<p>Nash's theorem says that every finite game has a NE in mixed strategies, but here <em>mixed</em>, coming without the qualifier <em>strictly</em>, implies the weak version that includes pure strategies. So your statement is indeed false and your game is a counterexample.</p>

CC BY-SA 4.0
2022-02-23T02:16:59.833Z
2022-02-23T02:16:59.833Z
VARulle
<p>Your very example shows the statement is false.</p>

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2022-02-23T02:16:37.827Z
Could you explain it, please? Is p=1 and (1-p)=0 the strictly mix strategy? As it says the probability must be strictly positive, I suppose (1-p) can't be 0, so p can't be 1. I'm lost.
Constanza Zepeda
CC BY-SA 4.0
2022-02-23T03:20:46.273Z
"All finite games have at least one Nash equilibrium in strictly mixed strategies, as long as there is no player with a single pure strategy" is false, for the very reason you have stated.
Walrasian Auctioneer
CC BY-SA 4.0
2022-02-23T01:48:07.203Z
2022-02-23T01:48:07.203Z
Walrasian Auctioneer
<p>I have the following statement which I have been said it is false, but I don't understand why:</p>
<p><em>&quot;All finite games have at least one Nash equilibrium in <strong>strictly mixed strategies</strong>, as long as there is no player with a single pure strategy&quot;.</em></p>
<p>It confuses me because the Nash theorem (1950) says that every finite game has a mixed strategy equilibrium. But if I make a simple game 2x2 as in the image, I don't find any mix strategy. Or does playing &quot;a&quot; with p=1 is the mix strategy?</p>
<p><a href="https://i.sstatic.net/GocpR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GocpR.png" alt="enter image description here" /></a></p>
<p>P.S. A strictly mixed strategy is a mixed strategy that assigns strictly positive probability to at least 2 pure strategies.</p>

CC BY-SA 4.0
2022-02-23T00:55:05.280Z
Do you know how to solve a game? Try making (or look up) some simple 2x2 examples and see if the above statement is true or not.
Walrasian Auctioneer
<game-theory><nash-equilibrium><dominant-strategy>
Nash equilibrium in strictly mixed strategies
Constanza Zepeda
# Nash equilibrium in strictly mixed strategies

(Asked by: username_0 on 2022-02-23)

I have the following statement which I have been said it is false, but I don't understand why:

*"All finite games have at least one Nash equilibrium in **strictly mixed strategies**, as long as there is no player with a single pure strategy".*

It confuses me because the Nash theorem (1950) says that every finite game has a mixed strategy equilibrium. But if I make a simple game 2x2 as in the image, I don't find any mix strategy. Or does playing "a" with p=1 is the mix strategy?

[![enter image description here](https://i.sstatic.net/GocpR.png)](https://i.sstatic.net/GocpR.png)

P.S. A strictly mixed strategy is a mixed strategy that assigns strictly positive probability to at least 2 pure strategies.

**Tags:** game-theory, nash-equilibrium, dominant-strategy
> **Comments:**
> **username_1**: Do you know how to solve a game? Try making (or look up) some simple 2x2 examples and see if the above statement is true or not.  

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## 2 answers

### Answer by username_2 on 2022-02-23. Score: 2

Nash's theorem says that every finite game has a NE in mixed strategies, but here *mixed*, coming without the qualifier *strictly*, implies the weak version that includes pure strategies. So your statement is indeed false and your game is a counterexample.


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### Answer by username_1 on 2022-02-23. Score: 0

Your very example shows the statement is false.

> **Comments:**
> **username_0**: Could you explain it, please? Is p=1 and (1-p)=0 the strictly mix strategy? As it says the probability must be strictly positive, I suppose (1-p) can't be 0, so p can't be 1. I'm lost.  
> **username_1**: "All finite games have at least one Nash equilibrium in strictly mixed strategies, as long as there is no player with a single pure strategy" is false, for the very reason you have stated.  

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