The probability of guessing all 6 balls in a single lottery is 1 in (50 ncr 6). So, the probability of losing is 1 - 1/(50 ncr 6). The probability of losing every time is (1 - 1/(50 ncr 6)^(n_games), where n_games = 365 * 1000. Therefore, the probability of winning at least one game is (1 - (1 - 1/(50 ncr 6)^(n_games)).
But the main point was the methodology. (1 - (1 - chance_of_sequence)^(n_draws)).
Couldn't there be 1 2 3 4 5 6 AND 2 3 4 5 6 7 AND 3 4 5... Doesn't this give you 45?
You get 14 possible draws if the order the balls are drawn in matters (14-20 being the highest, with 20 drawn last), and 20 permutations if it does not (since any straight above 20-25 is not possible). The math is much different for drawing if the order matters though. There also would probably be some funky stuff going on for the higher straights where order doesn’t matter, since for a 20-25 straight the 20 ball must be the special ball. For a 19-24 straight either 19 or 20 must be, etc. Really you’re looking at calculating “Chance that the first five balls can create a straight with a number between 1-20, and then a 1/20 chance that straight actually happens.”
There's 6 possible triple dice combinations and there are 6^3 possible dice combinations, so it's 6/(6^3).
If you wanted to know the odds of rolling 3 dice in order, you could roll: 1, 2, 3 OR 2, 3, 4 OR 3, 4, 5 OR 4, 5, 6 - which is 4/(6^3) - which is not 6c3.
Why is it different with the lottery? Or did I get the dice wrong?
Or are you calculating that the balls can be drawn in any order?