I cannot say with impunity what RJOh meant, but this is how I interpret his comment. My observations about about the 3,819,816 combinations of "regular" numbers for the MegaMillions 5/56. My data is a little out of date: 827 drawings from 6/24/2005 through 5/24/2013.
If I select a combination (for my ticket) that exactly matches a previous drawing, I would exclude it because there is only an 827-in-3,819,816 chance (0.02%) that the next drawing would repeat a previous drawing.
It's not that it cannot happen; all 5-tuple combinations are possible. Moreover, the chance of a duplicate will increase over time. But it's still very unlikely, to say the least.
What about 4-tuples (quads)?
Each ticket and drawing has C(5,4) = 5 quads. And there are C(56,4) = 363,175 unique quads total.
Empirically, among the past 827 drawings, there have been 4095 unique quads and 20 quads that appeared twice -- 4135 in all (= 827*5). So the current probability of a duplicate quad is 20-in-4135 (0.48%).
Again, that probability will increase over time. But since we have seen only about 1.12% of all 4-tuples (4115 / 367,290), I would guess that it will take some time for the probability of duplication quads to become significant.
So I would exclude a selection that includes a quad that has appeared in any previous drawing.
What about 3-tuples (triples)?
Each ticket and drawing has C(5,3) = 10 triples. And there are C(56,3) = 27,720 unique triples overall.
Empirically, we have see about 26% of all triples (7145 / 27,720) in the first 827 drawings. The breakdown is: 6133 triples appearing once, 905 triples appearing twice, 101 appearing 3 times, and 6 appearing 4 times. And again, the frequency of "duplicates" (including more than 2 times) will increase over time.
Also empirically, in the past year, 88% of the drawings have had 1 to 4 duplicate triples -- that is, triples that appeared in previous drawings. And 56% of the drawings have had 2 or 3 duplicates.
So I might (operative word) exclude a selection that has more than 4 duplicate triples (for now). I might even favor selections that have 2 or 3 duplicate triples (for now).
What about 2-tuples (pairs)?
I'll spare you the computational details. There are C(56,2) = 1540 pairs, and we have seen them all but one in the past 827 drawings. Each ticket and drawing has C(5,2) = 10 pairs. And in the past year, 94% of the drawings has 10 duplicate pairs -- that is, pairs that appeared in previous drawings; and the remainder have had 9 duplicate pairs.
So there is no point in trying to exclude drawings with duplicate pairs; it cannot be done. In fact, most drawings will have 10 duplicate pairs now. (One future drawing should have only 9 duplicate pairs. ;->)