"Fair" game in Williams In David Williams' Probability with Martingales, $\exists$ this exercise.
What's fair about a fair game?
Let $X_i$, i = 1, 2, ... be indp RVs s.t. $X_i = i^2 - 1$ with prob $1/i^2$ and $-1$ with prob $1-1/i^2$.
I find it clear that $E(X_n)=0$. However, I do not understand why:
if $S_n = \sum_{i=1}^{n} X_i$, then $S_n/n \to -1$ a.s.
My attempt:
Proving $S_n/n \to -1$ a.s. is equivalent to proving $P(S_n/n = -1 \ i.o.) = 1$ or $P(S_n/n = -1 \ ev.) = 1$. I think?
If so, I just take the first one $P(S_n/n = -1)$, sum it over n's,hope it's infinity and use Borel-Cantelli 2...after proving the events $S_n/n = -1$ are independent, of course.
$P(S_n/n = -1)$
$=P(\sum_{i=1}^{n} X_i = -n)$
$=P(\bigcap_{i=1}^{n} X_i=-1)$ I think?
$=\Pi_{i=1}^{n} P(X_i=-1)$ by independence
$=0$ if I plug in i=1.
Help please?
 A: The idea is that, since the series $\sum\limits_nP(X_n\ne-1)$ converges, Borel-Cantelli lemma (the simple one) indicates that, almost surely, there exists some finite $N$ such that $X_n=-1$ for every $n\geqslant N$. Every such sequence $(X_n)$ is such that $S_n/n\to-1$ hence $P(S_n/n\to-1)=1$.
The strong law of large numbers does not apply, although $(X_n)$ is independent and $E(X_n)=0$ for every $n$, because the distribution of $X_n$ depends on $n$.
You wrote:

Proving $S_n/n \to -1$ a.s. is equivalent to proving $P(S_n/n = -1 \ i.o.) = 1$ or $P(S_n/n = -1 \ ev.) = 1$. I think?

Really, no. 
A general positive result in this vein is that $[Z_n\to -1]=[Z_n = -1 \ \text{eventually}]$ when every $Z_n$ is integer valued (but, even then, $[Z_n\to -1]$ is not the same event as $[Z_n = -1 \ \text{infinitely often}]$...). In your setting, the random variables $Z_n=S_n/n$ are not integer valued hence the inclusion $[Z_n = -1 \ \text{eventually}]\subset[Z_n\to -1]$ holds (it always does) but the other inclusion fails. As a matter of fact, $P(Z_n\to -1)=1$ but $[Z_n = -1 \ \text{eventually}]=\varnothing$ because $[Z_n=-1]=\varnothing$ for every $n$ because $P(X_1=-1)=0$ hence $Z_n\geqslant-1+2/n$ almost surely.
