<p>In mixed strategies NE, you choose to act (based on that you know the payoffs of the second player) based on the probability that makes the other player indifferent. Still the strategy is not defined as a typical pure strategies case, rather than you mention that A will randomize with probability x and B will randomize(between his actions) with probability y.
Try to read Felix Munoz Garcia  game theory book, it very good with examples and strays from scratch to advanced.<img src="https://i.sstatic.net/mCFkN.jpg" alt="enter image description here"></p>

CC BY-SA 4.0
2019-01-16T15:54:39.893Z
This does not answer the OP's question.
Herr K.
CC BY-SA 4.0
2019-01-16T11:04:10.213Z
2019-01-16T11:04:10.213Z
J.T
<p>No, it's not an equilibrium.</p>

<p>If the column player uses the uniform mixed strategy 
<span class="math-container">$$\sigma_C = (1/4, 1/4, 1/4, 1/4),$$</span> 
then the four actions available to the row player have the values <span class="math-container">$-1/4$</span>, <span class="math-container">$+1/4$</span>, <span class="math-container">$-1/4$</span>, and <span class="math-container">$+1/4$</span>, respectively. These values are not identical, so the uniform strategy
<span class="math-container">$$\sigma_R = (1/4, 1/4, 1/4, 1/4)$$</span>
is not a best response for the row player. Specifically, a strategy <span class="math-container">$\sigma_R$</span> can only by a best response to <span class="math-container">$\sigma_C = (1/4, 1/4, 1/4, 1/4)$</span> if it assigns probability 0 to the first and third row.</p>

CC BY-SA 4.0
2019-01-17T22:00:11.633Z
2019-01-17T22:00:11.633Z
Mathias
<p>I have a  simple payoff matrix defined here: <a href="https://prnt.sc/m6rp5m" rel="nofollow noreferrer">https://prnt.sc/m6rp5m</a></p>

<p><strong>My question is: If both players play all 4 strategies with 1/4 probability, does that lead to nash equilibrium?</strong> </p>

<p>I can't quite figure this out. I know how to check if pure stratigies lead to nash equilibrium. You assume the row/columns player sticks to a strategy and check if the other player is incentivized to play differently.</p>

<p>I'd love a pointer or two in the right direction.</p>

CC BY-SA 4.0
2019-01-14T15:28:37.450Z
Just as in the case with pure strategy NE, you assume the row/column player sticks to playing all four actions with probability 1/4, and see if the other player is incentivized to play something other than randomizing over the four actions with equal probability.
Herr K.
<game-theory>
Mixed Strategies -- Does this lead to Nash equilibrium?
nz_21
# Mixed Strategies -- Does this lead to Nash equilibrium?

(Asked by: username_0 on 2019-01-14)

I have a simple payoff matrix defined here: <https://prnt.sc/m6rp5m>

**My question is: If both players play all 4 strategies with 1/4 probability, does that lead to nash equilibrium?**

I can't quite figure this out. I know how to check if pure stratigies lead to nash equilibrium. You assume the row/columns player sticks to a strategy and check if the other player is incentivized to play differently.

I'd love a pointer or two in the right direction.

**Tags:** game-theory

> **Comments:**
> **username_1**: Just as in the case with pure strategy NE, you assume the row/column player sticks to playing all four actions with probability 1/4, and see if the other player is incentivized to play something other than randomizing over the four actions with equal probability.  

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## 2 answers

### Answer by username_2 on 2019-01-16. Score: 0

In mixed strategies NE, you choose to act (based on that you know the payoffs of the second player) based on the probability that makes the other player indifferent. Still the strategy is not defined as a typical pure strategies case, rather than you mention that A will randomize with probability x and B will randomize(between his actions) with probability y.
Try to read Felix Munoz Garcia game theory book, it very good with examples and strays from scratch to advanced.![enter image description here](https://i.sstatic.net/mCFkN.jpg)

> **Comments:**
> **username_1**: This does not answer the OP's question.  

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### Answer by username_3 on 2019-01-17. Score: 0

No, it's not an equilibrium.

If the column player uses the uniform mixed strategy
$$\sigma\_C = (1/4, 1/4, 1/4, 1/4),$$
then the four actions available to the row player have the values $-1/4$, $+1/4$, $-1/4$, and $+1/4$, respectively. These values are not identical, so the uniform strategy
$$\sigma\_R = (1/4, 1/4, 1/4, 1/4)$$
is not a best response for the row player. Specifically, a strategy $\sigma\_R$ can only by a best response to $\sigma\_C = (1/4, 1/4, 1/4, 1/4)$ if it assigns probability 0 to the first and third row.


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