<p>I assume that you are calculating a mixed strategy for the row player so that the column player is indifferent between choosing $A$ and $B$.  And the reason why you get a negative probability is that the row player cannot make the column player indifferent by choosing, say strategy $C$, with a positive probability $\in [0,1]$.  This is because the column player (always) strictly prefers $B$.  </p>

<blockquote>
  <p>Side-note: To avoid such confusion in the future, try to see whether you can apply math to calculate a mixed strategy to a game.  Here you clearly cannot (because of the above argument), and that is why math is giving you a ''weird'' answer.  Namely, math yields an answer to a question: What would be a <strong>hypothetical</strong> probability with which the row player chooses, say, strategy $C$ to make the column player indifferent?  </p>
</blockquote>

CC BY-SA 4.0
2018-06-24T11:28:19.187Z
2018-06-24T11:28:19.187Z
Green.H
<p>Hints: </p>

<ul>
<li><p>As you say, B strictly dominates A for the second player.  You can draw a conclusion in terms of Nash equilibria about the second player</p></li>
<li><p>If the second player plays B then C and D produce the same result for the first player.  You can draw a conclusion in terms of Nash equilibria about the first player</p></li>
</ul>

<p>If you draw the sensible conclusions, you may be able to see why your calculations for a unique mixed strategy Nash equilibrium is unlikely to work </p>

CC BY-SA 4.0
2018-06-21T06:46:14.683Z
2018-06-21T06:46:14.683Z
Henry
<p>When I do the basic calculations for mixed probability, I get that the Column player always plays B. However, I am getting a negative probability for the row. Any help is appreciated.</p>

<p><a href="https://i.sstatic.net/1SBjO.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1SBjO.png" alt="Question"></a></p>

CC BY-SA 4.0
2018-06-21T06:44:51.393Z
Perhaps there is something wrong with the underlying assumption you are making when you are calculating the mixed Nash equilibrium?
Giskard
CC BY-SA 4.0
2021-05-12T11:52:39.387Z
You are correct that Player 2 always chooses $B$ as it is a dominant strategy. Maybe if you show us the calculations that you made, we can spot the mistake.
tdm
<game-theory><nash-equilibrium><probability>
Find all of the Pure and Mixed Strategy Nash Equilibria
Vraj Shroff
# Find all of the Pure and Mixed Strategy Nash Equilibria

(Asked by: username_0 on 2018-06-21)

When I do the basic calculations for mixed probability, I get that the Column player always plays B. However, I am getting a negative probability for the row. Any help is appreciated.

[![Question](https://i.sstatic.net/1SBjO.png)](https://i.sstatic.net/1SBjO.png)

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

> **Comments:**
> **username_1**: Perhaps there is something wrong with the underlying assumption you are making when you are calculating the mixed Nash equilibrium?  
> **username_2**: You are correct that Player 2 always chooses $B$ as it is a dominant strategy. Maybe if you show us the calculations that you made, we can spot the mistake.  

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## 2 answers

### Answer by username_3 on 2018-06-24. Score: 1

I assume that you are calculating a mixed strategy for the row player so that the column player is indifferent between choosing $A$ and $B$. And the reason why you get a negative probability is that the row player cannot make the column player indifferent by choosing, say strategy $C$, with a positive probability $\in [0,1]$. This is because the column player (always) strictly prefers $B$.

> Side-note: To avoid such confusion in the future, try to see whether you can apply math to calculate a mixed strategy to a game. Here you clearly cannot (because of the above argument), and that is why math is giving you a ''weird'' answer. Namely, math yields an answer to a question: What would be a **hypothetical** probability with which the row player chooses, say, strategy $C$ to make the column player indifferent?


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### Answer by username_4 on 2018-06-21. Score: 0

Hints:

* As you say, B strictly dominates A for the second player. You can draw a conclusion in terms of Nash equilibria about the second player
* If the second player plays B then C and D produce the same result for the first player. You can draw a conclusion in terms of Nash equilibria about the first player

If you draw the sensible conclusions, you may be able to see why your calculations for a unique mixed strategy Nash equilibrium is unlikely to work


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