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.

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?

No. If only one number out of 39 was drawn then there would be a 1 in 39 chance of of correctly guessing that number. Since there are 5 numbers that are drawn there is a 5 in 39 chance. When 5 of 39 numbers are drawn then 5/39ths of the numbers are drawn. 5/39ths is 12.82%, so there is a 12.82% chance that picking one number randomly will result in picking one of the 5 numbers that are drawn. The flip side is that if you decide that a particular number won't be drawn the chance that you will be correct is 87.18%, not 97%. You aren't eliminating 12.82% of the combinations with a 3% chance of choosing incorrectly. You're eliminating 12.82% of the combinations with a 12.82% chance of choosing incorrectly, so it's a wash.

As BobP notes, you can increase your chances of winning if a certain result occurs in exchange for an reduced chance of winning if that result doesn't occur. For a simple analogy imagine playing pick 3 with only even numbers. If the winning number is even you have a 1 in 500 chance of winning instread of 1 in 1000. Since there is only a 50% chance that the winning number will be even your overall chance of winning is .50 * 1 in 500, which is 1 in 1000. There is no net gain.

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: May 28, 2008, 9:49 pm - IP Logged

Quote: Originally posted by GASMETERGUY on May 28, 2008

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?

I did write a routine that calculates the average and median rate of occurances of the numbers in the winning combinations and compare that to the number of time they hit in a most recent period and color them accordinly, 0=blue, 2=red, 3 or more=yellow and everything else=green. When ever all of the winning numbers are one color and I calculate all the numbers that could be in that group, they're 80% of the total number pool so there's no real advantage. I keep checking hoping that one time there will be a group small enough that I could cover a winning combination in ten or twenty lines.

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

Dump Water Florida United States Member #380 June 5, 2002 3102 Posts Offline

Posted: May 28, 2008, 11:35 pm - IP Logged

Cough !!! You might want to take a second look at playing with only five even digits and how many straight numbers one can make with any five digits. When the winning number is even and you are playing an even number you have a 1 in 65 chance of winning $500.

United States Member #5599 July 13, 2004 1184 Posts Offline

Posted: May 29, 2008, 12:05 am - IP Logged

Hi,

Below is a listing for how many combinations there are for different amounts of lottery number pools. I believe the math to be correct, but wouldn't mind a double check. Like my initial example, you'll find 575757 combinations for a 5/39 game. If you eliminate one specific number from all the combinations you'll find 501942 combinations, for what becomes a 5/38 game. So, if you eliminate approximately 50% of the numbers accurately, then you would be playing say a 5/20 game with 15504 combinations or a 5/19 game with 11628 combinations.

I would have you note that this method reduces combination on a different scale than using combinations to reduce from the total amount combinations. I guess you could say that reduction methods using combinations is linear, while reduction methods using individual numbers is non-linear.

Sounds fairly good huh. But, KY Floyd is 100% correct when he says "There is no such thing as a free lunch." If you elect to elimate one number that ends up being picked, then every subsequent combination you bet will be unable to win the 5 out of 5 prize. To make matters worse, look at Maddogs challenges. Everyone, including myself, is having a tough time getting enough numbers right for a payoff.

The upside is, when you do select correclty there is a dramatic reduction in combinations.

My suggestion is to use this system first. Only eliminate those numbers which you deem to have the lowest probabilty of occuring. Then use the usual combination reductions methods to obtain your final picks.

Anyway, this completes what I want to say about the method I started this thread with.

Are ther any others methods anyone would like to suggest?

NUM

COMB

5

1

6

6

7

21

8

56

9

126

10

252

11

462

12

792

13

1287

14

2002

15

3003

16

4368

17

6188

18

8568

19

11628

20

15504

21

20349

22

26334

23

33649

24

42504

25

53130

26

65780

27

80730

28

98280

29

118755

30

142506

31

169911

32

201376

33

237336

34

278256

35

324632

36

376992

37

435897

38

501942

39

575757

40

658008

41

749398

42

850668

43

962598

44

1086008

45

1221759

46

1370754

47

1533939

48

1712304

49

1906884

50

2118760

51

2349060

52

2598960

53

2869685

54

3162510

55

3478761

56

3819816

57

4187106

58

4582116

59

5006386

60

5461512

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

Even a blind squirrel will occasioanlly find an acorn.

