2 + 0.5 \cdot 1\right) + 0.5 \cdot \left(0.5 \cdot 3 + 0.5 \cdot 0\right) = 1.5$</span></p>
<p>If <strong>A</strong> plays a <strong>pure strategy</strong> for <em>1</em>, <span class="math-container">$p=1$</span>, Payoff is = <span class="math-container">$q \cdot 2 + (1-q) \cdot 1 = 1.5$</span></p>
<p>If <strong>A</strong> plays a <strong>pure strategy</strong> for <em>2</em>, <span class="math-container">$p=0$</span>, Payoff is = <span class="math-container">$q \cdot 3 + (1-q) \cdot 0  = 1.5$</span></p>
<ul>
<li><strong>B</strong> plays <em>1</em> with probability q = 0.5 and <em>2</em> with probability 1-q = 0.5</li>
</ul>
<p>Payoff for <strong>B</strong> <span class="math-container">$ = q \cdot \left(p \cdot 2 + (1-p) \cdot 3\right) + (1-q) \cdot \left(p \cdot 1 + (1-p) \cdot 0\right) = 1.5$</span></p>
<p>If <strong>B</strong> plays a <strong>pure strategy</strong> for <em>1</em>, <span class="math-container">$q=1$</span>, Payoff is = <span class="math-container">$p \cdot 2 + (1-p) \cdot 3 = 1.5$</span></p>
<p>If <strong>B</strong> plays a <strong>pure strategy</strong> for <em>2</em>, <span class="math-container">$q=0$</span>, Payoff is = <span class="math-container">$p \cdot 1 + (1-p) \cdot 0  = 0.5$</span></p>
<p>For both players:</p>
<ul>
<li>Playing <em>1</em>, <em>2</em>, or the mixed strategy p = 0.5, q = 0.5 results in the same payoff 1.5</li>
<li>Neither player has an incentive to deviate to a <strong>pure strategy</strong>.</li>
</ul>
<p>The mixed strategy profile p = 0.5, q = 0.5 <strong>is not a Nash equilibrium</strong>, as players can achieve a better payoff by choosing different <strong>mixed strategies</strong>.
This demonstrates that the claim &quot;<em>a player will always do strictly better by deviating to a pure strategy</em>&quot; is not true in general.</p>

CC BY-SA 4.0
2024-11-16T13:57:53.607Z
There are mistakes in your calculations. Please correct them. When B plays pure strategy of $1$, payoff is $2p+1(1-p)=1.5$ given that $p=0.5$. And when B plays pure strategy of $2$, payoff is $3p+0(1-p)=1.5$ given that $p=0.5$. Also observe that the mixed strategy profile you are considering is a mixed strategy Nash equilibrium. So this is not a valid counterexample.
Amit
CC BY-SA 4.0
2024-11-16T13:45:28.433Z
2024-11-16T13:45:28.433Z
marco tognoli
<p>Suppose <span class="math-container">$(\sigma_1,\sigma_2,\ldots, \sigma_N)$</span> is a mixed strategy profile that is not a Nash equilibrium. Therefore, there exists a player <span class="math-container">$i\in\{1,2,\ldots,N\}$</span> such that <span class="math-container">$\sigma_i$</span> is not a Best response to <span class="math-container">$\sigma_{-i}$</span>. Since <span class="math-container">$\sigma_i$</span> is not the Best Response to <span class="math-container">$\sigma_{-i}$</span>, so there exists either a pure strategy <span class="math-container">$a_i$</span> of player <span class="math-container">$i$</span> such that the expected utility of player <span class="math-container">$i$</span> satisfy <span class="math-container">$U_i(a_i,\sigma_{-i})&gt;U_i(\sigma_i,\sigma_{-i})$</span>, or there exists another mixed strategy <span class="math-container">$\sigma_i'$</span> of player <span class="math-container">$i$</span> such that the expected utility satisfy <span class="math-container">$U_i(\sigma_i',\sigma_{-i})&gt;U_i(\sigma_i,\sigma_{-i})$</span>. In the latter case, when there exists a mixed strategy <span class="math-container">$\sigma_i'$</span>, observe that <span class="math-container">$U_i(\sigma_i',\sigma_{-i})$</span> is the weighted average of the expected utility of actions in its support. So there must be at least one action <span class="math-container">$a_i'$</span> in the support of <span class="math-container">$\sigma_i'$</span> such that <span class="math-container">$U_i(a_i',\sigma_{-i})\geq U_i(\sigma_i',\sigma_{-i})&gt;U_i(\sigma_i,\sigma_{-i})$</span>. Therefore, we have shown that:  If we consider a mixed-strategy profile of N-person game that is not a Nash Equilibrium then there exist a player <span class="math-container">$i$</span> who will be strictly better off by deviating to some pure strategy.</p>

