<pre><code>                    Player 2
           +---+-------+-------+-------+
           |   |   A   |   B   |   C   |
           +---+-------+-------+-------+
Player 1   | A | (1,1) | (0,0) | (0,0) | 
           | B | (0,0) | (2,1) | (1,2) |
           | C | (0,0) | (1,2) | (2,1) |
           +---+-------+-------+-------+
</code></pre>

<p>If we allow $3\times 3$ games, then above is an example of the game that has exactly one pure strategy Nash equilibrium $(A, A)$ and at least one mixed strategy Nash Equilibrium.</p>

CC BY-SA 3.0
2017-07-02T08:05:09.490Z
Thanks. If we restrict ourselves to the 2x2 case (which, I should have said, it is the case I am working with), does the mixed strategy eq disappear?
luchonacho
CC BY-SA 3.0
2017-07-02T10:03:14.943Z
The first line of the question is "Consider an arbitrary 2x2 simultaneous game with complete information. "
Giskard
CC BY-SA 3.0
2017-07-01T22:37:26.300Z
2017-07-01T22:37:26.300Z
Amit
<p>I was also thinking about this today and found the following counterexample. 
Take the game </p>

<p><span class="math-container">$$
    \begin{matrix}
       &amp; L &amp; R \\
    T &amp; 2,2 &amp; 0,2 \\
    B &amp; 0,2 &amp; 1,1 \\
    \end{matrix}
$$</span></p>

<p>(T,L) is the unique pure strategy NE, but there is a NE in which player 1 plays T and player 2 plays L with probability <span class="math-container">$p\geq 1/3$</span>.
I think it is true though that there will be no completely mixed NE in a game with a unique pure NE.</p>

CC BY-SA 4.0
2019-03-09T19:04:21.097Z
2019-03-09T19:04:21.097Z
user509037
<p>In general:
<br> If you draw the numbers of the payoff matrix from continuous i.i.distributions then with probability 1 the game will have an odd number of equilibrium points. (Meaning the sum total of pure and mixed equilibria.) 
See <a href="https://mindyourdecisions.com/blog/2014/10/07/game-theory-tuesdays-why-most-games-have-an-odd-number-of-nash-equilibria/" rel="nofollow noreferrer">here</a> or <a href="https://link.springer.com/article/10.1007/BF01737572" rel="nofollow noreferrer">here</a>.</p>

<p>If your game is 2x2 this means that with probability 1 it will have 1 or 3 equilibria. It can only have 2 equilibria if the game is 'degenerate', which would rely on the existence of weakly dominated strategies. An example is</p>

<pre><code>                   Player 2
           +---+-------+-------+
           |   |   A   |   B   |
           +---+-------+-------+
Player 1   | A | (1,1) | (0,0) |
           | B | (0,0) | (0,0) |
           +---+-------+-------+
</code></pre>

<p>Here both (A,A) and (B,B) are Nash-equilibria. There are no other equilibria, as A is better than B if the opponent plays A with non-zero probability. 
<br> (B,B) is not a trembling-hand perfect equilibrium, but that would be an additional requirement.</p>

<p>As the existence of weakly dominated strategies is necessary for exactly 2 NE and in a 2x2 game that makes mixed equilibria impossible, this means that you cannot have exactly 1 pure and 1 mixed NE.</p>

CC BY-SA 3.0
2017-07-02T08:15:13.170Z
Thanks. So, to summarise, a 2x2 game with a strongly dominated strategy (my example) will have only one pure-strategy NE, and because of oddity of eq, it has no mixed-strategy NE. Yet, a 2x2 game with a weekly dominated strategy might have two pure-strategy NE and zero mixed-strategy NE, because the oddity condition does not hold? So, the answer to my question (where I assume 1 NE, and so it seems I am imposing a strongly dominated strategy) is "yes, because of the proof of oddity of NE in strongly-dominated strategy". Do you think a best-response diagram acts as a sort of proof of this?
luchonacho
CC BY-SA 3.0
2017-07-02T09:58:52.480Z
I think graphical proofs are rarely good because it is hard to tell if you are omitting some special cases. 1 pure NE in a 2x2 game clearly implies the presence of a dominant strategy, yes. On oddity: You really should read the paper I linked to, it is very precise, more precise than I can be in a comment.
Giskard
CC BY-SA 4.0
2020-09-30T19:02:08.963Z
"generic" games have odds number of equilibria, others may not.
High GPA
CC BY-SA 3.0
2017-07-01T22:07:17.887Z
2017-07-04T06:37:59.833Z
2017-07-04T06:37:59.833Z
Giskard
<p>Consider an arbitrary 2x2 simultaneous game with complete information. Say that the model has only one pure-strategy Nash equilibrium. For example (first pay-off refers to Player 1): </p>

<pre><code>                    Player 2
           +---+-------+-------+
           |   |   A   |   B   |
           +---+-------+-------+
Player 1   | A | (1,2) | (2,1) |
           | B | (3,3) | (1,1) |
           +---+-------+-------+
</code></pre>

<p>Here, (3,3) is the only pure-strategy Nash equilibrium. The reason there is just one is, apparently, because one of the players have a dominant strategy (Player 2 always prefers A). If we change this, we would get two pure-strategy equilibria.</p>

<p>Abstracting from the fact that the pure-strategy equilibrium is also a mixed strategy one with probability 0/100%, <strong>does any simple game with just one pure-strategy equilibrium have no mixed strategy equilibrium?</strong> Is there a formal proof of this?</p>

<p>I educated guess is that there is no mixed strategy equilibrium. The "proof" I can think of is a best response plot. Basically, one player has a dominant strategy which means that for any probability, her response is a vertical/horizontal line at 0 or 1 (depending how probability is defined). Thus, regardless of how the other player's best response line looks like (i.e. where the indifference probability is), it will only cross the other player's line in one point. That point is the pure-strategy equilibrium.</p>

<p>PS: the adjective "simple" is to avoid more complex game scenarios like repetition, cooperation, incomplete information, etc. </p>

<game-theory><nash-equilibrium>
Can a game with a unique pure strategy Nash equilibrium also have a mixed strategy equilibria?
luchonacho
# Can a game with a unique pure strategy Nash equilibrium also have a mixed strategy equilibria?

