Should a Faraday cage block a radio's signal? Today, I tried creating a very basic Faraday cage by surrounding a radio with two baking trays made out of iron. It didn't seem to affect the radio's signal (AM was being used, not FM).
In theory, should what I used block the radio's signal?
I can't remember how thick the baking trays were. There was some gaps totalling a few square centimetres, because the two baking trays were not identical. I didn't attempt to "Ground" the baking trays.
 A: I'll assume that you were using a radio tuned to a 1Mhz frequency ($\omega = 6.3\times 10^6$ s$^{-1}$) and that the radio was completely enclosed inside $t=3\,$mm of pure iron.
There are two important effects to consider. (i) How much power is reflected from the iron surface. (ii) How much of the transmitted power makes it through the iron.
To figure this out we need the properties of iron; a conductivity $\sigma = 10^7$ S/m, a relative permittivity $\epsilon_r \simeq 1$ and a relative permeability $\mu_r \simeq 10^4$ (for 99.9% pure iron).
First we check whether iron works as a good conductor at these frequencies by noting that $\sigma/{\epsilon_r \epsilon_0 \omega} = 1.8\times 10^{11}$; i.e. $\gg 1$ and therefore a good conductor.
The modulus of the impedance of a conductor is given by $\eta_{\rm Fe} = (\mu_r \mu_0 \omega / \sigma)^{1/2} = 0.089$ $\Omega$.
So, now the relevant equations are:
Electric field transmission at the air/iron interface (assuming normal incidence)
$$\frac{E_t}{E_i} = \frac{2 \eta_{\rm Fe}}{\eta_0 + \eta_{\rm Fe}} \simeq 2\frac{\eta_{\rm Fe}}{\eta_0}\, ,$$.
where $\eta_0 = 377$ $\Omega$.
The EM waves then propogate into the metal but are exponentially attenuated on a scale defined by the "skin depth" $\delta = (2/\mu_r \mu_0 \sigma \omega)^{1/2} = 1.59 \times 10^{-6}\,$m.
Thus, after traversing a thickness $t$, the E-field is attenuated by $\exp(-t/\delta)$.
Finally the wave emerges through the iron/air interface on the other side and we use the transmission formula again but with the labels swapped on the impedance values.
Hence the ratio of the net transmitted electric field to the incident electric field is given approximately by
$$ R = 2 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, 2 \frac{\eta_0}{\eta_0 + \eta_{\rm Fe}} = 4 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, .$$
For the numbers I've assumed $R \simeq 0$ because the wave traverse $>1000$ skin depths to get through the iron! The transmitted power is $\propto R^2$.
So my conclusion is that enclosing within 3mm of pure iron would certainly block AM radio. 
How might this not work? Perhaps the iron you used is very impure and the permeability is orders of magnitudes lower? If $\mu_r =1$ then $\eta_{\rm Fe} = 0.00089\,\Omega$, $\delta = 1.59\times 10^{-4}\,$m. Thus 3mm is still 18 skin depths. The conductivity I assumed is unlikely to be much lower, so I'm a bit confused as to why it wouldn't work.
The demo I use in my lectures is wrapping a mobile phone in aluminium foil. In principle, this is much more marginal because though the frequencies are higher, the foil thickness is much lower, but it certainly works.
A: AM radio is in a band from about 500kHz to 1500kHz which corresponds to wavelengths from about 200m to 600m, vastly longer than your baking trays. This affects the manner of the interaction between the waves and the trays, and how much the trays will attenuate the signal.
For the sake of comparison, your microwave oven is a Faraday cage; it effectively confines some 800W to 1000W of microwave power. A microwave oven emitter typically operates around 2GHz or so, corresponding to a wavelength on the order of a few centimeters. That's much smaller than the cavity, The holes in the window screening are very much smaller than a wavelength - practically invisible to the wave. This relationship in sizes allows the cage to effectively reflect the microwave energy, preventing the escape of all but a tiny amount of energy. 
AM receivers are designed to work on miniscule signals. Any local broadcasts would have to be attenuated considerably before the receiver is no longer able to work with it. An improvised cage might prevent the radio from receiving distant stations, but it would take something more carefully constructed to effectively block a stronger, more local signal.
A: "There was some gaps totalling a few square cm"
A rule of thumb is that if the screen isn't airtight those radio waves will get in - or out
