Inference from Nash Equilibrium Violates Total Probability This is connected to a question I asked at Cross-Validated.
Given a game:
          k
        x   y
    x: 1*,2|1*,1*
j
    y: 1*,1*|2,3

The best choices are marked with stars. Inference from the Nash Equilibrium tells me that $p(A_j=x)=\frac{2}{3}$, $p(A_k=y)=\frac{1}{2}$, and $p(A_k=y|A_j=x)=1$, where $A_i$ is the action of person $i$. However, this inference does not satisfy total probability because $$p(A_k=y) = \sum_x' p(A_k=y|A_j=x')p(A_j=x') \geq p(A_k=y|A_j=x)p(A_j=x)=\frac23.$$ Thus, I'm obviously inferring wrongly from the Nash Equilibrium. How can I correctly find the probabilities I wish to find given the Nash Equilibrium?
 A: You seem to be badly confused about the basic meaning of a game like this (or perhaps what a Nash equilibrium is).  Here's how it's played.  Player $j$ and player $k$ each secretly choose to move either $x$ and $y$ (and they may make this choice randomly).  Then they reveal their moves and receive payoffs according to the table.
So, what strategies are $j$ and $k$ allowed to use?  They don't have much flexibility, since they have to make their choices before they've seen the other player's move.  The only choice they have to make is the probability with which they choose each move.  So $j$ can choose to move $x$ with probability $p$, say (this is what you called $P(A_j=x)$), and $y$ with probability $1-p$.  Meanwhile, $k$ can choose to move $x$ with probability $q$ (what you called $P(A_k=x)$), and $y$ with probability $1-q$.  Crucially, these events $A_k=x$ and $A_j=x$ are independent, because $j$ and $k$ make their random choices without knowing the other player's move yet.  In particular, $P(A_k=y|A_j=x)$ must always be the same as $P(A_k=y)$.
Now, what is a Nash equilibrium?  It's a choice of strategies for each player (i.e., values of $p$ and $q$) such that neither player could do better (in expected value) by changing their strategy while the other player's strategy remains the same.  This game actually has infinitely many Nash equilibria.  To analyze it, let's start by looking from the perspective of $j$.  Note that $x$ is always a better move for $j$ than $y$, unless $k$ is always going to pick $x$ (i.e., $q=1$) in which case $j$'s choice does not matter.  This means that in any Nash equilibrium with $q\neq 1$, $p$ must be $1$ since moving $x$ is strictly better than moving $y$ for $j$.  If $j$ is moving $x$, then the best choice for $k$ is to move $y$.  This gives the first equilibrium $(p,q)=(1,0)$, where $j$ always moves $x$ and $k$ always moves $y$.
That's not the only Nash equilibrium, though, because we can only conclude that $p$ must be $1$ if $q\neq 1$.  If $q=1$, then $j$'s choice is irrelevant (for $j$) and so all strategies are equally good for $j$.  So, we can have a Nash equilibrium $(p,1)$ for any value of $p$ such that $k$'s best move is $x$, given that $j$ moves $x$ with probability $p$.  A bit of calculation shows that this is true as long as $p\leq 2/3$.  So for any $c\leq 2/3$, $(p,q)=(c,1)$ is also a Nash equilibrium.
A: Any intermediate level microeconomics text will do the job. Moreover, game theory itself involves probability, and in some context (e.g. when you have incomplete information), you do consider Bayes-Nash equilibria. Here is what I used when I learned micro. This text is more advanced.
