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.
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.