Tensor Decomposition Consider a tensor product 
$$ V^{\otimes n} = \underbrace{V\otimes\cdots\otimes V}_{n} $$
where $V$ is a vector space over $\mathbb R$, $\dim V = m$ , hence $\dim V^{\otimes n} = m^n$ .
So every $A \in V^{\otimes n}$ can be represented as
$$A = \sum_{i=1}^r a^i_1 \otimes a^i_2 \ldots \otimes a^i_n, \;\;\; a_i \in V $$
in a non-unique way. Taking $R$ to be minimum $r$ among all the possible decompositions of A.
$$R = \min \left \{ r : A = \sum_{i=1}^r a^i_1 \otimes a^i_2 \ldots \otimes a^i_n, \;\;\; a_i \in V \right \}$$
How many tensors have certain $R$ ? How many tensors have $R=1$? Or $R = m^n$ ? What is the typical $R$ (mean, median mean, the most probable), what is the distribution?
IMPORTANT How should I imagine (picture) tensors for which $R$ is (near) maximum? What hinders them from decomposition?
Maybe there are some experimental data. I'm mostly interested in high $m$'s and $n$'s, though every answer is welcome.
 A: If $V$ would have been a vector space over $\mathbb C$ instead, there is only one value of $R$ where the set of tensors having rank $R$ has non-zero (Lebesgue) measure (this single value of $R$ is called the generic rank). As Yrogirg says, this $R$ is expected to be 
$$\left\lceil \frac{m^n}{mn - m + 1}\right\rceil.$$
However, this is not always the case. For example, over $\mathbb C^3 \otimes \mathbb C^3 \otimes \mathbb C^3$ the generic rank is 5.
Over $\mathbb R$ the situation is more complicated and we can have multiple values of $R$ where the set of tensors having rank $R$ has non-zero measure. These $R$ are called typical ranks. For example, in $\mathbb R^2 \otimes \mathbb R^2 \otimes \mathbb R^2$ both 2 and 3 are typical ranks (and 3 is the maximal rank). Of course, in the case of $\mathbb R^n \otimes \mathbb R^n$ the single typical rank is $n$.
Determining the typical ranks for tensors over $\mathbb R$ is an open question, and it's mostly third order tensors which have been studied. A way of determining the minimal typical rank over $\mathbb R$ numerically is described in 
P. Comon, J.M.F. ten Berge, L. De Lathauwer, J.Castaing
(2009), Generic and typical ranks of multi-way arrays, Linear Algebra and its Applications, 430, 2997-3007.
A: Well, I've found some sort of conjecture/rule of thumb, that the "expected rank" for both (?) complex and real tensors is almost everywhere
$$\frac{m^n}{mn - m + 1}$$
At least that's how I understood it. It seems that this estimation could be obtained rather trivially, just by counting the number of degrees of freedom, but I couldn't understand that $-m+1\;$ part.
Check "Tensor Decompositions, Alternating Least Squares and other Tales." by P. Comon, X. Luciani and A. L. F. de Almeida for the details. In addition, it is a good starting to point for people interested in tensor decomposition and Comon's webpage features software (MATLAB codes) for tensor decomposition.
