Concentric shells of charge What exactly happens when two concentric shells, of different radii carrying different charge say $q$ and $Q$, are connected through a conducting wire?
Will the potential of both the shells be the same?
Won't all the charges gather around the bigger shell?
 A: 
Will the potential of both the shells be the same?

Yes. This is the definition of an ideal conductor. 

Won't all the charges gather around the bigger shell?

Yes, they will. Contrary to what I said in the comments, they will gather up on the surface of the bigger sphere. To show this, consider the electric flux through a spherical surface between the two spheres. This is given by:$$\int\vec E\cdot d\vec A=q_{enc}$$where the enclosed charge $q_{enc}$ is the charge on the inner sphere. Now, due to the spherical symmetry of the charge distribution, the electric field has to point radially outwards and have the same value on the sphere. This gives $$|\vec E(r)|=\frac{q_{enc}}{4\pi r^2}.$$
Now, $$\Delta V=\frac{q_{enc}}{4\pi}\int _r ^R\frac{dr}{r^2}=\frac{q_{enc}}{4\pi}(\frac{1}{r}-\frac{1}{R}).$$
Now, since $\Delta V=0$ and $r\neq R$, $q_{enc}=0$. So, all charges end up on the outer shell.
Note:


*

*I employed the same principle as in my comments to conclude it: calculate $\Delta V$ for the configuration and then set it to zero. My mistakes were to assume non-applicable formulas of $\Delta V$.

*You were right in your intuition. 

*This process of connecting the two spheres  doesn't conserve energy. The field outside the outer sphere is not changed by this action but inside it drops to zero instantly. So, $\int E(x)^2 d^3x$ is not the same before and after the charge rearrangement. Which it should, because,  the energy conservation equation in this case is $\frac{dU}{dt}=-\vec E\cdot \vec j$ and according to the configuration, $\vec j$ must have a radially outward component, along the direction of $\vec E$.

