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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.

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The normal-form of the game looks like this:

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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.

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  • $\begingroup$ Perhaps this should be in math.SE? $\endgroup$ – Mozibur Ullah Dec 21 '20 at 7:59
  • $\begingroup$ I don't get the downvotes. $\endgroup$ – Michael Greinecker Dec 21 '20 at 22:04
  • $\begingroup$ @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 $\endgroup$ – Justin Malik Dec 23 '20 at 0:48
  • $\begingroup$ @MichaelGreinecker Me neither. Maybe my question wasn't as clear as I wanted it to be $\endgroup$ – Justin Malik Dec 23 '20 at 0:49
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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." Econometrica: Journal of the Econometric Society (1991): 909-924.

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  • $\begingroup$ 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. $\endgroup$ – Justin Malik Dec 23 '20 at 0:45
  • $\begingroup$ 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. $\endgroup$ – Justin Malik Dec 23 '20 at 0:56

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