What is the intution behind the ping-pong lemma? Is the ping-pong lemma a difficult characterization of free groups? Or is it just me? Does someone have a nice intuition about its idea or should I carry on staring at the statement?
 A: The basic idea is very simple and nice. Suppose that $A$ and $B$ are matrices, or more generally maps, such that $Av$ and $Bw$ are different for all $v$ and $w$. The intuition is that $A$ maps in one direction and $B$ in the other ($v$ and $w$ being balls crossing the "ping-pong table").
Now suppose you have two words in $A$ and $B$ that are equal, say
$$
ABBABA=AAABBB.
$$
Let's multiply by $A^{-1}$ and $B^{-1}$ as needed to strip off any leading identical letters; we get
$$
BBABA=AABBB.
$$
Well, then with $w=BABAx$ and $v=ABBBx$ (for any ping-pong ball $x$) we get $Av=Bw$, which was supposed to be impossible. Done!
A: I'm an outsider, so maybe what I'm saying is silly. I would just think about the Tits alternative application since that really gives all the intuition.  To apply it---and we'll just look at the complex case---you should produce two matrices $A$ and $B$ that have the following properties. $A$ has a dominant eigenvalue $\lambda$ (i.e., the eigenspace of $A-\lambda I$ is one-dimensional and all other eigenvalues of $A$ have modulus strictly less than $|\lambda|$) with corresponding eigenvector $v$ and $A^{-1}$ has a dominant eigenvalue $\mu$ with coresponding eigenvector $w$.  Similarly, let's suppose that $B$ and $B^{-1}$ have these dominant eigenvectors with corresponding eigenvectors $v'$ and $w'$.  
Then if we consider $v,w,v',w'$ as being points in projective space then the dominant eigenvalue hypotheses say that there are open disjoint neighbourhoods of $v,w,v',w$, say $U_v, U_w, U_{v'}, U_{w'}$ such that there is some $N$ such that for $n\ge N$ we have: $A^n$ maps each of these neighbourhoods into $U_v$, $A^{-n}$ maps all these neighbourhoods into $U_{w}$, $B^n$ maps each of these neighbourhoods into $U_{v'}$, and $B^{-n}$ maps all these neighbourhoods into $U_{w'}$.  
Now I think of it as being like a finite-state machine with four states labelled $U_v, U_w, U_{v'}, U_{w'}$ and an alphabet $A^N, A^{-N}, B^N, B^{-N}$ with transitions as described before; e.g., if you are in state $U_w$ and you read $A^N$ then you transition to state $U_v$ (and we'll read words right-to-left).  
Now you can see that the group generated by $A^N$ and $B^N$ is free since if words $W$ and $W'$ on $A^N, A^{-N}, B^N, B^{-N}$ are the same then when we let $W$ act on $U_v$ it will send it to $U_v$ if and only if the first letter of $W$ is $A^N$; to $U_w$ iff the first letter of $W$ is $A^{-N}$, etc.  Thus we see that if $W=W'$ then the first letters of $W$ and $W'$ must be the same.  Now we cancel and keep going.  Well, maybe this isn't what you wanted, but I tried my best and maybe we're both better off because of this.  Palin out.  
