because it looks interesting but I don't get it. thanks. . . . . . . . . . . . . . . . . . . . . . . . .
Roulette System "3 Sets Of Numbers"
1). We will be working with 3 sets of numbers.
2). Those sets are: zero to ten ( 0 to 10 ) eleven to twenty ( 11 to 20 ) and twenty-one to thirty ( 21 to 30 )
4). Using a simple reduction technique, we are going to reduce those sets until one set has only six numbers remaining.
5). We will play those six numbers for a maximum of six spins.
6). When one of the six numbers strikes, we discard that set of six, and proceed to find another six, using the same tequnique.
7). We use flat bets only.
8 We ignore all the numbers above 30.
9). THE METHOD.
For simplicity sake: ALL the roulette numbers can be represented as > 1. 2. 3. 4. 5. 6. 7. 8. 9. 0.
(1 is 1. 11. 21. 31. . . . 2 is 2. 12. 22. 32. . . . . . . . . . . . . . . . . . . . 0 is 0. 10. 20. 30. etc etc. )
Counting back from (and including) the last number spun, we proceed through the previously spun numbers in sequence, one at a time.
For every 1 we also reduce a 0 (and vice-versa) for every 0 we also reduce a 1
For every 2 we also reduce a 9 (and vice- versa) for every 9 we also reduce a 2
For every 3 we also reduce a 8 (and vice- versa) for every 8 we also reduce a 3
For every 4 we also reduce a 7 (and vice- versa) for every 7 we also reduce a 4
For every 5 we also reduce a 6 (and vice- versa) for every 6 we also reduce a 5
To simplify the above table:
1 2 3 4 5
0 9 8 7 6
Example: You wish to find six numbers to play from spin number 8.
Spin numbers: 5. 0. 30. 23. 8. 11. 35.
First reduce 11 and 20
Next reduce 8 and 3
Next reduce 23 and 28
Next reduce 30 and 21.
You now have your six numbers to play from spin number 8. 22. 29. 24. 27. 25. 26.
Example: You wish to find six numbers to play from spin number 11.
Spin numbers: 12. 30. 0. 5. 21. 25. 35. 36. 22. 28.
First reduce 28 and 23
Next reduce 22 and 29
You now have your six numbers to play from spin number 11. 21. 30. 24. 27. 25. 26.
Example: You wish to find six numbers to play from spin number 12:
Spin numbers: 29. 4. 29. 0. 6. 2. 3. 28. 16. 18. 22.
First reduce 22 and 29
Next reduce 18 and 13
Next reduce 16 and 15
You now have your six numbers to play from spin number 12:
11. 20. 12. 19. 14. 17.
10). Clear Profit per Hit for single unit bets.
Spin number Profit
1 30
2 24
3 18
4 12
5 6
6 break -even
No-hit a loss of 36 units.
11). The one exception to the above figures is.
When playing the first set of numbers and one of the selections is 0 or 10.
You play 7 numbers instead of 6
Example: You wish to play 6 numbers from spin number 5.
Spin numbers: 20. 19. 8. 5.
First reduce 5 and 6
Next reduce 8 and 3
You now have seven numbers to play 1. 0. 10. 2. 9. 4. 7. for the next five spins.
For a Clear profit per Hit for this situation.
Spin number Profit
1 29
2 22
3 15
4 8
5 1
No-Hit a loss of 35 units
12). As you can see it is a simple task to find the numbers to bet.
Using this simple set of digits:
1 2 3 4 5
0 9 8 7 6
You count back from the last number spun, or from the last number to hit.
Proceed to reduce using the formula provided.
Practice with a set of spins.
It soon becomes very easy to instantly identify the six numbers to bet.
13). Bank: and notes:
I would recommend a minimum bankroll of $200.
The first target would be to double ones bank.
When that is achieved, pause.
Resume some time after with a similar target,
Again, with a bank of $200, aim to double it.
Set your own limits and stick to them.
I will not attempt to explain for now why I believe it works;
I believe it exploits a mathematical anomaly.
To refresh:
Always count back from the number just spun; or the number that just hit.
Reduce the sets of numbers as described and as soon as one as six remaining
Play those numbers.
Play until one hits.
Play for a maximum of six spins.
As soon as one hits; discard that set and proceed to get another six.
