I will do the case where the material is homogeneous and isotropic, $\rho = \sigma^{-1}$ is a constant proportional to the identity matrix. We are interested in the steady state, where none of our variables depend on time.
We have
$\nabla \times E = 0$ from Faradays law and, $\nabla\cdot J = 0$ from the equation of continuity, where $J$ is the current density. Ohm's law tell us that $J=\sigma E$. Taking the divergence of Ohm's Law we get $\nabla \cdot E = 0$. Therefore in steady state $E$ must be a divergence-free, curl-free function.
This means that the potential $\phi$ ($\nabla\phi =E$) obeys Laplace's equation,
$\nabla^2 \phi = 0$.
To solve this we need the appropriate boundary conditions which are as follows. At the boundary where your resistor connects to a lead the potential $\phi$ must be the same value as the potential on the lead. At the boundary of your resistor not connected to a lead you can't have current flowing out, so the appropriate condition is $\nabla\phi\cdot n = 0$, where $n$ is the surface normal. This is sufficient to determine a single solution to Laplace's equation.
Laplace's equation is a very nice and friendly equation and there is a lot of material on numeric and analytic solutions available, although the mixed boundary conditions will be annoying.
Once you have your solution to Laplace's equation you need the total current. To get that, pick any cross sectional surface $S$ of your resistor, and using $J = \sigma E = \sigma\nabla\phi$, we get the total current I is equal to
$I = \sigma\int dS\cdot\nabla\phi$.
Then you can use $R = V/I$ to get an effective resistance. The case with a non homogeneous or non isotropic material is similar, you just end up with a different equation from Laplace's equation, which may be a little more annoying to solve.
I can't imagine a dough based system will need this level of precision though :). For anything dough based probably just assuming its a cylinder and using $R = \rho L /A$ will get you close enough no? I always thought of dough physics as really being an order of magnitude sort of game, like astronomy.
Your assumption that Ohm's law is fully accurate for the stun gun + human circuit isn't correct. A stun gun uses a capacitor to store charge and the capacitor is constantly being recharged to deliver a series of high-voltage pulses.
A capacitor has a finite amount of charge. Once it's charged, that's it, it can never deliver more charge than that until it is recharged. The measure of the stun gun voltage is when it's an open circuit. Basically the voltage is telling you how strong the electric field is but it isn't telling you something that's very useful for Ohm's law.
The graph of a charging and then discharging capacitor looks something like this:
The only thing the resistance of your body can affect is how quickly the discharge occurs. It's not possible to deliver more charge than is stored in the capacitor and it is the capacitance that is the primary safety mechanism.
So the voltage of the open circuit at the peak is not really a meaningful measure of the total amount of current that will flow per pulse.
Regarding your option 1: it isn't a lie but yes, you're somewhat right anyways (voltage drops on discharge).
Regarding your option 2: the pulses duration is closely related to the capacitance of the capacitor so yes, this is somewhat right too.
Regarding your option 3: I'm sure there is some amount of "skin effect" but I'm not sure how much.
I should also mention that sometimes stun guns DO kill.
Best Answer
Think what the short will do: it will melt the hair drier parts and the live side of the incoming circuit will connect with the water and the water through the taps and supports of the tub to the ground. You will be holding something sparking and in contact with the water, all wet. So unless the fuse goes off there is great danger the sparks will include you too because they will be erratic. i.e. your second scenario
comes fast on the tracks of the first.
5 amperes is a lot of current. Not to forget the surprise element for the person being stupid enough to use a hairdrier in the bath. People are over 50% or so water after all, in good paths like blood vessels.
btw I have seen sparks from a 220 volt meter for the house (provided by the electricity company) in which rain water had run through, and the whole meter gauge was sparking and melting everything resembling in sparks those melting clocks of Salvatore Dali, flowing down the wall.