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: May 29, 2008, 12:49 am - IP Logged

Quote: Originally posted by GASMETERGUY on May 28, 2008

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?

FILE :OHIO ROLLING CASH5 COUNT OCCURRENCES OF NUMBERS IN FILE

Tried your suggestion and there were 21 distribution patterns. Nothing that I could use to pick combinations in a unique way.

1. A B B C C =176 2. A A B B C =171 3. A A B C C =139 4. A B C C C =118 5. A B B B C = 99 6. A A A B C = 98 7. A A B B B = 52 8. A A A B B = 48 9. B B B C C = 44 10. B B C C C = 44 11. A A C C C = 43 12. A A A C C = 33 13. A A A A B = 24 14. A A A A C = 24 15. A B B B B = 22 16. B B B B C = 17 17. A C C C C = 16 18. B C C C C = 15 19. B B B B B = 5 20. C C C C C = 3 21. A A A A A = 2

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

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: May 29, 2008, 11:32 am - IP Logged

Having separated numbers in winning combinations into groups of high/low/medium, active/normal/inactive, hot/normal/cold and other ways and seeing all the different mixes has less than 25 distribution patterns and are were dominated by nine or fewer patterns that looked similar regardless of the type of mix convinces me that it almost impossible to eliminate a small group of numbers that will not have any of the winning numbers most of the time.

In the Ohio Rolling Cash5 file which I sighed for this thread, 51% of the time at least one number from the previous drawing was in the winning combinations and 10% of time it was two or more. With its payout of $1 for a match2, $10 for a match3 and $300 for a match4 had I simply replayed the previous winning combination every drawing, I would have won back $582 of the $1,192 it would have cost to play.

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

I did write a routine that calculates the average and median rate of occurances of the numbers in the winning combinations and compare that to the number of time they hit in a most recent period and color them accordinly, 0=blue, 2=red, 3 or more=yellow and everything else=green. When ever all of the winning numbers are one color and I calculate all the numbers that could be in that group, they're 80% of the total number pool so there's no real advantage. I keep checking hoping that one time there will be a group small enough that I could cover a winning combination in ten or twenty lines.

I am please to noted that you took the time to look. While the results were disappointing, having that knowledge is better than not having it.

I once had a chemistry professor in college who told us that 99% of all the experiments turned out to be failures. Had someone documented those failures, future chemists would not waste their time re-inventing the wheel, or the experiment in this case.

I think LP would benefit if there was a forum for "failed" attemts. There might be a gem or two of benefit to someone else.

Hey, that would make a good poll. "How many failures have you experienced while developing your method?"

United States Member #13130 March 30, 2005 2171 Posts Offline

Posted: May 29, 2008, 2:45 pm - IP Logged

How would you define "failure"?

I have tried systems that had the winner but were too expensive to play. (haven't we all?) I have had systems that didn't work well with other systems (non-confirming). Usually, afterwards, I had created some code that I could use in other places.

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

Kentucky United States Member #32652 February 14, 2006 7295 Posts Offline

Posted: May 29, 2008, 6:35 pm - IP Logged

Quote: Originally posted by time*treat on May 29, 2008

How would you define "failure"?

I have tried systems that had the winner but were too expensive to play. (haven't we all?) I have had systems that didn't work well with other systems (non-confirming). Usually, afterwards, I had created some code that I could use in other places.

LP has a 2if5 wheel with 25 combinations that uses all the numbers in a 5/39 game. Granted playing $25 with a win expectation of $1 doesn't look all that great, but it costs nothing to experiment with the wheel. The wheel is unbalanced; 1 number will appear 7 times, 6 numbers 4 time, 30 numbers 3 times, and 2 of the numbers will appear twice. It might be easier to break this wheel down and create a system that raises the win expectation than using a system that is more expensive to play.

