This analysis is spot-on under the assumption of 1000 lotteries per day, although your asterisks are getting trashed by the markdown.

To the other folks ending up with some wild results, there is a basic checksum on probability: If you compute the probability of an event happening at greater than 100% you've borked something up.

Bug: They drew 5 balls from one pool of 50 with an independent draw from another smaller pool of 20, so you need ~~50 ncr 6~~ (50 ncr 5) * 20, not 50 ncr 5.

Nit: I would rephrase your answer that there is a 0.85% chance that it happens one or more times in any given year. There remains a (vanishingly small) chance that it happened on every single random draw during the year.

Wait. Why is it 50c5? If you want to know how many ways you can roll triples with 3 dice, it isn't 6c3, right?

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?

Balls drawn in any order, without replacement.

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)).

Got it. 50c6 is the total combinations. But there's more than 1 combination of 6 ascending balls, right? Why is it 1 in (50c6) instead of 45 in (50c6)?

Oh, sure. There's a number of different suspect or convenient sequences out there. All of the evenly-spaced sequences could be considered suspicious if you go broadly enough. A detail I tried to add back in up-thread: The sixth ball is from a separate pool of 20 balls. So 50c5 * 20 is the total number of possible draws, and there are 14 directly in-order sequences.

But the main point was the methodology. (1 - (1 - chance_of_sequence)^(n_draws)).

Thanks! Why are there only 14 in-order sequences??

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?

As the person above said, one of the balls is restricted to only 1-20. This ball is drawn last.

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.”