Try out the formula above. Read the threads. Build some lights and measure them. It will all work out. What you are stating is a common misconception, like I said.
Why do you mention a toothpick sized reflector? It obviously needs to be wider than the die of the LED as you also mention.
Example:
Lets say you have two identical LEDs with a luminance of 200cd/mm2. Each has a die with an area of 1mm2.
Lets also say that you have two parabolic reflectors. Both have an LED hole with a diameter of 10mm. Both have a center depth of 20mm. Both have an aluminium coating with 90% reflectance of visible light. Both have an ar-coated glass lens with 96% transmittance (typical chinese flashlight quality) on top.
Reflector A has an outer diameter of 20mm.
Reflector B has an outer diameter of 50mm.
Now you want to find out how far you can see at night with these two lights (the ANSI range of a flashlight). To calculate this distance you need the luminous intensity [candela]. To calculate luminous intensity you need to calculate the frontal surface area of both reflectors.
Reflector A:
refl_a_area = area_circle - area_led_hole
= (20mm / 2)2 * pi - (10mm / 2)2 x pi
= 314.2mm2 - 78.5mm2
= 235.7mm2
refl_a_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity
= 200cd/mm2 * 235.7mm2 * 0.96 * 0.9
= 40,729cd
Note here that cd is the same as lux@1m.
refl_a_ansi_range = sqrt ( refl_a_lum_intensity / 0.25 Lux )
= sqrt (40,729lux / 0.25lux)
= 403.6m
Reflector B:
refl_a_area = area_circle - area_led_hole
= (50mm / 2)2 * pi - (10mm / 2)2 * pi
= 1963.5mm2 - 78.5mm2
= 1885mm2
refl_b_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity
= 200cd/mm2 * 1885mm2 * 0.96 * 0.9
= 325,728cd
refl_b_ansi_range = sqrt ( refl_b_lum_intensity / 0.25 Lux )
= sqrt (325,728lux / 0.25lux)
= 1141.5m
As you can see, the depth of the reflector has no effect on the results.