Combinations & % Error Challenge

Reply #45

RJOh — Sat, 14 Jun 2008 08:42:18 GMT

Even with all the winning numbers in the mix, it's unlikely most of them will be in the same combination. When you limit the number of lines you play you either has to eliminate some numbers or some combinations, it's usually after the drawing you wished you had eliminated the other.

Reply #44

DownandDirty — Sat, 14 Jun 2008 06:55:13 GMT

How many times have you eliminated/filtered a set of numbers; only to have them come in soon after? Happened with last nights 5/39 draw for me(Thurs. draw). I chose to delete this combination I had played on Weds. the night before. 4 of 5 came in. Not the first time or last.

My method includes putting as many numbers in play that you can..with the least expenditure. I try to include most or all 39 in my nightly selections.

This is a very informative thread. Data needs to be combined with o... [ More ]

Reply #43

RJOh — Sat, 14 Jun 2008 05:54:39 GMT

That's like suggesting the best way to avoid the high price of gasoline is to park all your vehicles, even the lawn mower and garden tractor and start walking.

Reply #42

time*treat — Wed, 11 Jun 2008 00:37:59 GMT

These threads always seem to get a bit away from the original topic.

To go back to the beginning, I find you can eliminate a block (pair or triple) of numbers with more confidence than any individual number ('tho' I follow a cold-numbers-stay-cold rule similar to the one described by GASMETERGUY). But, that may be just my preference. It's a bit more complex code-wise but it still reduces the total combos.

I think it depends on if you start with a blank canvas and try to fill in the winnin... [ More ]

Reply #41

KY Floyd — Mon, 09 Jun 2008 16:48:55 GMT

The best way to avoid filtering out the winning combination is as easy as can be. Just don't filter any combinations that are random.

Reply #40

RJOh — Mon, 09 Jun 2008 15:02:59 GMT

The trouble with filtering is even winning combinations look bad until they win and once they've won they look bad again, in fact some players filter out past winners totally from their list of combinations to play. Unless you can find some qualities or characteristics that are unique about combinations at the time they win, they're likely to be filtered out when it's their time to win.

These qualities or characteristics are most likely dynamic, constantly changing as the list of recent winne... [ More ]

Reply #39

time*treat — Sat, 07 Jun 2008 17:48:16 GMT

The way I grade some of my filters is to see how they perform based on number of drawings back. I can select a block size of N drawings-per-block and X number-of-blocks and then see the percentage of success over N *X drawings, in groups of N. That tells me if a filter is any good and for how long. The result is a simple chart with multiples of N in col 1 and a % in col 2. I can't say I have the best combination of filters, but for certain ones, I have a standard way of comparing their effective... [ More ]

Reply #38

RJOh — Sat, 07 Jun 2008 17:41:03 GMT

Getting the Biggest Bang For Your Buck depends on the game payouts too. I compared different games that I follow by calculating the amount I could expect to win if I played 10 lines during 100 drawings. For my calculations I used the winning numbers from the 10 previous drawings during the last 100 drawings. I used these numbers because they were probably as random as any I could have picked.

OHIO CLASSIC LOTTO 6/49 TOTAL MATCHES

MATCH 0 = 441

MATCH 1 = 388

MATCH 2 = 152

MATCH 3

Reply #37

JKING — Sat, 07 Jun 2008 14:49:10 GMT

Hi,

Yes time*treat, I write code.

The point of calculating the combinations for each filter type is to select the best combination of filters. Basically getting the Biggest Bang For Your Buck . That would be reducing the most combinations possible with the least amount of error. Sometimes I feel that I am the only one that understands that error is multiplicative. And that you can filter yourself out of winning anything if you don't pay attention to the amount of error in your system.... [ More ]

Reply #36

RJOh — Fri, 06 Jun 2008 22:39:56 GMT

My system uses a RNG to pick numbers which are compared with previous drawings before being checked using the filters parameters since I'm trying to avoid using combinations that have already won.

