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! 
 A: 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.
A: 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. 
