Battle of the Sexes, Mixed Strategy BNE below is a question pertaining to finding mixed strategies BNE:


So for this question I know how to calculate the probabilities at which each player should be indifferent.
Alice will be indifferent in the Mean and Nice case when Pr(pub)= 5/6 and Pr(cafe) = 1/6)
Given there is a 50/50 shot of being mean of nice, bob can either play Pub always when nice and 2/3 when mean, or 2/3 when nice and always when mean.
Mean Bob will be indifferent when Alice plays Pr(Cafe) = 1/6 and Pr(pub) = 5/6. Nice Bob will be indifferent when Alice plays Pr(Cafe) = 5/6 and Pr(Pub) = 1/6.
However, after this point I am pretty stuck. I am not sure how to combine these two to determine the mixed-strategy equilibrium. I have the answer, (below) but I do not know how to interpret it.
 
Thank you! 

It is useful to think of this as a game with three agents: $A$ (Alice), $B_m$ (Bob mean), and $B_n$ (Bob nice), each trying to best reply to what the (relevant) opponents do conditional on the available information.
Consider the strategy profile $(C,CP)$. This is an equilibrium because each agent is best replying: by choosing $C$, Alice gets an expected payoff $(1/2)5+(1/2)2=2.5$ against the $(1/2)0+(1/2)3=1.5$ she would get by choosing $P$. Similarly, by choosing $C$ against Alice's $C$, $B_m$ gets 3 instead of 2 from deviating to $P$. The same holds for $B_n$. You can check that at least one agent would prefer to deviate in any other pure-strategy profile.
Consider mixed-strategy profiles. Alice is indifferent (and hence willing to randomise) if she expects $C$ with probability $1/6$ and $P$ otherwise. The choice of $C$ comes from the choices of the two types of Bob, who are equally likely. So, if we denote by $p_i$ (for $i=m,n$) the respective probability with which $B_i$ plays $C$, Alice expects to see $C$ with overall probability $p=(1/2)p_m+(1/2)p_n$. Alice is indifferent if $p=1/6$ so we need to have $B_m$ and $B_n$ independently choosing their mixed strategies so that $p=(1/2)p_m+(1/2)p_n =1/6$. 
On the other hand, an agent $B_i$ is willing to randomise (and set $p_i \not\in \{0,1\}$), only if he is indifferent. $B_m$ is indifferent only if $A$ plays $C$ with probability $1/6$, whereas in this case $B_n$ strictly prefers $P$. This yields $p_m=2/3$ and $p_n=0$. This gives the first equilibrium in mixed strategies in your solution. $B_n$ is indifferent only if $A$ plays $C$ with probability $5/6$, whereas in this case $B_m$ strictly prefers $P$. This yields $p_n=2/3$ and $p_m=0$. This gives the second equilibrium in mixed strategies in your solution.
