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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 i

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)).

onlyrealcuzzo 6y ago
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)?
- brandmeyer 6y ago
  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)).
  - onlyrealcuzzo 6y ago
    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?
    - NineStarPoint 6y ago
      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.”

onlyrealcuzzo 6y ago

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)?

brandmeyer 6y ago
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)).
- onlyrealcuzzo 6y ago
  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?
  - NineStarPoint 6y ago
    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.”