Let $g, r, a, b$ be positive integers. In Friedman's urn model we have an urn with $r$ red and $g$ green balls in it. In each step we take one ball out of urn, register its color and return it to the urn. Additionally, we put $a$ more balls of this color and $b$ more balls of the other color.

Let $X_n$ be the relative amount of green balls in the box after $n$-th step. It can be proven (e.g. Richard Durrett, *Probability: Theory and Examples*, pages 254-255) that
$$\lim_{n\to\infty}X_n = \frac12\quad\text{a.e.}$$
If one thinks about this statement for a second it would most probably strike him as extremely counter-intuitive. The proof cited above make use of square integrable martingales, and unfortunately doesn't seem to give *intuitive* explanation of this phenomena. 
I'm looking for an explanation which would explain on some heuristic level why this result is in some sense logical. An idea for more intuitive proof would also definitely be helpful.

I would also like to note that if we take $b=0$ the model becomes well-known Polya-Eggenberger urn model for which we have
$$\lim_{n\to\infty}X_n\sim B\left(\frac{g}{a}, \frac{r}{a}\right).$$