# Mixed Strategies in Bayes Nash Equilibrium (Bayesian Battle of the Sexes). Shouldn't it depend on $p$?

I have a question about calculating mixed strategies in a Bayes Nash Equilibrium in a simple 2-player bimatrix game. To demonstrate the issue, consider Bayesian Battle of the Sexes.'' Suppose P1 faces a P2 whose type is unknown with probability $$p$$. An example is below in which $$p=.5$$.

Here's what's surprising to me. If you calculate the mixing strategy for P1 by making sure that P2 is indifferent, you wind up with P1 mixing strategies that don't depend at all on $$p$$. This seems strange to me.

If P2's type is more clear, shouldn't the optimal mixed strategy shift to anticipate this?

End of question. Addendum: The game above apparently has two mixed strategy EQMs:

1. P1 mixes $$(\frac{1}{3},\frac{2}{3})$$, P2 left mixes (0,1), P2 right ($$\frac{2}{3},\frac{1}{3}$$).
2. P1 mixes $$(\frac{2}{3},\frac{1}{3})$$, P2 left mixes ($$\frac{2}{3},\frac{1}{3}$$), P2 right (0,1).

Like I said, this doesn't seem to depend on $$p$$ (the probability of facing the first type).

Also: Shouldn't this game have an odd # of EQM? I thought this was a rule of thumb. I am contemplating the idea that there's a third equilibrium that mixes the two above.

Given game also has one pure-strategy Nash equilibrium besides the two you mentioned: Row player plays $$C$$, Column player (left) plays $$C$$ and right plays $$H$$ i.e. $$C,H$$.
Also, when player 1 chooses a mixed strategy in order to make player 2 indifferent, that does not necessarily mean that player 1 is choosing optimally. It is only when player 2 makes player 1 indifferent, then the mixed strategy chosen by 1 becomes optimal. In fact as $$p$$ gets close to $$1$$ (not necessarily equals 1), given that player 2 (left) chooses $$H$$ and player 2 (right) chooses $$(\frac{2}{3}, \frac{1}{3})$$, player 1 will not choose $$(\frac{1}{3},\frac{2}{3})$$, since player 1 is no longer indifferent between $$C$$ and $$H$$. Playing $$C$$ will give row player a payoff of $$\frac{20}{3}(1-p)$$, while playing $$H$$ gives him $$\frac{10p}{3}+\frac{5}{3}$$ and these are only equal when $$p=\frac{1}{2}$$.
Suppose player 1 chooses a mixed strategy such that the left type of player 2 is indifferent between C and H. There is a unique mixed strategy that does this. The same mixed strategy, however, does NOT make the right type of player 2 indifferent. Loosely speaking, this is simply because the preferences of the left type are exactly those of the right type switched around. Thus it is impossible for player 1 to make both types indifferent simultaneously. Therefore, the probability $$p$$ plays no role when choosing a strategy such that at least one of them is indifferent.