CC BY-SA 4.0
2024-11-12T17:16:21.790Z
Dear Amit, I believe that if you repeat the reasoning for all N players, you come to the conclusion that pure strategies are better for everyone! So, all players deviating to pure strategy get the best payment. In conclusion, they have reached an equilibrium point made of pures strategies. This is equivalent to saying that all equilibrium points are constituted by pure strategies. All Nash equilibria would consist of pure strategies and this is absolutely FALSE!
marco tognoli
CC BY-SA 4.0
2024-11-12T17:30:53.253Z
Dear Marco, We cannot conclude from my argument that all equilibrium are always in pure-strategies. There are mistakes in your argument.
Amit
CC BY-SA 4.0
2024-11-13T12:28:56.467Z
@marcotognoli Consider matching pennies 
\begin{array}{c|cc}
      & \text{Heads} & \text{Tails} \\
\hline
\text{Heads} & (1, -1) & (-1, 1) \\
\text{Tails} & (-1, 1) & (1, -1) \\
\end{array}
There is only a mixed-strategy equilibrium where both players choose their actions with a probability of 0.5 each, and there is no pure-strategy equilibrium. Notice, however, that given Player 1's equilibrium mixed strategy, Player 2 finds both actions to be best responses, including any mix of the two.
Amit
CC BY-SA 4.0
2024-11-13T12:55:14.270Z
@marcotognoli In other words, in Matching Pennies, given that Player 1 plays a mixed strategy of choosing each action with a probability of 1/2, Player 2 is indifferent between playing the pure strategy Heads and the pure strategy Tails, as both yield the same expected utility of 0. Therefore, Player 2 has no reason to avoid playing a mixed strategy of choosing each action with a probability of 1/2, which will also result in an expected payoff of 0.
Amit
CC BY-SA 4.0
2024-11-13T22:24:24.377Z
@Amit I think you got a brilliant answer. In my humble opinion you are right. However, if others get a contradictory answer, I might think again. Many thanks for your help!
High GPA
CC BY-SA 4.0
2024-11-14T08:31:57.497Z
Thanks for your example Amit, the property "can improve its payoff by deviating to a pure strategy" is sometimes true and sometimes false, depending on the structure of the game. It is not valid as a general statement, because there are games (like Matching Pennies you have mentioned) in which improvement is possible only through mixed strategies. Q.E.D. which was to be demonstrated: the implication
"If a strategy profile is not a Nash equilibrium, then there exists a player i who can improve his payoff by deviating towards a pure strategy" is not true in general.
marco tognoli
CC BY-SA 4.0
2024-11-15T11:56:32.090Z
@marcotognoli You write "because there are games (like Matching Pennies you have mentioned) in which improvement is possible only through mixed strategies." Please give an example of a mixed strategy profile that is not a NE for the Matching Pennies game and for which neither player can improve their payoff by deviating to a pure strategy.
smcc
CC BY-SA 4.0
2024-11-15T15:33:58.947Z
@smcc Thanks for the suggestion, I provide a counterexample to what has been debated. We have to find a mixed strategy profile that: it is not a Nash equilibrium and no player can improve his payoff by deviating to a pure strategy. The profile (p = 0.6, q = 0.6) is not a Nash equilibrium and no player can improve his payoff by deviating towards a pure strategy. This is a specific case in which the property that a player "can improve by deviating towards a pure strategy" is false, despite being far from a Nash equilibrium.
marco tognoli
CC BY-SA 4.0
2024-11-15T15:46:47.260Z
@marcotognoli You have not provided a counterexample. If $p$ and $q$ are the probabilities of each player playing heads, then player $1$'s expected payoff from heads is $0.6(1)+0.4(-1)=0.2$ and the expected payoff from tails is $0.6(-1)+0.4(1)=-0.2$. Clearly player $1$ would be better off deviating to the pure strategy of heads than mixing over both heads and tails.
smcc
CC BY-SA 4.0
2024-11-12T05:06:53.037Z
2024-11-12T06:03:44.550Z
2024-11-12T06:03:44.550Z
Amit
<p>Suppose that in one strategy profile of N-person game, everyone is playing a mixed strategy, and it is not in Nash Equilibrium.</p>
<p>Does this imply that there exist a player i who will be strictly better off by deviating to some pure strategy?</p>

CC BY-SA 4.0
2024-11-11T16:34:16.270Z
Yes, the implication is true
Amit
CC BY-SA 4.0
2024-11-11T20:46:52.723Z
In general, pure strategies are not better than mixed ones, so the implication is not true. If pure strategies were better then we should find pure strategies at the equilibrium points. Pure strategies are particoular case of mixed strategies. Player i-th could be better off simply playing the mixed strategy and adopting the max strategy while other N-1 players play mixed strategy far to equilibrium.
marco tognoli
<microeconomics><game-theory>
If one profile of mix strategy is not Nash, then one player will deviate to a pure strategy?
High GPA
# If one profile of mix strategy is not Nash, then one player will deviate to a pure strategy?

