Yes, if you have 1,000,000 nodes and there are at least 499,994,500,020 edges out of a possible 499,999,500,000 (ie. a bit more than 99.999% of them), then yes you can conclude the diameter of the graph is no more than 6. But no, simply counting the number of edges is not particularly useful. You need to make further assumptions about the graph to get a useful bound.
Been getting into LoL recently, and while I agree it's very polished (I really like the character designs and the free-to-play model is a good one), the one area that Valve could improve on is making it more newbie-friendly. While LoL may be more newbie-friendly than DoTA, that's like saying Venus is less hot than the sun. It's still not a very hospitable place. It's basically a full-time job to get up to speed with all the acronyms, jargon, and conventions. You join your first match of LoL and your teammate says something like: "I'm going jungle Amumu with an AP Sunfire build so I can tank the carry in the lane with my ult when they ping." and then they get mad at you when you have no idea what they're saying. Don't get me wrong, I do like the game, it's just really hard for beginners like me.
Compare the steep learning curve of TFC where you have to master grenade jumping with every class just to compete with how easy it is to pick up TF2. So hopefully Valve will do something similar with DoTA, and make it accessible.
Doesn't she ever learn? Fox's paranoia is always proven justified in the end.
That's the Hales-Jewett theorem. The density Hales-Jewett(3) is a little different. Suppose you're putting X's on a 3x3x...x3 (n-dimensional) grid. Such a grid has 3^n points. What the theorem says is that if you've put X's on at least, say, 1% of the 3^n points, then you must have made a line somewhere if n is large enough. You can replace the 1% with whatever fraction you please, but that will change how large n has to be. No, the theorem doesn't state exactly how large n has to be, but already it's a challenging problem.
Terry tao's blog has an explanation that's about as simple as you're going to find. (At least one that actually explains the math without handwaving).