Electrical resistance of two conductors of equal surface area in contact with another?
That one's a bit harder than it seems.

So, first of all, let's assume the conductivities of the two metals are in the same order of magnitude, so that a simplification that "all significant current is carried by the lower-resistance wire" isn't justified.

In a first approach, you could model the wires as sum of small resistors that are contacted regularly:

       x·₁   x·₁   x·₁  x·₁
A ---+-[===]-+-[===]-+-[===]-+-[===]-+--
     |       |       |       |       |
B ---+-[===]-+-[===]-+-[===]-+-[===]-+--
       x·₂   x·₂  x·₂   x·₂


With conductor A having the specific resistance (ohm per meter) ₁, and B having ₂. Each resistor than has a resistance x· ; x is the small wire length between contacts.

We set the total length \$X= N\cdot\Delta x\$. A single "parallel element" has the resistance

$$\begin{align}
R_{\Delta x} &amp;= (\Delta x\cdot\rho_1)||(\Delta x\cdot\rho_2)\\
&amp;= \frac{\Delta x\cdot\rho_1\,\cdot\,\Delta x\cdot\rho_2}{\Delta x\cdot\rho_1+\Delta x\cdot\rho_2}\\
&amp;=\frac{(\Delta x)^2\rho_1\rho_2}{\Delta x\cdot(\rho_1+\rho_2)}\\
&amp;=\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2}
\end{align}$$

The total resistance of A||B then becomes:

$$\begin{align}
R_X &amp;= N\cdot R_{\Delta x}\\
&amp;= N\frac{\Delta x\,\rho_1\rho_2}{\rho_1+\rho_2}&amp;\text{with $N=\frac{X}{\Delta x}$:}\\
&amp;= X\frac{\rho_1\rho_2}{\rho_1+\rho_2}
\end{align}$$

In other words, it's the same formula as if you put two resistors in parallel, just applied to the specific resistances (times the length of the conductors).

Notice that this construct will also be subject to the Seebeck Effect, which means that if you have a temperature difference between the ends of your composite wire, you might see a current flowing – a small one.