# Rule for Number of Strategies in Bayesian Games

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.

• Bayesian games is a pretty big category. Are you thinking only of games were each player only makes one decision simoultaneosly with other players? May 27, 2021 at 7:43
• "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? May 27, 2021 at 7:44

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.
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, $$S=A^T$$.