If light bounces off the moon anyway, why would anyone bother with a mirror? But EGs theory on moon phases also make this seem unlikely.
There are several reasons the mirror is needed. I can list these two: 1) To ensure the return intensity is high enough to detect. 2) To ensure the angle of the return matches the angle of the incidence.
1) If sunlight can maintain enough intensity to bounce off and reach earth, I'm sure a high powered laser would.
2) Surely a sweep with a laser, will by the law of averages produce an earth bound return?
Thank you for your well wishes by the way. Its been a rotten day, but everyone has been supportive. On the bright side, its going to give me a lot more time to convince you of your errors on FES. 
1) Show your math. Are you saying James is wrong about the bioluminescence of the Moon?
2) No, the law of averages doesn't say that.
You're welcome.
1) The maths is irrelevant. REr's can't have it both ways. Either the sun is intense enough to bounce light off and therefore so is a laser.
http://en.wikipedia.org/wiki/LuxSun 32,000-130,000 lux
You can buy a little laser with 200,000 lux yourself.
http://www.dealextreme.com/details.dx/sku.5100Finally this does not discredit bioluminescence. The moon can be shiny and biolumiescent. You still get a return regardless of what those pesky moon mites are up to.
A laser would be powerful enough not to need a mirror.
2) Well you would agree that firing a laser randomly at the moon would make the resulting beam bounce off in all directions? Well one of those directions will be back to earth, where you could get your reading.
@Berny - I suspect you are trying to pull the wool over my eyes with the "Yes the mirror is there". And being as you want to drag shrimp back into this, I believe they were described as being highly intelligent. maybe they built the mirror to comb their antennae with. And fried road-kill is probably also full of nutrients but I won't be eating that either.
Even with a mirror, and a very high powered laser, very few photons are reflected back to the detector. It's not even the power you have to worry about, it's the spreading of the beam.
I don't think so. Getting any part of the beam back is going to be enough. The time is measured by pulses. And a few photons will be enough.