Combinations & % Error Challenge

Hi,

   The Challnge is to reduce the most combinations with the least amount of error. The game type will be 5/39. Obviously, any lessons learned would be applicable to MegaMillions, Powerball, and CA Superlotto.

   To kick things off, consider the following:

   1) There are 575757 combinations in 5/39.

   2) By eliminating 1 number, any number, there is a 1 in 39 percent chance that you are wrong ( 3%).

   3) Assuming the one number you eliminated isn't picked, then you are playing a 5/38 game with 501942 combinations.

   4) Therefore, 73815 combinations can be eliminated with only a 3% error,

   All input, comments, and contributions are welcome.

Thanks.

"2) By eliminating 1 number, any number, there is a 1 in 39 percent chance that you are wrong ( 3%).

That's not how it works.  There's a 1 in 39 chance that any given number will be the first number drawn, but if it isn't the first number there's a 1 in 39 chance it will be the 2nd number drawn. And so on. Since 5 numbers are drawn there's a 5 in 39 chance that any given number will be 1 of the 5. That means there's a 5/39 or 12.82% chance that you'll be wrong. Not coincidentally, the 73,815 combinations that include that number account for 12.82% of the possible combinations in a 5/39 game.

There is no such thing as a free lunch.

Hi,

   Thanks for your input.

   If you would have included the third item, then it is exaclty how it works.

3) Assuming the one number you eliminated isn't picked, then you are playing a 5/38 game with 501942 combinations.

   The readers can decide which logic they think is correct.

   Ky....What is your best combination reduction method with the least amount of error?  Any input that builds for a solution is appreciated.

You are headed in the right direction, because, it's just as important to find a combination by determining what it ISN'T as determining what it IS.

Would the CA game be used as the standard?

Hi,

   It's okay to use California's Fantasy 5 as a standard. But, any 5/39 lottory would work. I am not picky about the methods submitted. They can be a pure mathematical approach or based on historical data.

  Thanks for asking.

             FILE :OHIO ROLLING CASH5 
     COUNT OCCURRENCES OF NUMBERS IN FILE

      10/04/04 TO 05/26/08  1191 RECORDS

   1. 150   11. 174   21. 175   31. 161
   2. 137   12. 147   22. 133   32. 149
   3. 168   13. 153   23. 159   33. 159
   4. 152   14. 150   24. 155   34. 152
   5. 145   15. 172   25. 134   35. 164
   6. 138   16. 162   26. 146   36. 160
   7. 171   17. 138   27. 154   37. 140
   8. 152   18. 150   28. 136   38. 165
   9. 137   19. 166   29. 149   39. 136
  10. 148   20. 148   30. 170   

Ohio Rolling Cash5 is a 5/39 game and when you divide the 39 numbers in to 3 groups of 13 numbers where A=1-13, B=14-26 and C=27-39, there are only 20 distribution patterns that have matched 5 during the 1191 drawings.

   1. A  A  B  C  C =168
   2. A  B  B  C  C =164
   3. A  A  B  B  C =143
   4. A  A  A  B  C =110
   5. A  B  B  B  C =109
   6. A  B  C  C  C =105
   7. B  B  B  C  C = 54
   8. A  A  A  B  B = 53
   9. A  A  C  C  C = 47
  10. B  B  C  C  C = 46
  11. A  A  A  A  B = 31
  12. A  A  A  C  C = 30
  13. A  A  B  B  B = 28
  14. A  B  B  B  B = 24
  15. B  C  C  C  C = 24
  16. A  A  A  A  C = 20
  17. B  B  B  B  C = 16
  18. A  C  C  C  C = 15
  19. C  C  C  C  C =  3
  20. B  B  B  B  B =  1
 
Combinations that haven't followed one of these distribution pattern have never matched five so I eliminate them from my picks.  You can use other groups to eliminate combinations but 3 groups of 13 seem to be the most clearly defined.

You and the readers can think anything they want, but thinking something is correct doesn't mean it is. The lottery is about math, not logic. Your logic in #3 is correct, but you're making an assumption that may or may not be correct. With math there are no assumptions and you have to get everything right.  The chance that your assumption is correct is 87.18%, and you've left 87.18% of the combinations in play. There's also a 12.82% chance that your assumption is wrong, in which case it's absolutely certain that the winning combination will be among the 12.82% that you couldn't possibly have chosen, because they all contain the number you eliminated. Once you know whether or not your assumption was correct it's not even a game of chance. At that point it's a done deal with 5 numbers that were picked and 34 that weren't. The problem is that you can't know ahead of time whether or not any given number will be drawn.

As I said, there's no such thing as a free lunch. Anything that reduces the number of combinations you consider also reduces the chance that those combinations will include the winning combination, and both reductions are the same. That means that  the amount of the error doesn't matter. The reality is that you're going to play a certain number of combinations, regardless of how you choose the numbers. If you play 0.001% of the possible combinations you have a 0.001% chance of winning.

Average number of occurances of each number in a 5/39 matrix drawn 1,191 times is 152.6923, meaning each of the numbers should have averaged 152.6923 appearances. On the extremes on RJOh's list, 21 has appeared 175 times, 22 has appeared 133 times. Neither is statistically out of line, given the limited number of drawings versus the total possibilities. Nice work, though.

You're right of course, but that's not the point of the exercise.

The idea is to exchange the chance to win in all draws for a better
chance in some draws by giving up the chance to win other draws.

When we choose the strategy not to play the possible repeat
numbers from the prior draw and are correct roughly 50% of the
time we've improved our chance of winning  in half the draws in
exchange for giving up the chance of a jackpot in the other draws.

The fact that cannot be denied is odds have been improved in
some draws. Yes it's an even exchange, what else could it be
in a closed number universe?

BobP