You might be interested in [the article][1] by David A. Freedman on Friedman's urn.


>D. Ornstein has obtained this very intuitive proof that $(W_n + B_n)^{-1}W_n$ converges
to $1/2$ with probability 1 for $b > 0$. Suppose first $a > b$. If $0 \leq  x \leq 1$ and
$$\mathbb P\left\{\limsup \frac{W_n}{W_n + B_n} \leq x\right\} = 1,$$ 
by an easy variation of the Strong Law, with probability 1, in $N$ trials there will be at most $Nx + o(N)$ drawings of a white ball; so at least $N(1 - x) - o(N)$ drawings of black. Therefore, with probability 1, $\limsup (W_n+ B_n)^{-1}B_n$ is bounded above by
$$\lim\limits_{N\to\infty}\frac{a[Nx + o(N)] + b[N(1 - x) - o(N)]}{N(a + b)}=\frac{b+(a-b)x}{a+b}.$$
Starting with $x = 1$ and iterating,
$$\mathbb P\left\{\limsup \frac{W_n}{W_n + B_n} \leq \frac{1}{2}\right\} = 1$$
follows. Interchange white and black to complete the proof for $a > b$. If $a < b$, and 
$$\mathbb P\{\limsup (W_n + B_n)^{-1}W_n \leq x\} = 1,$$
then a similar argument shows 
$$\mathbb P\left\{\limsup\frac{B_n}{W_n + B_n} < \frac{a+(b-a)x}{a + b}\right\}=1$$
The argument proceeds as before, except both colors must be considered simultaneously.


  [1]: http://www.jstor.org/stable/2238205?seq=2