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! 

Here is a way to think about it. Let $a_c=Pr($alice plays cafe$)$ and $b_c^i=Pr($ bob of type i plays cafe$)$ where $i=\{nice(n), mean(m)\}$. 
The only way Alice will play a mix strategy is if she is indifferent given Bob's actions, i.e
$$
\underbrace{\frac{1}{2}\left[5b_c^n+2(1-b_c^n)\right] + \frac{1}{2}\left[5b_c^m+2(1-b_c^n)\right]}_{\text{Expected payoff if Alice plays Cafe}} 
$$
$$
= \underbrace{\frac{1}{2}\left[0b_c^n+3(1-b_c^n)\right] + \frac{1}{2}\left[0b_c^m+3(1-b_c^n)\right]}_{\text{Expected payoff if Alice plays Pub}} 
$$
or simply
$$
b_c^n+b_c^m=\frac{1}{3}.
$$
The above expression gives us a relationship between how the different types of Bob must randomize in order to leave Alice indifferent. We don't need both types to randomize in order for Alice to be indifferent but at least one of Bob's types must. 
The nice type is indifferent if:
$$
\underbrace{3a_c+0(1-a_c)}_{\text{Nice plays Cafe}}=\underbrace{2a_c+5(1-a_c)}_{\text{Nice plays Pub}}
$$
which happens only if $a_c=\frac{5}{6}$. Nice Bob strictly prefers Pub if $a_c<\frac{5}{6}$, and strictly prefers Cafe if $a_c>\frac{5}{6}$.
The mean type is indifferent if:
$$
\underbrace{0a_c+3(1-a_c)}_{\text{Mean plays Cafe}}=\underbrace{5a_c+2(1-a_c)}_{\text{Mean plays Pub}}
$$
which happens only if $a_c=\frac{1}{6}$. Mean Bob strictly prefers Cafe if $a_c<\frac{1}{6}$, and strictly prefers Pub if $a_c>\frac{1}{6}$. 
Now we can construct out mixed strategy BNE. One is Alice mixes with $a_c=\frac{5}{6}$, Nice Bob mixes with $b_c^n=\frac{1}{3}$ and mean Bob plays Pub ($b_c^m=0$). In the expression given by the question, this is Alice playing $5/6C+1/6P$, and Bob playing $1/3 CP+2/3PP$.
Another is where Alice mixes with $a_c=\frac{1}{6}$, Nice Bob plays pub for sure ($b_c^n=0$) and mean Bob mixes $b_c^m={1}{3}$. In the expression given by the question, this is Alice playing $1/6C+5/6P$, and Bob playing $1/3 PC+2/3PP$.
There is no equilibrium where both types of Bob mix simultaneously. 
Hope this helps. 
