<p>The game is indeed an <span class="math-container">$8\times 4$</span>-game. A strategy formally specifies what to do at each information set, including those information sets that can never occur under the strategy. In that sense, a strategy is not simply a &quot;plan of action.&quot; This point has been famously made in (see Section 2):</p>
<p>Rubinstein, Ariel. &quot;<a href="http://arielrubinstein.tau.ac.il/papers/37.pdf" rel="nofollow noreferrer">Comments on the interpretation of game theory.</a>&quot;
Econometrica: Journal of the Econometric Society (1991): 909-924.</p>

CC BY-SA 4.0
2020-12-23T00:45:27.047Z
Thank you, it makes much more sense now. I might have been too quick. Tadelis also explains it very good in this book Game Theory: An Introduction. If future readers of this of this post are interested it is definition 8.3 in section 8.2. The paper you have linked to was very helpful. Once again thanks.
Justin Malik
CC BY-SA 4.0
2020-12-23T00:56:30.053Z
Further, for some reason I forgot that there actually is a formula in Tadelis which makes it easier to calculate all pure strategies. If player $i$ has $k > 1$ information sets, the first with $m_1$ actions from which to choose, the second $m_2$, and so on until $m_k$. Then the total number of pure strategies player $i$ has is $|S_i| = m_1 \times m_2 \times \cdots \times m_k$. In my case it would be $k=3$ and player 1 has 2 actions in each information set. Thus, $|S_1| = 2 \times 2 \times 2 = 8$. This can be done for player 2 as well.
Justin Malik
CC BY-SA 4.0
2020-12-21T06:33:02.603Z
2020-12-21T06:33:02.603Z
Michael Greinecker
<p>This might be a stupid question but please bear with me. I'm trying to solve this game but I'm in doubt on how to represent the strategy profile of the game. The game looks like this in extensive-form. I've marked the path with red that show the best actions of the players when it is their turn to move. The game is a game of perfect information.</p>
<p><a href="https://i.sstatic.net/zYgQw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zYgQw.png" alt="enter image description here" /></a></p>
<p>The normal-form of the game looks like this:</p>
<p><a href="https://i.sstatic.net/QOvNn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QOvNn.png" alt="enter image description here" /></a></p>
<p>From the normal-form we can see that there are 4 Nash equilibria of the game. Using backward induction I've come to the conclusion that <span class="math-container">$(df,\neg rr)$</span> is the unique subgame-perfect Nash equilibrium. My concern arises when we take a look at player 2. His pure strategy set would be <span class="math-container">$S_{2} = \{rr, r\neg r, \neg rr, \neg r \neg r\}$</span>. In my normal-form representation of the game I've written the pure strategies of player 1 as <span class="math-container">$S_{1} = \{df, d\neg f, \neg df, \neg d \neg f\}$</span>. For some reason I'm more tempted to write player 1's pure strategy set as <span class="math-container">$S_{1} = \{dff, df\neg f, d\neg ff, d\neg f \neg f,\neg dff, \neg df \neg f, \neg d \neg ff, \neg d \neg f \neg f\}$</span> and we would have a 8x4 matrix-representation of the game. Thus, the unique subgame-perfect Nash equilibrium would be <span class="math-container">$(df\neg f, \neg rr)$</span>. Can somebody shed some light and help me with my doubts? Thanks in advanced.</p>

CC BY-SA 4.0
2020-12-21T07:59:07.743Z
Perhaps this should be in math.SE?
Mozibur Ullah
CC BY-SA 4.0
2020-12-21T22:04:53.123Z
I don't get the downvotes.
Michael Greinecker
CC BY-SA 4.0
2020-12-23T00:48:27.987Z
@MoziburUllah I thought it would be more sensible to post it here. But you are right it could just as well be posted in math.SE
Justin Malik
CC BY-SA 4.0
2020-12-23T00:49:52.543Z
@MichaelGreinecker Me neither. Maybe my question wasn't as clear as I wanted it to be
Justin Malik
<game-theory><nash-equilibrium><extensive-games>
Subgame-perfect Nash equilibrium perfect information
Justin Malik
# Subgame-perfect Nash equilibrium perfect information

(Asked by: username_0 on 2020-12-21)

This might be a stupid question but please bear with me. I'm trying to solve this game but I'm in doubt on how to represent the strategy profile of the game. The game looks like this in extensive-form. I've marked the path with red that show the best actions of the players when it is their turn to move. The game is a game of perfect information.

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

The normal-form of the game looks like this:

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

From the normal-form we can see that there are 4 Nash equilibria of the game. Using backward induction I've come to the conclusion that $(df,\neg rr)$ is the unique subgame-perfect Nash equilibrium. My concern arises when we take a look at player 2. His pure strategy set would be $S\_{2} = \{rr, r\neg r, \neg rr, \neg r \neg r\}$. In my normal-form representation of the game I've written the pure strategies of player 1 as $S\_{1} = \{df, d\neg f, \neg df, \neg d \neg f\}$. For some reason I'm more tempted to write player 1's pure strategy set as $S\_{1} = \{dff, df\neg f, d\neg ff, d\neg f \neg f,\neg dff, \neg df \neg f, \neg d \neg ff, \neg d \neg f \neg f\}$ and we would have a 8x4 matrix-representation of the game. Thus, the unique subgame-perfect Nash equilibrium would be $(df\neg f, \neg rr)$. Can somebody shed some light and help me with my doubts? Thanks in advanced.

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

> **Comments:**
> **username_1**: Perhaps this should be in math.SE?  
> **username_2**: I don't get the downvotes.  
> **username_0**: @username_1 I thought it would be more sensible to post it here. But you are right it could just as well be posted in math.SE  
> **username_0**: @username_2 Me neither. Maybe my question wasn't as clear as I wanted it to be  

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## 1 answers

### Answer by username_2 on 2020-12-21. Score: 4 | Accepted answer

The game is indeed an $8\times 4$-game. A strategy formally specifies what to do at each information set, including those information sets that can never occur under the strategy. In that sense, a strategy is not simply a "plan of action." This point has been famously made in (see Section 2):

Rubinstein, Ariel. "[Comments on the interpretation of game theory.](http://arielrubinstein.tau.ac.il/papers/37.pdf)"
Econometrica: Journal of the Econometric Society (1991): 909-924.

> **Comments:**
> **username_0**: Thank you, it makes much more sense now. I might have been too quick. Tadelis also explains it very good in this book Game Theory: An Introduction. If future readers of this of this post are interested it is definition 8.3 in section 8.2. The paper you have linked to was very helpful. Once again thanks.  
> **username_0**: Further, for some reason I forgot that there actually is a formula in Tadelis which makes it easier to calculate all pure strategies. If player $i$ has $k > 1$ information sets, the first with $m_1$ actions from which to choose, the second $m_2$, and so on until $m_k$. Then the total number of pure strategies player $i$ has is $|S_i| = m_1 \times m_2 \times \cdots \times m_k$. In my case it would be $k=3$ and player 1 has 2 actions in each information set. Thus, $|S_1| = 2 \times 2 \times 2 = 8$. This can be done for player 2 as well.  

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