Observe the record of spins following. The ones highlighted are hits.
17. 22. 10. 25. 24. 21. 17. 23. 22 . 20. 2. 1. 34. 35. 0. 14. 36. 5. 0. 20. 30. 23. 8. 11. 35. 2 7.
You wish to bet on spin number 5.
Counting back from and including spin 4.
25. reduce 25 and 26
10. reduce 10 and 0 and 5 and 6
22. reduce 22 and 29.
We now have the numbers to bet from spin number 5:
21. 30. 23. 28. 24. 27.
Spin number 5 24 HIT
You want 6 numbers to bet from spin 6.
Counting back from and including spin 5.
24. reduce 24 and 27
25. reduce 25 and 26
You now have the six numbers to bet spin 6 onwards.
21. 30. 22. 29. 23. 28.
Spin number 6 21 HIT
You want 6 new numbers to bet from spin 7 on.
Counting back from and including spin 6.
21 reduce 21 and 30
24 reduce 24 and 27
We now have six numbers to take from spin 7 on:
22. 29. 23. 28. 25. 26.
Spin number 8 23 HIT
You want 6 new numbers to bet from spin 9 on.
Counting back from and including spin 8.
23 reduce 23 and 28
17 reduce 17 and 14
21 reduce 21 and 30
We now have six numbers to bet frome spin 8 on:
21. 30. 22. 29. 25. 26.
Spin number 9 22 HIT
You want 6 new numbers to bet from spin 10 on:
Counting back from and including spin number 9:
22 reduce 22 and 29
23 reduce 23 and 28
We now have six numbers to take from spin 10 on:
21. 30. 24. 27. 25. 26.
Spin number 15 0 LOSS
You want 6 new numbers to bet from spin 16 on:
Counting back from and including spin number 14.
(ignore all the numbers 31. 32. 33. 34. 35. 36. )
0 reduce 1 and 10
2 reduce 2 and 9
We now have the six numbers to take from spin 15 on:
3. 8. 4. 7. 5. 6.
Spin number 18 5 HIT
You want 6 new numbers to bet from spin 19 on:
Counting back from and including spin 18:
5 reduce 5 and 6
(ignore all the numbers 31. 32. 33. 34. 35. 36. )
14 reduce 14 and 17
0 reduce 0 and 1 and 10
We now have the numbers to take from spin 19 on:
2. 9. 3. 8. 4. 7.
Spin number 24 8 HIT
You now want 6 new numbers to bet from spin 25 on:
Counting back from and including spin 24:
8 reduce 8 and 3
23 reduce 23 and 28
30 reduce 30 and 21
We now have the numbers to take from spin 25 on:
22. 29. 24. 27. 25. 26.
Spin number 28 27 HIT.
I agree with Red. I have read this system numerous times, and have never been able to understand exactly how the author reduces the numbers. If anyone knows what the heck he's talking about, please help.
Thanks,
G.
I think I got it. Let me try to explain it with my poor English.
As he points first, we divide the layout in 3 10-number sets: 0-10, 11-20, 21-30 excluding numbers from 31 to 36.
Then we consider cadences (numbers that end with identical digit)
0,10, 20, 30
1, 11, 21, 31 etc.
Then we go through the sequence of numbers that just has been spun in order to cross off numbers from each set. We look for the first set in which there has been exactly six numbers left. We use the following rules:
When there is a previous number ending in 0 we cross off that number and any other numbers ending in 0 and 1 from the same set. And vice versa if there is number ending in 1 we cross off any number ending in 1 and 0 from the same set
the same rule applies accordingly to:
previous number ending in 2, we cross off numbers ending in 2 and 9 from the same set;
3 <-> 8,
4 <-> 7
5 <-> 6
Example: #17 is a previous number. We consider the set 11-20 and we cross off numbers 17, 14. There are now 8 numbers left in this set. We need to cross off two more numbers in order this set to qualify.
Example: #10 is a previous number. We cross off 0, 1, 10. There are now 7 numbers left in this set (2, 3 , 4, 5, 6, 7, 8, 9) and we play them for the next 6 spins according to exception 11).
Now look at his examples. I hope this helps a little.
P. S the confusion comes from the word "reduce" the author uses. Change it with discard and all become clear.
thank you chovek, I think your explanation will help a lot. will try and follow and test it just for the heck of it. I wonder if there is any logical rationale to this system?
