<p>The short answer is that a strategy must include plans of action in information sets that cannot be reached if the strategy is played because this is necessary to answer questions arising if you put yourself into the other players' shoes and ask what they would do if you chose differently. The longer answer is that this is somewhat problematic, though. Ariel Rubinstein has a good discussion of these issues in <a href="https://arielrubinstein.tau.ac.il/papers/37.pdf" rel="nofollow noreferrer">this 1991 comment</a>.</p>

CC BY-SA 4.0
2022-10-10T14:35:27.580Z
2022-10-10T14:35:27.580Z
VARulle
<p>Question from Strategy by Joel Watson:
<a href="https://i.sstatic.net/1v78Z.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1v78Z.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/ZStdY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZStdY.png" alt="enter image description here" /></a></p>
<p>I understand that the answer should be &quot;not hire is not a strategy&quot; but I don't see the problem with it being one. What's the point of keeping {NFF', NFR', NRF', NRR'} in Manager's strategy space instead of just N when we know that NFF' etc. cannot happen anyway? Isn't it inefficient in terms of storage too? I read the author's explanation:</p>
<blockquote>
<p>One reason for this is that our study of rationality will explicitly
require the  evaluation of players’ optimal moves starting from
arbitrary points in a game.  This evaluation is connected to the
beliefs that players have about each other.  For example, in the game
depicted in Figure 3.1(b), player l’s optimal choice at  his first
information set depends on what he thinks player 2 would do if put on
the move. Furthermore, to select the best course of action,
perspicacious player  2 must consider what player 1 would do at his
second information set. Thus,  player 2 must form a belief about
player l’s action at the third node. A belief is  a conjecture about
what strategy the other player is using; therefore, player l’s
strategy must include a prescription for his second information set,
regardless of  what this strategy prescribes for his first information
set.</p>
</blockquote>
<p>But NFF' is not even an &quot;arbitrary point&quot; in the game. Can anyone explain why keep 4 instead of 1 element in the strategy space in this case?</p>

CC BY-SA 4.0
2022-10-03T19:24:56.887Z
Does this answer your question? [When action and strategy differ in game theory](https://economics.stackexchange.com/questions/15747/when-action-and-strategy-differ-in-game-theory)
Bayesian
CC BY-SA 4.0
2022-10-04T10:57:04.753Z
NF', NF, NR, and NR' are not strategies of the Manager. These should be NFF', NFR', NRF', and NRR'.
VARulle
CC BY-SA 4.0
2022-10-10T08:37:58.837Z
@VARulle thanks for the correction. Fixed it.
learner
<game-theory>
What's the point of keeping impossible strategies in strategy space?
learner
# What's the point of keeping impossible strategies in strategy space?

(Asked by: username_0 on 2022-10-03)

Question from Strategy by Joel Watson:
[![enter image description here](https://i.sstatic.net/1v78Z.png)](https://i.sstatic.net/1v78Z.png)
[![enter image description here](https://i.sstatic.net/ZStdY.png)](https://i.sstatic.net/ZStdY.png)

I understand that the answer should be "not hire is not a strategy" but I don't see the problem with it being one. What's the point of keeping {NFF', NFR', NRF', NRR'} in Manager's strategy space instead of just N when we know that NFF' etc. cannot happen anyway? Isn't it inefficient in terms of storage too? I read the author's explanation:

> One reason for this is that our study of rationality will explicitly
> require the evaluation of players’ optimal moves starting from
> arbitrary points in a game. This evaluation is connected to the
> beliefs that players have about each other. For example, in the game
> depicted in Figure 3.1(b), player l’s optimal choice at his first
> information set depends on what he thinks player 2 would do if put on
> the move. Furthermore, to select the best course of action,
> perspicacious player 2 must consider what player 1 would do at his
> second information set. Thus, player 2 must form a belief about
> player l’s action at the third node. A belief is a conjecture about
> what strategy the other player is using; therefore, player l’s
> strategy must include a prescription for his second information set,
> regardless of what this strategy prescribes for his first information
> set.

But NFF' is not even an "arbitrary point" in the game. Can anyone explain why keep 4 instead of 1 element in the strategy space in this case?

**Tags:** game-theory
> **Comments:**
> **username_1**: Does this answer your question? [When action and strategy differ in game theory](https://economics.stackexchange.com/questions/15747/when-action-and-strategy-differ-in-game-theory)  
> **username_2**: NF', NF, NR, and NR' are not strategies of the Manager. These should be NFF', NFR', NRF', and NRR'.  
> **username_0**: @VARulle thanks for the correction. Fixed it.  

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## 1 answers

### Answer by username_2 on 2022-10-10. Score: 1

The short answer is that a strategy must include plans of action in information sets that cannot be reached if the strategy is played because this is necessary to answer questions arising if you put yourself into the other players' shoes and ask what they would do if you chose differently. The longer answer is that this is somewhat problematic, though. Ariel Rubinstein has a good discussion of these issues in [this 1991 comment](https://arielrubinstein.tau.ac.il/papers/37.pdf).


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