So this is how I come to these odds. Please correct me if I'm wrong. Let's use a very simplified example.
Say you have only 3 balls. The first ball has a 1:3 chance of being right. Now you cannot choose that ball again so the second pick has a 1:2 chance of being right. Multiply the odds of both picks being right 1:3 x 1:2 = 1:6. With 3 balls, there are 6 possible outcomes. 1 and 2, 1 and 3, 2 and 1, 2 and 3, 3 and 1, and 3 and 2. As you can see, there are reversed but similar picks. That is the difference in the odds I stated and the odds you stated. If 1 and 25 and 25 and 1 are both paying regardless of which is drawn first, the odds are dramatically reduced. If the order of the drawn balls matters, the odds are 1:69 × 1:68 = 1:4,692.
So it comes down to wether it matters if the order of the drawn two numbers counts or not. I picked 8 and 13. Will I win if the draw is 13 then 8 or must it be 8 then 13?
I apologize if I come off being Debby Downer. I just like math.
I kind of think you're reading too much into it. It does not matter what order the balls are drawn. Maybe that is the difference.
Math has never been my strong suit, but I think the contest will produce a winner if we get enough entries. If not, we'll keep running it until somebody does win. And I am confident that somebody will win.
