<p>There’s a pure strategy and one mixed strategy with probability distribution (0.5,0.5,0) and (0.5,0.5,0) for each player in this game.</p>
<p>But I was wondering if this example fails the fact that there are almost always odd numbers of equilibriums in one game.</p>
<p>secondly, is there an easier way to obtain the mixed strategy equilibrium if not putting a zero on every single action of a player and then calculating the probability distribution for 9 cases?(cuz let’s say we disregard action U for player 1 and L for player 2, then we consider the now 2x2 matrix and calculate probabilities) I saw in Joel Watson’s textbook where he considers a mix of two actions of a player and its combination could happen to be dominant over the remaining action. Then he just eliminates the dominated action and reduces the matrix. Does this work here?</p>
<p>If I put p1 for L, p2 for M, and p3 for R, and set up the equation for expected payoff of player 1, the equation does not have a solution(q1=q2=-q3). So why is it that we still can find one mixed-strategy equilibrium when it clearly does not satisfy the equation?</p>
<p>Is there anything special about this payoff matrix? I know its symmetric, but are there any special theories about these matrices that would help solve the question faster?</p>
<p>Any help would be appreciated!</p>
<p><a href="https://i.sstatic.net/YE6dA.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YE6dA.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-09-10T17:07:21.027Z
$((0.5,0.5,0),(0.5,0.5,0))$ is not an equilibrium; if Column mixes $L$ and $M$ with 50-50%, the expected payoff of Row from paying $U$ or $M$ is $-1/2$, but they could attain the higher payoff of 0 by playing $D$.
Giskard
CC BY-SA 4.0
2022-09-11T01:25:04.397Z
@Giskard but if you mix L and M with 1/2 and 1/2, you're putting a zero on R and D, why should we worry about the payoff of playing D?
sdvjdla
CC BY-SA 4.0
2022-09-11T08:22:01.473Z
In equilibrium all players play a best response to the others given strategies. I have shown that for this strategy of the column player, the row player is not giving their best response, thus, this is not an equilibrium.
Giskard
CC BY-SA 4.0
2022-09-12T07:30:27.863Z
Also, the payoff matrix is not symmetric.
VARulle
<microeconomics><game-theory><general-equilibrium>
Finding all symmetric mixed strategy equilibria in a finite game
sdvjdla
# Finding all symmetric mixed strategy equilibria in a finite game

(Asked by: username_0 on 2022-09-10)

There’s a pure strategy and one mixed strategy with probability distribution (0.5,0.5,0) and (0.5,0.5,0) for each player in this game.

But I was wondering if this example fails the fact that there are almost always odd numbers of equilibriums in one game.

secondly, is there an easier way to obtain the mixed strategy equilibrium if not putting a zero on every single action of a player and then calculating the probability distribution for 9 cases?(cuz let’s say we disregard action U for player 1 and L for player 2, then we consider the now 2x2 matrix and calculate probabilities) I saw in Joel Watson’s textbook where he considers a mix of two actions of a player and its combination could happen to be dominant over the remaining action. Then he just eliminates the dominated action and reduces the matrix. Does this work here?

If I put p1 for L, p2 for M, and p3 for R, and set up the equation for expected payoff of player 1, the equation does not have a solution(q1=q2=-q3). So why is it that we still can find one mixed-strategy equilibrium when it clearly does not satisfy the equation?

Is there anything special about this payoff matrix? I know its symmetric, but are there any special theories about these matrices that would help solve the question faster?

Any help would be appreciated!

[![enter image description here](https://i.sstatic.net/YE6dA.jpg)](https://i.sstatic.net/YE6dA.jpg)

**Tags:** microeconomics, game-theory, general-equilibrium

> **Comments:**
> **username_1**: $((0.5,0.5,0),(0.5,0.5,0))$ is not an equilibrium; if Column mixes $L$ and $M$ with 50-50%, the expected payoff of Row from paying $U$ or $M$ is $-1/2$, but they could attain the higher payoff of 0 by playing $D$.  
> **username_0**: @username_1 but if you mix L and M with 1/2 and 1/2, you're putting a zero on R and D, why should we worry about the payoff of playing D?  
> **username_1**: In equilibrium all players play a best response to the others given strategies. I have shown that for this strategy of the column player, the row player is not giving their best response, thus, this is not an equilibrium.  
> **username_2**: Also, the payoff matrix is not symmetric.  

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## 0 answers