(Asked by: username_0 on 2024-11-11)

Suppose that in one strategy profile of N-person game, everyone is playing a mixed strategy, and it is not in Nash Equilibrium.

Does this imply that there exist a player i who will be strictly better off by deviating to some pure strategy?

**Tags:** microeconomics, game-theory
> **Comments:**
> **username_1**: Yes, the implication is true  
> **username_2**: In general, pure strategies are not better than mixed ones, so the implication is not true. If pure strategies were better then we should find pure strategies at the equilibrium points. Pure strategies are particoular case of mixed strategies. Player i-th could be better off simply playing the mixed strategy and adopting the max strategy while other N-1 players play mixed strategy far to equilibrium.  

---

## 2 answers

### Answer by username_1 on 2024-11-12. Score: 3 | Accepted answer

Suppose $(\sigma\_1,\sigma\_2,\ldots, \sigma\_N)$ is a mixed strategy profile that is not a Nash equilibrium. Therefore, there exists a player $i\in\{1,2,\ldots,N\}$ such that $\sigma\_i$ is not a Best response to $\sigma\_{-i}$. Since $\sigma\_i$ is not the Best Response to $\sigma\_{-i}$, so there exists either a pure strategy $a\_i$ of player $i$ such that the expected utility of player $i$ satisfy $U\_i(a\_i,\sigma\_{-i})>U\_i(\sigma\_i,\sigma\_{-i})$, or there exists another mixed strategy $\sigma\_i'$ of player $i$ such that the expected utility satisfy $U\_i(\sigma\_i',\sigma\_{-i})>U\_i(\sigma\_i,\sigma\_{-i})$. In the latter case, when there exists a mixed strategy $\sigma\_i'$, observe that $U\_i(\sigma\_i',\sigma\_{-i})$ is the weighted average of the expected utility of actions in its support. So there must be at least one action $a\_i'$ in the support of $\sigma\_i'$ such that $U\_i(a\_i',\sigma\_{-i})\geq U\_i(\sigma\_i',\sigma\_{-i})>U\_i(\sigma\_i,\sigma\_{-i})$. Therefore, we have shown that: If we consider a mixed-strategy profile of N-person game that is not a Nash Equilibrium then there exist a player $i$ who will be strictly better off by deviating to some pure strategy.

> **Comments:**
> **username_2**: Dear Amit, I believe that if you repeat the reasoning for all N players, you come to the conclusion that pure strategies are better for everyone! So, all players deviating to pure strategy get the best payment. In conclusion, they have reached an equilibrium point made of pures strategies. This is equivalent to saying that all equilibrium points are constituted by pure strategies. All Nash equilibria would consist of pure strategies and this is absolutely FALSE!  
> **username_1**: Dear Marco, We cannot conclude from my argument that all equilibrium are always in pure-strategies. There are mistakes in your argument.  
> **username_1**: @marcotognoli Consider matching pennies  \begin{array}{c|cc}       & \text{Heads} & \text{Tails} \\ \hline \text{Heads} & (1, -1) & (-1, 1) \\ \text{Tails} & (-1, 1) & (1, -1) \\ \end{array} There is only a mixed-strategy equilibrium where both players choose their actions with a probability of 0.5 each, and there is no pure-strategy equilibrium. Notice, however, that given Player 1's equilibrium mixed strategy, Player 2 finds both actions to be best responses, including any mix of the two.  
> **username_1**: @marcotognoli In other words, in Matching Pennies, given that Player 1 plays a mixed strategy of choosing each action with a probability of 1/2, Player 2 is indifferent between playing the pure strategy Heads and the pure strategy Tails, as both yield the same expected utility of 0. Therefore, Player 2 has no reason to avoid playing a mixed strategy of choosing each action with a probability of 1/2, which will also result in an expected payoff of 0.  
> **username_0**: @Amit I think you got a brilliant answer. In my humble opinion you are right. However, if others get a contradictory answer, I might think again. Many thanks for your help!  
> **username_2**: Thanks for your example Amit, the property "can improve its payoff by deviating to a pure strategy" is sometimes true and sometimes false, depending on the structure of the game. It is not valid as a general statement, because there are games (like Matching Pennies you have mentioned) in which improvement is possible only through mixed strategies. Q.E.D. which was to be demonstrated: the implication "If a strategy profile is not a Nash equilibrium, then there exists a player i who can improve his payoff by deviating towards a pure strategy" is not true in general.  
> **username_3**: @marcotognoli You write "because there are games (like Matching Pennies you have mentioned) in which improvement is possible only through mixed strategies." Please give an example of a mixed strategy profile that is not a NE for the Matching Pennies game and for which neither player can improve their payoff by deviating to a pure strategy.  
> **username_2**: @smcc Thanks for the suggestion, I provide a counterexample to what has been debated. We have to find a mixed strategy profile that: it is not a Nash equilibrium and no player can improve his payoff by deviating to a pure strategy. The profile (p = 0.6, q = 0.6) is not a Nash equilibrium and no player can improve his payoff by deviating towards a pure strategy. This is a specific case in which the property that a player "can improve by deviating towards a pure strategy" is false, despite being far from a Nash equilibrium.  
> **username_3**: @marcotognoli You have not provided a counterexample. If $p$ and $q$ are the probabilities of each player playing heads, then player $1$'s expected payoff from heads is $0.6(1)+0.4(-1)=0.2$ and the expected payoff from tails is $0.6(-1)+0.4(1)=-0.2$. Clearly player $1$ would be better off deviating to the pure strategy of heads than mixing over both heads and tails.  

