We prove that from a topological point of view most numbers fail to be normal in a spectacular way. For an integer $N \geq 2$ and $x \in [0, 1]$, let $x = \sum^\infty_{n=1} \varepsilon_{N,n}(x)/N^n$, where $\varepsilon_{N, n}(x) \in \{0, 1,\ldots, N - 1\}$ for all $n$, denote the unique non-terminating $N$-adic expansion of $x$. For a positive integer $n$ and a finite string ${\bf i} = i_1\cdots i_k$ with entries $i_j \in \{0, 1,\ldots, N - 1\}$, we write $$ \Pi_N (x,{\bf i};n) = \frac{|\{1 \le i \le n | \varepsilon_{N,i}(x) = i_1, \ldots, \varepsilon_{N,i+k-1} (x) = i_k\}|}{n} $$ for the frequency of the string ${\bf i}$ among the first $n$ digits in the $N$-adic expansion of $x$, and let $\Pi_N^k(x;n) \,{=}\, (\Pi_N(x,{\bf i};n))_{\bf i}$ denote the vector of frequencies $\Pi_N(x,{\bf i};n)$ of all strings ${\bf i} = i_1\cdots i_k$ of length $k$ with entries $i_j \in \{0, 1,\ldots, N-1\}$. We say that a number $x$ is extremely non-normal if each shift invariant probability vector in ${\mathbb R}^{N^k}$ is an accumulation point of the sequence $(\Pi_N^k(x;n))_n$simultaneously for all $k$ and all bases $N$, and we denote the set of extremely non-normal numbers by ${\mathbb E}$, i.e. \begin{eqnarray*} &&{\mathbb E} = \bigcap_N \bigcap_k \big\{x \in [0,1]|\, {\rm each}\, {\bf p} \in \Gamma^k_N\\ &&\quad\hbox{is an accumulation point of the sequence } (\Pi_N (x, {\bf i};n))_n\big\}, \end{eqnarray*} where $\Gamma^k_N$ denotes the simplex of shift invariant probability vectors in ${\mathbb R}^{N^k}$. Our main result says that ${\mathbb E}$ is a residual set, i.e. the complement $[0,1]\backslash {\mathbb E}$ is of the first category. Hence, from a topological point of view, a typical number in [0, 1] is as far away from being normal as possible. This result significantly strengthens results by Maxfield and Schmidt. We also determine the Hausdorff dimension and the packing dimension of ${\mathbb E}$.