<p>Consider a type space <span class="math-container">$\mathcal T$</span> of size <span class="math-container">$T$</span>, an action space <span class="math-container">$\mathcal A$</span> of size <span class="math-container">$A$</span>, and the corresponding set of pure strategies <span class="math-container">$\mathcal S$</span> of size <span class="math-container">$S$</span>. By definition, a pure strategy is a mapping from <span class="math-container">$\mathcal T$</span> to <span class="math-container">$\mathcal A$</span>, implying that <span class="math-container">$\mathcal S=\mathcal A^\mathcal T$</span>. Therefore, <span class="math-container">$S=A^T$</span>.</p>

CC BY-SA 4.0
2021-05-28T13:14:08.680Z
2021-05-28T13:14:08.680Z
VARulle
<p>Consider a type space <span class="math-container">$\mathcal T=\{1,2,\dots,T\}$</span> and an action space <span class="math-container">$\mathcal A=\{1,\dots,A\}$</span>. With <span class="math-container">$A=T=2$</span>, you correctly found that there are <span class="math-container">$S=4$</span> different pure strategies mapping <span class="math-container">$\mathcal T \to \mathcal A$</span>. Now fix <span class="math-container">$A$</span> and add one more type, so <span class="math-container">$T=3, A=2$</span>. Type 1 and 2 still have the same number of different actions (4), and the strategy can be completed by two different actions by type 3. That is, <span class="math-container">$S= 4\cdot2=8$</span>. If you added a fourth type, these <span class="math-container">$8$</span> combinations can again be completed to a full strategy with 2 different actions, i.e., <span class="math-container">$S=16$</span>... and so on such that <span class="math-container">$S=2^T$</span>.</p>
<p>Next, go back to <span class="math-container">$A=T=2$</span>, fix <span class="math-container">$T$</span> and add actions. You see that type 1 has <span class="math-container">$A$</span> actions that can each be paired with <span class="math-container">$A$</span> actions by type 2, giving you <span class="math-container">$S=A^2$</span>.</p>
<p>Combining your insights you find <span class="math-container">$S=A^T$</span>. In simultaneous-move games, <span class="math-container">$\mathcal A$</span> is just the set of simultaneous actions. In more general games, <span class="math-container">$\mathcal A$</span> contains all complete plans of actions for each contingency in the game.</p>

CC BY-SA 4.0
2021-05-27T10:33:05.230Z
2021-05-27T10:33:05.230Z
Bayesian
<p>Is there a general rule for finding the number of strategies (denote as <span class="math-container">$S$</span>) for each player in a Bayesian game? I think it's related to the number of types (denote as <span class="math-container">$T$</span>) and the number of actions (denote as <span class="math-container">$A$</span>). From the few examples I've come across, I've noticed the following:</p>
<ul>
<li>2 actions, 2 types, 4 strategies</li>
<li>2 actions, 1 type, 2 strategies</li>
<li>3 actions, 1 type, 3 strategies</li>
</ul>
<p>This could mean that <span class="math-container">$S=T \times A$</span> but it could also be <span class="math-container">$S=A^{T}$</span>.</p>
<p>Any clarification would be appreciated.</p>

CC BY-SA 4.0
2021-05-27T07:43:41.307Z
Bayesian games is a pretty big category. Are you thinking only of games were each player only makes one decision simoultaneosly with other players?
Giskard
CC BY-SA 4.0
2021-05-27T07:44:46.183Z
"This could mean that $S=T \times A$ but it could also be $S=A^T$." If you understand what a *strategy* is, then it seems like you could settle this by creating and examining a game with 2 actions, 3 types or 3 actions, 2 types?
Giskard
<game-theory><self-study><bayesian-game>
Rule for Number of Strategies in Bayesian Games
Wivaviw
# Rule for Number of Strategies in Bayesian Games

(Asked by: username_0 on 2021-05-27)

Is there a general rule for finding the number of strategies (denote as $S$) for each player in a Bayesian game? I think it's related to the number of types (denote as $T$) and the number of actions (denote as $A$). From the few examples I've come across, I've noticed the following:

* 2 actions, 2 types, 4 strategies
* 2 actions, 1 type, 2 strategies
* 3 actions, 1 type, 3 strategies

This could mean that $S=T \times A$ but it could also be $S=A^{T}$.

Any clarification would be appreciated.

**Tags:** game-theory, self-study, bayesian-game
> **Comments:**
> **username_1**: Bayesian games is a pretty big category. Are you thinking only of games were each player only makes one decision simoultaneosly with other players?  
> **username_1**: "This could mean that $S=T \times A$ but it could also be $S=A^T$." If you understand what a *strategy* is, then it seems like you could settle this by creating and examining a game with 2 actions, 3 types or 3 actions, 2 types?  

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## 2 answers

### Answer by username_3 on 2021-05-27. Score: 3 | Accepted answer

Consider a type space $\mathcal T=\{1,2,\dots,T\}$ and an action space $\mathcal A=\{1,\dots,A\}$. With $A=T=2$, you correctly found that there are $S=4$ different pure strategies mapping $\mathcal T \to \mathcal A$. Now fix $A$ and add one more type, so $T=3, A=2$. Type 1 and 2 still have the same number of different actions (4), and the strategy can be completed by two different actions by type 3. That is, $S= 4\cdot2=8$. If you added a fourth type, these $8$ combinations can again be completed to a full strategy with 2 different actions, i.e., $S=16$... and so on such that $S=2^T$.

Next, go back to $A=T=2$, fix $T$ and add actions. You see that type 1 has $A$ actions that can each be paired with $A$ actions by type 2, giving you $S=A^2$.

Combining your insights you find $S=A^T$. In simultaneous-move games, $\mathcal A$ is just the set of simultaneous actions. In more general games, $\mathcal A$ contains all complete plans of actions for each contingency in the game.


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### Answer by username_2 on 2021-05-28. Score: 1

Consider a type space $\mathcal T$ of size $T$, an action space $\mathcal A$ of size $A$, and the corresponding set of pure strategies $\mathcal S$ of size $S$. By definition, a pure strategy is a mapping from $\mathcal T$ to $\mathcal A$, implying that $\mathcal S=\mathcal A^\mathcal T$. Therefore, $S=A^T$.


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