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### Answer by username_2 on 2024-11-16. Score: -2

Let give a counterexample showing that *player i who will be strictly better off by deviating to some pure strategy* is in general **false**
We are going to introduce a game for which

1. The chosen mixed strategy profile *is not a Nash equilibrium*;
2. *No player can improve his payoff by deviating to a pure strategy*.

Let's define the two-player game A and B, with the following two strategies (options): "1" and "2".
The payoff matrix is:
Let give a counterexample showing that *player i-th who will be strictly better off by deviating to some pure strategy* is in general **false**.

Let's define the two-player game **A** and **B** having the following two strategies (options): *1* and *2*.
The payoff matrix is

|  | (B: 1) | (B: 2) |
| --- | --- | --- |
| **(A: 1)** | (2, 2) | (1, 3) |
| **(A: 2)** | (3, 1) | (0, 0) |

If both players play the following mixed strategies:

* **A** plays *1* with probability p = 0.5 and *2* with probability 1-p = 0.5

Payoff for **A** $ = p \cdot \left(q \cdot 2 + (1-q) \cdot 1\right) + (1-p) \cdot \left(q \cdot 3 + (1-q) \cdot 0\right) = 0.5 \cdot \left(0.5 \cdot 2 + 0.5 \cdot 1\right) + 0.5 \cdot \left(0.5 \cdot 3 + 0.5 \cdot 0\right) = 1.5$

If **A** plays a **pure strategy** for *1*, $p=1$, Payoff is = $q \cdot 2 + (1-q) \cdot 1 = 1.5$

If **A** plays a **pure strategy** for *2*, $p=0$, Payoff is = $q \cdot 3 + (1-q) \cdot 0 = 1.5$

* **B** plays *1* with probability q = 0.5 and *2* with probability 1-q = 0.5

Payoff for **B** $ = q \cdot \left(p \cdot 2 + (1-p) \cdot 3\right) + (1-q) \cdot \left(p \cdot 1 + (1-p) \cdot 0\right) = 1.5$

If **B** plays a **pure strategy** for *1*, $q=1$, Payoff is = $p \cdot 2 + (1-p) \cdot 3 = 1.5$

If **B** plays a **pure strategy** for *2*, $q=0$, Payoff is = $p \cdot 1 + (1-p) \cdot 0 = 0.5$

For both players:

* Playing *1*, *2*, or the mixed strategy p = 0.5, q = 0.5 results in the same payoff 1.5
* Neither player has an incentive to deviate to a **pure strategy**.

The mixed strategy profile p = 0.5, q = 0.5 **is not a Nash equilibrium**, as players can achieve a better payoff by choosing different **mixed strategies**.
This demonstrates that the claim "*a player will always do strictly better by deviating to a pure strategy*" is not true in general.

> **Comments:**
> **username_1**: There are mistakes in your calculations. Please correct them. When B plays pure strategy of $1$, payoff is $2p+1(1-p)=1.5$ given that $p=0.5$. And when B plays pure strategy of $2$, payoff is $3p+0(1-p)=1.5$ given that $p=0.5$. Also observe that the mixed strategy profile you are considering is a mixed strategy Nash equilibrium. So this is not a valid counterexample.  

---

