Puckdropper writes:
That makes a LOT more sense. I'll have to breadboard it to see what
actually happens...

Even if the switch has a non-zero resistance, the voltage across it
and the light will be so low that the bulb won't have enough power to
produce any visible light.

But to simplify...

Consider each bulb has a 10 ohm resistance, and it's a 10 volt
battery. One light = one amp. Two lights in series = 20 ohms = 0.5
amps, which each light sees. Two lights in parallel = 5 ohms = 2
amps, split between the lights = 1 amp each. Assume the switch is
0.01 ohms (they're usually much less).

So with one bulb in parallel with the switch, and a second in series
with those two, the MOST current you'll get is 1 amp (just the second
light alone limits it to that). If the switch is closed, 1 amp
through it is 0.01 volts. That's the most voltage that will be across
the switch/light combo. 0.01 volts divided by 10 ohms (the first
light) gives 0.001 amps (1 milliamp), far less than the 1 amp it's
expecting.

To figure exact values, consider:

V1 V2
*---\/\/\/\----+-----\/\/\/\/----+ 0v
-- R1 | R2 |
I +-----\/\/\/\/----+
R3 |
---

R2 and R3 in parallel give 1/(1/R2+1/R3) ohms. Let's call this R23.

V1 V2
*---\/\/\/\----+-----\/\/\/\/----+ 0v
-- R1 R23 |
I |
|
---

Total resistance between V1 and 0v (ground) is thus R1+R23.

Current is V1/(R1+R23). Call this I.

Voltage at V2 (relative to 0v) is I*R23

Current through R2 is V2/R2.

Current through R3 is V2/R3.

For our simplified example, rounded to three sig digits...

V=10 R1=10 R2=10 R3=0.01

R23 = 1/(1/10 + 1/0.01) = 0.00999 ohms
I = 10/(10+0.00999) = 0.999 amps
V2 = 0.999 * 0.00999 = 0.00998 volts
Current through R2 = 0.00998 / 10 = 0.000998 amps
Current through R3 = 0.00998 / 0.01 = 0.998 amps