Comparing Static Frictions In this figure, which of the static frictional forces will be more?

My aim isn't to solve this particular problem but to learn how is static friction distributed . Since each of the rough-surfaces are perfectly capable of providing the $-1N$ horizontal frictional force but why don't they ? This is kind of ambiguity that who will provide a bigger share in total static friction. And as the surface have different $\mu$, so we can't even invoke symmetry.

The contact forces with two blocks are $N_1 = m_1 g + m_2 g$ for the bottom block (to the floor) and $N_2 = m_2 g$ for the top block (to the 1st block).
The available traction is $F^\star_1 = \mu_1 (m_1+m_2)\,g$ and $F^\star_2 = \mu_2  m_2\, g$ or
$$ \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}\mu_1 m_1\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$
The balance of horizontal forces is
$$ \boxed{ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix}} $$
where $P_1$, $P_2$ are any applied forces on the blocks (in your case $P_1=1N,\; P_2=0N$) and $F_1$, $F_2$ are the friction forces. Here comes the fun part:
Assume blocks are sticking and solve for the required friction $F_1$, $F_2$ when $\ddot{x}_1=\ddot{x}_2=0$
$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} = \begin{bmatrix}1&-1\\0&1\end{bmatrix} ^{-1} \begin{pmatrix}P_1\\P_2\end{pmatrix} = \begin{pmatrix}P_1+P_2\\P_2\end{pmatrix} $$
Find the cases where required friction exceeds traction
$$ \begin{pmatrix}F_1\\F_2\end{pmatrix}_{stick} > \begin{pmatrix}F_1^\star\\F_2^\star\end{pmatrix} = \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} $$
For those cases set $F_i = F_i^\star$ otherwise set $\ddot{x}_i = \ddot{x}_{i-1}$ and solve the balance of horizontal forces.
Example 1, All slipping:
$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\\mu_2 m_2\,g\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_2\end{pmatrix} $$
to be solved for $\ddot{x}_1$ and $\ddot{x}_2$
Example 2, All sticking:
$$\begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}F_1\\F_2\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$$
to be solved for $F_1$ and $F_2$
Example 3, Bottom slipping, top sticking:
$$ \begin{pmatrix}P_1\\P_2\end{pmatrix} - \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{pmatrix}\mu_1 (m_1+m_2)\,g\\ F_2\end{pmatrix} = \begin{pmatrix}m_1 \ddot{x}_1\\m_2 \ddot{x}_1\end{pmatrix} $$
to be solved for $F_2$ and $\ddot{x}_1$.
The matrix $A=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$ is the connectivity matrix, and it can be expanded if you have more blocks. See my full solution here of similar problem in more detail: https://physics.stackexchange.com/a/79182/392
