RJOh wrote: ``Had I played and picked 31 as my key number, my odds of picking the other five winning numbers would have been 1:1,712,304``.
Yes, if you could know a priori that 31 would be picked. But looking forward, you cannot know that. So your true odds are the probability of matching a combination with 31 (1/1,712,304) times the probability that one of those combinations is drawn (1,712,304/13,983,816). That is 1:13,983,816, the same probability as matching all 6 randomly. So your odds are not improved by choosing a combination with 31.
RJOh wrote: ``Odds would have been even better had I avoided any combination of four that had already hit.``
Again yes, if you could know a priori that the next drawing would not include any previously drawn quads (combinations of 4). But looking forward, you cannot know that. In fact, RJOh wrote later: ``Each combination of six has 15 combinations of fours of which OCL have used 14,521 of the 211,876 possible`` in 1000 previous drawings. So there have been 479 repeated quads drawn. Certainly a small percentage; so I agree that it seems prudent to steer clear of previously-drawn quads. Nevertheless, your true odds are the probability of matching a combination with no previously-drawn quads (limited to combinations with 31 to include the previous requirement) times the probability that one of those combinations is drawn. The result should be 13,545/13,983,816, about 1:1032, the same probability as matching any 4 (and not more) randomly. Again, your odds would not be improved.
(PS: Well, the 1:1032 odds do not include the limitation of combinations with 31. I had added "limited to combinations 31 to include the previous requirement" as an after-thought.)
Frankly, I find it difficult to calculate that conditional probability mathematically. I would verify it by writing a program to actually count the two component probabilities. "The exercise is left to the student".