I recently added a feature that stops the program when ever it picks a combination that has all the winning numbers of a previous drawing and the program seldom stops. This tells me that the odds of picking a winning combination to filter out is the same as picking a winning combination without filt... [ More ]

Reply #35

time*treat — Fri, 06 Jun 2008 22:12:55 GMT

@JKING: Are you working your system on paper or by code? That will make a big difference in how sophisticated you can work these filters.

Reply #34

RJOh — Fri, 06 Jun 2008 03:15:35 GMT

@RJOh, have you ever changed the order the combos are filtered in?

I don't generate a bunch of combinations to be filtered later, so it makes no different. I filter the combinations as they are generated and once a combination is accepted, it blocks any other similar combinations. For example if the combination 1 2 3 4 5 makes it by the filter and is accepted then any other combinations with three or more of those numbers are rejected. So a combination like 2 4 5 9 10 would be rejected and i... [ More ]

Reply #33

time*treat — Fri, 06 Jun 2008 01:19:43 GMT

@RJOh, have you ever changed the order the combos are filtered in?

I wrote in another thread, the p5 is very sensitive to the initial conditions. You can have the big winner in your list but if it's too far down the list , you run out of budget before you get to it. Which (besides being very annoying) makes a very big difference in the profit/loss column.

@JKING, yes, the % of combos removed by those various filters can be calculated easily enough. You may find the totals to be disappoint... [ More ]

Reply #32

RJOh — Thu, 05 Jun 2008 23:14:27 GMT

Unless you narrow the parameters of each filters, most of the combinations you generate will pass through and narrowing them down and running previous winners will eliminate most of them.

I try to set filter parameters so that 70% of the previous winners would have passed and once a combination is accepted, it is used to block other combinations that passed the filters but match three or more of the numbers in one of the combinations already accepted and I record them in a reject file.

Whi... [ More ]

Reply #31

JKING — Thu, 05 Jun 2008 21:44:58 GMT

Hi,

Below are addition filtering methods available in commercial software packages. Quite a few of the following methods have been recycled constantly here at the LP ever since I joined. That is why my thinking is that there should be some standardized measure (ie- amount of combinations being reduce versus the amount of error being incurred.

Any thoughts?

Lottery Filter

Consecutives - A filter that eliminate combinations with N consecutive numbers. For example, the combination 21,2... [ More ]

Reply #30

JKING — Thu, 05 Jun 2008 20:17:55 GMT

Hi,

I thought this was very good input

from Starprose for Mega Millions. I haven't

had the time to figure out how many

combinations he is eliminating under

each heading yet.

Filter Name

From

To

Number of Draws

Percentage

Last Digit Check

1

3

50

100 %

First Digit Check

2

3

42

84 %

Same Digit Check

2

4

50

100 %

Prior Number Check

0

2

49

98... [ More ]

Reply #29

GASMETERGUY — Tue, 03 Jun 2008 01:07:26 GMT

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.... [ More ]

Reply #28

JKING — Fri, 30 May 2008 09:46:38 GMT

Hi,

By using an individual number method, for Pick 3, you get the following:

1-leaves 729 combinations -%error=27.1 (ie-eliminate the number 8 from all combinations being considered)

2-leaves 512 combinations-%error=48.8

3-leaves 343 combinations-%error=65.7

4-leaves 216 combinations-%error=78.4

5-leaves 125 combinations-%error=87.5

6-leaves 64 combinations-%error=93.6

7-leaves 27 combinations-%error=97.3

8-leaves 8 combinations-%error=99.2

9-leaves 1 combinations

Reply #27

BobP — Fri, 30 May 2008 07:15:55 GMT

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 Straig... [ More ]

Reply #26

KY Floyd — Fri, 30 May 2008 05:09:59 GMT

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 mo... [ More ]

Reply #25

RJOh — Thu, 29 May 2008 23:09:02 GMT

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 answe... [ More ]

Reply #24

Stack47 — Thu, 29 May 2008 22:35:15 GMT

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.