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: May 29, 2008, 7:09 pm - IP Logged

Hey, that would make a good poll. "How many failures have you experienced while developing your method?"....qoute:GASMETERGUY

If success is winning more than you spend every time you play then lottery players fail almost every time they play and that's the way the games are designed or else the states couldn't keep 50% of the sales for themselves and offer a large jackpot to the winner to attract more players.

When I test a theory, what ever happens is not a failure to me, the results answer a question I asked or disprove a myth that I read or heard. Almost any system can win something if you're willing to buy enough combinations. Beating the odds seem to be more a matter of luck than how one pick combinations, at least that's my theory until I or someone else proves other wise.

I'll continue to play combinations picked using a system because even when I lose I like to think I'm in control of my own destiny.

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

Cough !!! You might want to take a second look at playing with only five even digits and how many straight numbers one can make with any five digits. When the winning number is even and you are playing an even number you have a 1 in 65 chance of winning $500.

If that's directed at my post, I didn't say anything about playing pick 3 with 5 even digits. I said playing pick 3 only with even numbers, of which there are 500. Play 1 of them (straight) and your chance of winning is 1 in 500 if the winning number is even. FWIW, if you played even numbers boxed you could also win on odd numbers that have an even digit, such as 176 paying for 167, 617, 671, and 761.

If I was going to play numbers that can be made with (all) 5 even digits I can play a lot more than the 65 you list: 1: 000 2: 002 3: 004 4: 006 5: 008 6: 020 7: 022 024, 026, 028 8: 040 042 9: 044 046, 048 10: 060

Dump Water Florida United States Member #380 June 5, 2002 3102 Posts Offline

Posted: May 30, 2008, 3:15 am - IP Logged

Quote: Originally posted by KY Floyd on May 30, 2008

If that's directed at my post, I didn't say anything about playing pick 3 with 5 even digits. I said playing pick 3 only with even numbers, of which there are 500. Play 1 of them (straight) and your chance of winning is 1 in 500 if the winning number is even. FWIW, if you played even numbers boxed you could also win on odd numbers that have an even digit, such as 176 paying for 167, 617, 671, and 761.

If I was going to play numbers that can be made with (all) 5 even digits I can play a lot more than the 65 you list: 1: 000 2: 002 3: 004 4: 006 5: 008 6: 020 7: 022 024, 026, 028 8: 040 042 9: 044 046, 048 10: 060

You said: " For a simple analogy imagine playing pick 3 with only even numbers. If the winning number is even you have a 1 in 500 chance of winning instread of 1 in 1000. Since there is only a 50% chance that the winning number will be even your overall chance of winning is .50 * 1 in 500, which is 1 in 1000. There is no net gain."

Ok. Always amazing how we can see exactly what we want to see.

One interesting thing was I learned something about the Maybel Quik Wheeler. I entered Straight 0-2-4-6-8 and Force Doubles and got 65 combinations for my trouble.

Went back and unchecked Force Doubles and got 125. Got to stop doing things when I'm half asleep. Zzzzzzzzz

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

Posted: June 2, 2008, 9:07 pm - IP Logged

Quote: Originally posted by time*treat on May 29, 2008

How would you define "failure"?

I have tried systems that had the winner but were too expensive to play. (haven't we all?) I have had systems that didn't work well with other systems (non-confirming). Usually, afterwards, I had created some code that I could use in other places.

Failure is testing a theory only to discover the concept did not pick the winning numbers a sufficent number of times to make the financial effort profitable.

I agree most heartily with your statement about systems that were too expensive to play. These systems, in my humble but sagacious opinion, can be labeled "failure".

So what is "success"? Sucess is a system that is reasonable, that results in less than 100 number sets and is correct ( it predicts a winner) at least one time out of ten. (And if anyone has such a system, I would be more than willing to invest in your project.) I doubt such a system exists. I have no doubt that such system will be developed by someone at some time in the future. There is just too much research going on right now for such a system not to be found. We may not have PhD's or Nobel prizes but we have something most people do not have....we have imagination.

Feel free to tweak the above characteristics of "success". Success, like failure, means different things to different people. One man's failure is another man's success or something akin to that.