(Asked by: username_0 on 2017-07-01)

Consider an arbitrary 2x2 simultaneous game with complete information. Say that the model has only one pure-strategy Nash equilibrium. For example (first pay-off refers to Player 1): 
                    Player 2
           +---+-------+-------+
           |   |   A   |   B   |
           +---+-------+-------+
Player 1   | A | (1,2) | (2,1) |
           | B | (3,3) | (1,1) |
           +---+-------+-------+

Here, (3,3) is the only pure-strategy Nash equilibrium. The reason there is just one is, apparently, because one of the players have a dominant strategy (Player 2 always prefers A). If we change this, we would get two pure-strategy equilibria.
Abstracting from the fact that the pure-strategy equilibrium is also a mixed strategy one with probability 0/100%, does any simple game with just one pure-strategy equilibrium have no mixed strategy equilibrium? Is there a formal proof of this?
I educated guess is that there is no mixed strategy equilibrium. The "proof" I can think of is a best response plot. Basically, one player has a dominant strategy which means that for any probability, her response is a vertical/horizontal line at 0 or 1 (depending how probability is defined). Thus, regardless of how the other player's best response line looks like (i.e. where the indifference probability is), it will only cross the other player's line in one point. That point is the pure-strategy equilibrium.
PS: the adjective "simple" is to avoid more complex game scenarios like repetition, cooperation, incomplete information, etc.

**Tags:** game-theory, nash-equilibrium


---

## 3 answers

### Answer by username_2 on 2017-07-01. Score: 4 | Accepted answer

In general:
 If you draw the numbers of the payoff matrix from continuous i.i.distributions then with probability 1 the game will have an odd number of equilibrium points. (Meaning the sum total of pure and mixed equilibria.) 
See here or here.
If your game is 2x2 this means that with probability 1 it will have 1 or 3 equilibria. It can only have 2 equilibria if the game is 'degenerate', which would rely on the existence of weakly dominated strategies. An example is
                   Player 2
           +---+-------+-------+
           |   |   A   |   B   |
           +---+-------+-------+
Player 1   | A | (1,1) | (0,0) |
           | B | (0,0) | (0,0) |
           +---+-------+-------+

Here both (A,A) and (B,B) are Nash-equilibria. There are no other equilibria, as A is better than B if the opponent plays A with non-zero probability. 
 (B,B) is not a trembling-hand perfect equilibrium, but that would be an additional requirement.
As the existence of weakly dominated strategies is necessary for exactly 2 NE and in a 2x2 game that makes mixed equilibria impossible, this means that you cannot have exactly 1 pure and 1 mixed NE.

> **Comments:**
> **username_0**: Thanks. So, to summarise, a 2x2 game with a strongly dominated strategy (my example) will have only one pure-strategy NE, and because of oddity of eq, it has no mixed-strategy NE. Yet, a 2x2 game with a weekly dominated strategy might have two pure-strategy NE and zero mixed-strategy NE, because the oddity condition does not hold? So, the answer to my question (where I assume 1 NE, and so it seems I am imposing a strongly dominated strategy) is "yes, because of the proof of oddity of NE in strongly-dominated strategy". Do you think a best-response diagram acts as a sort of proof of this?  
> **username_2**: I think graphical proofs are rarely good because it is hard to tell if you are omitting some special cases. 1 pure NE in a 2x2 game clearly implies the presence of a dominant strategy, yes. On oddity: You really should read the paper I linked to, it is very precise, more precise than I can be in a comment.  
> **username_4**: "generic" games have odds number of equilibria, others may not.  

---

### Answer by username_1 on 2017-07-01. Score: 6

Player 2
           +---+-------+-------+-------+
           |   |   A   |   B   |   C   |
           +---+-------+-------+-------+
Player 1   | A | (1,1) | (0,0) | (0,0) | 
           | B | (0,0) | (2,1) | (1,2) |
           | C | (0,0) | (1,2) | (2,1) |
           +---+-------+-------+-------+

If we allow $3\times 3$ games, then above is an example of the game that has exactly one pure strategy Nash equilibrium $(A, A)$ and at least one mixed strategy Nash Equilibrium.

> **Comments:**
> **username_0**: Thanks. If we restrict ourselves to the 2x2 case (which, I should have said, it is the case I am working with), does the mixed strategy eq disappear?  
> **username_2**: The first line of the question is "Consider an arbitrary 2x2 simultaneous game with complete information. "  

---

### Answer by username_3 on 2019-03-09. Score: 4

I was also thinking about this today and found the following counterexample.
Take the game

$$
\begin{matrix}
& L & R \\
T & 2,2 & 0,2 \\
B & 0,2 & 1,1 \\
\end{matrix}
$$

(T,L) is the unique pure strategy NE, but there is a NE in which player 1 plays T and player 2 plays L with probability $p\geq 1/3$.
I think it is true though that there will be no completely mixed NE in a game with a unique pure NE.


---

