Alright, there seems to be a lot of confusion about this, so allow me to give a quick overview of the first week or two of a physics course on electricity.
Analyzing circuits in general is pretty complicated; even a simple circuit with just a resistor, inductor, and capactitor takes about a semester to fully understand. However, most home circuits can be modeled using just resistors, which makes them much, much easier to understand.
I'm going to assume you're familiar with the electrical concepts common for home-wiring: current (amperage), voltage, alternating current, etc.
This is what a typical circuit diagram looks like:
Let's first review what all these symbols mean.
This symbol is an AC (alternating current) generator, the kind of power supplied to homes. It would be equivalent to the left- and right-prongs in an outlet.
This is the symbol for a resistor, something which resists the flow of current. It can be used to model a lightbulb, a person touching a wire, or even the resistance of the wire.
This is the wire. We assume it has 0 resistance; if we need to model the resistance of the wire, we add an imaginary resistor to the circuit.
Note that the voltage across any two points on our ideal resistanceless wires will always be 0, seeming to indicate (by Ohm's Law, below) that there can never be any current across them (or rather, the current is 0/0). We ignore this fact as theoretical, since real wires will always have some resistance.
This symbol is called "ground," but it actually represents the neutral wire, not the third prong that an electrician would call "ground." If there is more than one ground in a circuit diagram, we assume they are all connected.
Analyzing these kinds of circuits only requires knowledge of three laws: Ohm's Law, Kirchoff's Current Law, and Kirchoff's Voltage Law.
Voltage = Current * Resistance
(aka. V = IR)
Thus, if you only know two of
(Voltage, Current, Resistance) for a single resistor, you can find the third.
Kirchoff's Current Law (KCL)
"The total current entering a wire-junction or device is equal to the total current leaving the junction/device"
If you think about the water-analogy of electricity and replace "current" with "water," this seems almost obvious. You could also call this "conservation of current."
Note that this means that current is not "used up" as many people seem to believe.
Kirchoff's Voltage Law (KVL)
"The sum of all voltages in a loop is 0"
This sounds complicated, but it's also really simple; it basically says that, if you sum all the voltage-increases and voltage-decreases in a loop, then when you get back to where you started you get 0. If you replace "voltage" with "pressure," this also makes sense with the water analogy.
(image taken from here)
Finally, let's analyze your circuit.
By KVL, the voltage across R2 = 0. By Ohm's law, this means the current through R2 = 0, so the light will not light. The diagram you gave in an answer...
is exactly correct, assuming R1 = 960Ω (the resistance of R2 is irrelevant).
There was concern in the comments above that this circuit isn't realistic, since it doesn't model the resistance of the wire. This is true; let's try analyzing a more realistic circuit.
What is the current through "You" in this circuit?
First, we find the current through the entire circuit. This can be done using only the three above laws and a lot of algebra (Physics students are taught a shortcut called equivalent resistance). Since this isn't an algebra class, I'll skip this step and just tell you the equivalent resistance is just slightly under 1002Ω, meaning the current drawn from the generator is about 120V/1002Ω ≈ 0.11976A (you are welcome to do the calculations yourself, however).
By KCL, we see that
(Current through generator)
= (Current through hot wire)
= (Current through device)
= (Current through You + Current through neutral wire)
Using Ohm's Law, that means the voltage DROP across the hot wire is
1Ω * 0.11976A = 0.11976V
and the voltage DROP across the device is
1000Ω * 0.11976A = 119.76V
Since the voltage INCREASE across our AC generator is 120V, by KVL this means that the voltage DROP across You (and across the neutral wire) is
120V - 119.76V - 0.11976V ≈ 0.12V
Finally, by Ohm's Law, this means the current across you will be about
0.12V/10kΩ = 0.012mA
If "You" in that diagram is actually you: according to this answer, the threshold for sensation of 60Hz AC is around 0.4mA. So although current would indeed flow through your body, you wouldn't feel it, even if you were standing in a tub of water.
In the case where "You" is actually an incandescent lightbulb, current will flow, but the light it produces will likely be so dim that you can't see it.
And if "You" is a florescent lightbulb like in your pictures, the ballast will prevent any current from flowing at all.
Finally, let's compare the above case to one where you touch the hot wire while the device is on and you are grounded:
Or the similar case where the device is off (or the neutral becomes disconnected!):
In both cases, the current through you would be just slightly less than 12mA. 12mA is enough to cause severe pain and contract your muscles so you couldn't let go of the wire. And if you're sweaty or have lower skin resistance that day, the current would be even greater, possibly enough to kill you (which would require around 100mA, according to the above link).