<p>Three-player games are notoriously tricky to analyze. The ideal way to display them would be a three-dimensional array of cells, each containing three payoffs. But this is difficult to write down on two-dimensional paper. So typically an <span class="math-container">$n\times m\times l$</span>-game is displayed as <span class="math-container">$l$</span> different <span class="math-container">$n\times m$</span>-matrices. In your example you might think about it in this way:</p>

<ul>
<li><p>Player 1 chooses the row (upper row, <span class="math-container">$A$</span>, or lower row, <span class="math-container">$B$</span>)</p></li>
<li><p>Player 2 chooses the column (left column, <span class="math-container">$A$</span>, or right column, <span class="math-container">$B$</span>) </p></li>
<li><p>Player 3 chooses the matrix (upper matrix, <span class="math-container">$A$</span>, or lower matrix, <span class="math-container">$B$</span>)</p></li>
</ul>

<p>Therefore, to highlight the best payoff of player 3, for each of the 4 choices of players 1 and 2 (for each of the 4 cells in the matrices) you have to <strong>compare the player-3-payoffs between the upper and the lower matrix</strong>. So in the <span class="math-container">$(A,A)$</span>-cell you highlight the <span class="math-container">$70$</span> in the upper matrix, since it is greater than the <span class="math-container">$60$</span> in the lower matrix. For the <span class="math-container">$(A,B)$</span>-cell you highlight <span class="math-container">$23$</span>, which is greater than <span class="math-container">$0$</span>, and so on. In the end, the <span class="math-container">$(A,A)$</span>-cell of the upper matrix (corresponding to the choice of <span class="math-container">$A$</span> by player 3) will have all three payoffs highlighted, and the same will be true for the <span class="math-container">$(B,B)$</span>-cell of the upper matrix (again corresponding to the choice of <span class="math-container">$A$</span> by player 3). So the pure NE are <span class="math-container">$(A,A,A)$</span> and <span class="math-container">$(B,B,A)$</span>.</p>

CC BY-SA 4.0
2020-03-04T12:57:28.423Z
Thanks @VARulle, that was really good explanation .

And it is written as (player1, player2, player3)?
soetirl13
CC BY-SA 4.0
2020-03-04T13:51:15.583Z
@soetirl13: Yes, that's the convention.
VARulle
CC BY-SA 4.0
2020-03-04T12:34:43.700Z
2020-03-04T12:34:43.700Z
VARulle
<p>I'm trying to solve this pure-strategy Nash equilibria of this game below: </p>

<p><a href="https://i.sstatic.net/aPaoD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aPaoD.png" alt="enter image description here"></a></p>

<p>I highlighted the best pay off for player 1 and 2. But I don't get it when it comes to player 3. The correct answer is (A) </p>

<p>I tried to solve it as a gaming tree. But still I don't get it. </p>

<p>How shall I think? </p>

<game-theory><nash-equilibrium>
Pure Nash Equilibria 3 players game
soetirl13
# Pure Nash Equilibria 3 players game

(Asked by: username_0 on 2020-03-04)

I'm trying to solve this pure-strategy Nash equilibria of this game below:

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

I highlighted the best pay off for player 1 and 2. But I don't get it when it comes to player 3. The correct answer is (A)

I tried to solve it as a gaming tree. But still I don't get it.

How shall I think?

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


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## 1 answers

### Answer by username_1 on 2020-03-04. Score: 2

Three-player games are notoriously tricky to analyze. The ideal way to display them would be a three-dimensional array of cells, each containing three payoffs. But this is difficult to write down on two-dimensional paper. So typically an $n\times m\times l$-game is displayed as $l$ different $n\times m$-matrices. In your example you might think about it in this way:

* Player 1 chooses the row (upper row, $A$, or lower row, $B$)
* Player 2 chooses the column (left column, $A$, or right column, $B$)
* Player 3 chooses the matrix (upper matrix, $A$, or lower matrix, $B$)

Therefore, to highlight the best payoff of player 3, for each of the 4 choices of players 1 and 2 (for each of the 4 cells in the matrices) you have to **compare the player-3-payoffs between the upper and the lower matrix**. So in the $(A,A)$-cell you highlight the $70$ in the upper matrix, since it is greater than the $60$ in the lower matrix. For the $(A,B)$-cell you highlight $23$, which is greater than $0$, and so on. In the end, the $(A,A)$-cell of the upper matrix (corresponding to the choice of $A$ by player 3) will have all three payoffs highlighted, and the same will be true for the $(B,B)$-cell of the upper matrix (again corresponding to the choice of $A$ by player 3). So the pure NE are $(A,A,A)$ and $(B,B,A)$.

> **Comments:**
> **username_0**: Thanks @username_1, that was really good explanation .  And it is written as (player1, player2, player3)?  
> **username_1**: @username_0: Yes, that's the convention.  

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