Let $(\epsilon_k)_{k\geq 0}$ be independently uniformly chosen at random from $\{-1,1\}$. If $R_n$ counts the number of real roots of $f_n(z)=\sum_{0\leq k\leq n}\epsilon_k z^k$ then is it true that, almost surely, $$\lim_{n\to \infty}\frac{R_n}{\log n}=\frac{2}{\pi}?$$

Let $(\epsilon_k)_{k\geq 0}$ be independently uniformly chosen at random from $\{-1,1\}$. If $R_n$ counts the number of real roots of $f_n(z)=\sum_{0\leq k\leq n}\epsilon_k z^k$ then is it true that, almost surely, $$\lim_{n\to \infty}\frac{R_n}{\log n}=\frac{2}{\pi}?$$

The answer is no: this almost-sure limit fails. This result was obtained first by others, who deserve the credit for the problem; the link is to an independent machine-checked proof by Star Fleet Math.