Pólya urn model ($X_n=B_n/(B_n+R_n),\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$)

Begin with positive counts of two colours. At each step select one ball uniformly and replace it together with one new ball of its colour. If the total initial count is $m$, the blue proportion satisfies $X_{n+1}-X_n=(I_{n+1}-X_n)/(m+n+1)$, where the next blue indicator has conditional mean $X_n$. Thus $X_n$ is a bounded martingale and converges almost surely and in $L^2$ to a random proportion. If uniforms generate the draws, the update from time $n$ to $n+1$ must use the fresh $U_{n+1}$ relative to ${\mathcal{F}}_n=\sigma (U_1,\dots ,U_n)$.

Both colour counts tend to infinity almost surely. For example, $X_n\ge B_0/(m+n)$, and independent events $U_{n+1}\le B_0/[2(n+1)]$ occur infinitely often by the Borel-Cantelli second lemma, forcing infinitely many blue additions.

The martingale increment formula gives $\mathbb{E}[(X_{k+1}-X_k{)}^2\mid {\mathcal{F}}_k]\le X_k/(m+k+1{)}^2$. Conditional orthogonality of later increments and $\mathbb{E}[X_k\mid {\mathcal{F}}_n]=X_n$ implyOn $X_{\infty }=0$ the squared difference is $X_n^2$, so its conditional probability is at most $1/[(m+n)X_n]=1/B_n\to 0$. Dominated convergence proves zero probability of that endpoint. The symmetric argument for red excludes $X_{\infty }=1$. No explicit beta law is needed for this proof.