Schematic: Flame Deflector

One
Hundred
Eighty
Two
FREAKING
Gnomeregan runs....
BUT IT FINALLY DROPPED!
Yep, it does still drop.
Got the Flash Bomb recipe on the 59th run, got it again around 136, 141 and 150-ish. Got several Craftsman's Monocles, and a bunch of other green recipes. Going in through the Workshop Entrance is a must - that way only takes 7 minutes for full in-and-out, and most of that is just running around corners and occasionally AoEing down the trash. Do 5 runs in 35 minutes, do something else for a half hour, come back and start over.

If your character is not maxed in all his basic faction reps, WEAR A TABARD. I went from high Friendly with Gilneas to 20573/20999 Revered, getting at least 150 rep per run (guild bonus included). If you want to grind max rep while farming recipes, leave some trash alive but aggroed before you take the elevator down into the Engineering Labs. You'll have a mob of trash waiting for you when you finish coming up the access tunnel from the Eng Labs. AoE it down, then jump into the channel and head towards Thermaplugg. You'll probably get netted a few times along the way, so that's the best time to hit your AoE, especially if it has a CD and isn't strong enough to kill all the robots in one use. On the way out, run up the right side of the channel to pick up that trash, then AoE it before you turn down the tunnel back to the Eng Labs. Taking these tiny extra steps will get you about 50 extra rep per run, which isn't a lot, but it's about 1/3 more than you'd get on a least-time run. As it'll only take an extra 30 seconds per run, that's a good return on your rep farming time.

Actually, that is wrong both in terms of math and logic. It is true that 0.6%/run * 33.3hrs * 5runs/hr = 100% but you can't add probabilities that way when trying to determine the chance of something happening at least once. Moreover, even if you could, then why would that represent the "average" time to get the drop? The way it's done is via this formula: 1 - (1 - 0.006)^N = 0.5 where N is the number of runs. That's basically saying the chance of NOT getting the drop is P = (1 - 0.006)^N for N tries. The chance of getting it is, then, 1 - P. Solving for N we get: N = 115.18 or 23.04

Just found this on the Lightninghoof auction house for 8 silver.

Make sure to wowhead recipes if you are unsure of its value guys :o