United States Member #5599 July 13, 2004 1196 Posts Offline

Posted: May 26, 2008, 6:48 pm - IP Logged

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.

You are a slave to the choices you have made. jk

Even a blind squirrel will occasionally find an acorn.

There is no elevator to success, you will have to take the stairs.

NY United States Member #23835 October 16, 2005 3544 Posts Offline

Posted: May 27, 2008, 1:00 am - IP Logged

"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.

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.

Would the CA game be used as the standard?

In neo-conned Amerika, bank robs you. Alcohol, Tobacco, and Firearms should be the name of a convenience store, not a govnoment agency.

United States Member #5599 July 13, 2004 1196 Posts Offline

Posted: May 27, 2008, 3:44 pm - IP Logged

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.

You are a slave to the choices you have made. jk

Even a blind squirrel will occasionally find an acorn.

There is no elevator to success, you will have to take the stairs.

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 don't need to buy more tickets, just buy a winning ticket *

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 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.

CA United States Member #2987 December 10, 2003 832 Posts Offline

Posted: May 28, 2008, 2:40 am - IP Logged

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.

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

Dump Water Florida United States Member #380 June 5, 2002 3134 Posts Offline

Posted: May 28, 2008, 2:57 am - IP Logged

Quote: Originally posted by KY Floyd on May 28, 2008

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.

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?

United States Member #5599 July 13, 2004 1196 Posts Offline

Posted: May 28, 2008, 2:31 pm - IP Logged

Hi,

First of all, thanks to everyone for their comments. Especially, RJOh for submitting a suggestion.

KY Floyd

It’s good to see your comments even if we end up agreeing to disagree. *L* Let me start by giving you a perspective from my point of view.

1)Because the lottery is random, the only sure fire way to win it is to bet all the combinations.

2)Because the lottery is random, there is currently no fool proof way to predict what the next numbers picked will be.

3)Any attempt to eliminate individual lottery numbers or combinations is an assumption. The lottery Post is a place where those assumptions are submitted, scrutinized, and measured for accuracy.

With that out of the way, lets go back to my initial suggestion.

Would you agree that if you select one number out of 39 different numbers that there is a 3% chance that you are either right or wrong? ( I’ll let you choose 1 different number every game for the next 100 draws and let you decide whether you where right 97% of the time or wrong 97% of the time) *L*

The best suggestion I have for you is not to think of beginnings of this system in the usual terms. It is and will be a little different than you are used to. Be patient, weigh it on its own merits, and you might possibly have a new tool to work with. And in the end, if your disagree and think it’s all hogwash, that’s fine too.

RJOh

I really like the way your data is organized. Very cool. I’m not sure the best way to quantify the amount of combinations eliminated versus error though.

You are a slave to the choices you have made. jk

Even a blind squirrel will occasionally find an acorn.

There is no elevator to success, you will have to take the stairs.

NASHVILLE, TENN United States Member #33372 February 20, 2006 1044 Posts Offline

Posted: May 28, 2008, 5:28 pm - IP Logged

In the process to eliminate one number from 39, I have been looking at the number that has not been drawn for awhile, or, to put it another way, the number furthest out.

What I am seeing is that this number, currently the number 5 in Tennessee's 5/39 game, has gone 27 draws without showing.

Be eliminating the number furthest out, we will be right more often than wrong.

I can also consider the number following the 5, in this case 6. So I could concievably eliminate two numbers.

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.

Impressive research, RJOH. I feel humbled by your work.

I see that you divide the 39 numbers into three groups in decending order. Have you compiled the three groups using a different standard? By that I mean something like taking the 13 most frequently drawn numbers over the life of the drawing to be in the "A" group, and the 13 least drawn numbers over the life of the crawing to be in the "C" group, and allowing the remaining numbers fall into the "B" group?