We start with a binary string of length $2a>0$ which contains exactly $a$ zeros and $a$ ones and we call it $L_{2a}$. For every whole number $n\ge2a$, we form $L_{n+1}$ by picking a random bit from $L_n$ with unifrom probability ($\frac{1}{n}$ for each bit in $L_n$) and concatenating it to $L_n$. So, if we picked $0$, then $L_{n+1}$ is just $L_0$ "plus" a zero or equivalently "plus" a one if we picked $1$ initially. Does the value $\frac{p_n}{n}$ where $p_n$ is the number of ones in $L_n$ converge to $0$ or $1$ almost always, i.e. with probability 1, as $n$ tends to $+\infty$?
EDIT: After some research, I vaguely insinuated some relations with brownian motions, but such where steps get smaller and smaller, enough so that the path converges. It looks like the ratio can converge to arbitrary number from $(0,1)$, but the interesting part now becomes that the distribution appears to be unifrom over that interval. At my first attempt, I thought the ratio has to converge to a border because there is always more probability to go there instead of towards the $\frac{1}{2}$ (center), but the expected value for the ratio after one and each step is the same ratio as before, so statistically there is no preference of direction in short-terms. Dispersion is of order $\frac{1}{n}$, so it should collapse the parabola in which the normal brownian path would stay.