Reply #23

time*treat — Thu, 29 May 2008 18:45:25 GMT

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.

Reply #22

GASMETERGUY — Thu, 29 May 2008 16:38:00 GMT

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 b... [ More ]

Reply #21

RJOh — Thu, 29 May 2008 15:32:52 GMT

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 f... [ More ]

Reply #20

RJOh — Thu, 29 May 2008 04:49:52 GMT

FILE :OHIO ROLLING CASH5

COUNT OCCURRENCES OF NUMBERS IN FILE

10/04/04 TO 05/28/08 1193 RECORDS

11. 176 31. 161 18. 151 37. 141

21. 175 36. 160 1. 150 17. 139

15. 172 23. 159 14. 150 6. 138

7. 171 33. 159 29. 149 2. 137

30. 170 24. 156 32. 149 9. 137

3. 169 13. 154 10. 148 28. 136

19. 166 27. 154 20. 148 39. 136

38. 165 4. 153 12. 147 25. 135

35. 164 8. 152 26. 146 22. 133

16. 162 34. 152 5. 145

A=11 21 15 07 30 03 19 38 35 16 31 36 23

B=33 24 13 27... [ More ]

Reply #19

JKING — Thu, 29 May 2008 04:05:17 GMT

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 com... [ More ]

Reply #18

BobP — Thu, 29 May 2008 03:35:57 GMT

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.

1: 000

2: 002

3: 004

4: 006

5: 008

6: 020

7: 022

8: 040

9: 044

10: 060

11: 066

12: 080

13: 088

14: 200

15: 202

16: 220

17: 222

18: 224

19: 226

20: 228

21: 242

22

Reply #17

RJOh — Thu, 29 May 2008 01:49:20 GMT

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 the... [ More ]

Reply #16

KY Floyd — Thu, 29 May 2008 00:18:31 GMT

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 on... [ More ]

Reply #15

JAP69 — Wed, 28 May 2008 23:24:42 GMT

What state would anyone like 5 numbers eliminated in?

I could get real lucky though and hit five of five.

Reply #14

JAP69 — Wed, 28 May 2008 23:17:03 GMT

I knock out five numbers from the grid 80% of the time that will not be drawn when I play my choices on the one line I play when I do play.

Reply #13

time*treat — Wed, 28 May 2008 22:58:27 GMT

@JKING: I've been working on this kill-one-number angle for a while. I think it has merit.

@GASMETERGUY: Farthest out is a good place to start.

Reply #12

GASMETERGUY — Wed, 28 May 2008 21:38:37 GMT

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

Reply #11

GASMETERGUY — Wed, 28 May 2008 21:28:43 GMT

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.

Reply #10

JKING — Wed, 28 May 2008 18:31:52 GMT

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.

Reply #9

BobP — Wed, 28 May 2008 06:57:32 GMT

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 cann... [ More ]

Reply #8

johnph77 — Wed, 28 May 2008 06:40:11 GMT

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.

Reply #7

KY Floyd — Wed, 28 May 2008 05:14:48 GMT

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 absolut... [ More ]

Reply #6

RJOh — Tue, 27 May 2008 20:15:54 GMT

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 3... [ More ]

Reply #5

JKING — Tue, 27 May 2008 19:44:19 GMT

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.

Reply #4

time*treat — Tue, 27 May 2008 18:09:12 GMT

Would the CA game be used as the standard

Reply #3

JADELottery — Tue, 27 May 2008 18:05:21 GMT

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.

Reply #2

JKING — Tue, 27 May 2008 13:12:14 GMT

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.

Reply #1

KY Floyd — Tue, 27 May 2008 05:00:45 GMT

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 co... [ More ]

Combinations & % Error Challenge

JKING — Mon, 26 May 2008 22:48:46 GMT

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... [ More ]