Quote from: chovek69 link=topic=330. msg2288#msg2288 date=1277548124
I think I got it. Let me try to explain it with my poor English.
As he points first, we divide the layout in 3 10-number sets: 0-10, 11-20, 21-30 excluding numbers from 31 to 36.
Then we consider cadences (numbers that end with identical digit)
0,10, 20, 30
1, 11, 21, 31 etc.
Then we go through the sequence of numbers that just has been spun in order to cross off numbers from each set. We look for the first set in which there has been exactly six numbers left. We use the following rules:
When there is a previous number ending in 0 we cross off that number and any other numbers ending in 0 and 1 from the same set. And vice versa if there is number ending in 1 we cross off any number ending in 1 and 0 from the same set
the same rule applies accordingly to:
previous number ending in 2, we cross off numbers ending in 2 and 9 from the same set;
3 <-> 8,
4 <-> 7
5 <-> 6
Example: #17 is a previous number. We consider the set 11-20 and we cross off numbers 17, 14. There are now 8 numbers left in this set. We need to cross off two more numbers in order this set to qualify.
Example: #10 is a previous number. We cross off 0, 1, 10. There are now 8 numbers left in this set (2, 3 , 4, 5, 6, 7, 8, 9). We need to cross off one more number to play them for according to exception 11).
Now look at his examples. I hope this helps a little.
P. S the confusion comes from the word "reduce" the author uses. Change it with discard and all become clear.
I am still having trouble following this system. below are 75 consecutive spins from a real double zeroe wheel. chovek (or anyone) is it possible you could show where and what the betting opportunities would be here as per this system and if they won or not? thanks so much.
33
21
15
24
35
28
11
32
4
3
21
35
17
8
2
26
11
28
15
14
4
14
27
25
22
6
3
12
31
30
24
28
9
25
17
11
22
8
1
25
27
14
5
0
18
16
16
1
4
21
22
36
4
6
5
35
22
33
2
25
20
21
16
9
16
25
9
36
4
9
25
10
12
16
6
I remember, I gave this confusing puzzle system to the old Steve's forum to clear up. Nobody answered in 3 months and I am still in clouds! I almost reduced my brain to walnut and still didn't find the answer.
Create Holy Grail is easier than explain that system.
hermes
???
in above sequence of numbers I started out crossing off the 33 and 28, then the 24 and 17, then the 26 and 15, and then I got lost what to do next and where the bet is. . .
Maybe we need to get Victor or TwoCatSam to look at this. They are usually pretty good at figuring things out.
Scooby Doo
chovek gets it. as I understand it we want to eliminate four of the numbers in any number group and then bet on the other six numbers in the group. my question is where is the first betting opportunity where you can do this in the above outcomes?
There are several ways to play this concept in a much simpler fashion. I will present a Matrix below that does Exactly the same but much more easy;)
Here's my Favourite Global Pies Constellations. The way I belive they fall consistently:
So You want to play each of the 3 Matrix separatly/simultaneously Like this:
Example:
Outcome:
4
17
16
x closing vertical line of 4 numbers
32 15 19
4 21 2 x closing horisontal line of 3 numbers
31 14 9
22 18 29
x closing vertical line of 4 numbers
25 17 34 x closing horisontal line of 3 numbers..
6 27 13
7 28 12
35 3 26
x closing vertical line of 4 numbers
36 11 30
8 23 10
5 24 16 x closing horisontal line of 3 numbers..
33 1 20
You all see where to put the marks that "Closes" the Lines of numbers.
Repeat this procedure until You have 3-12 numbers "Open" for play. Then bet.
Very simple. This is also refered to as the "Raindrop"-Method.
Cheers!
/Compa
thanks. unfortunately I have no idea what above is about. I still want to see how the system does on the sequence of numbers I showed. I will work on it. . .
Good Luck
Hello Red
I'm going to have a go at explaining this method,I will work on the numbers that you posted ;
Set out 3 tables like the author ;
0-1-2-3 11-12-13 21-22-23
4-5-6-7 14-15-16 24-25-26
8-9-10 17-18-19 27-28-29
20 30
Ok now every time a number is spun in those tables we cross it off with its corresponding partner.
Partners Table
For every 0 we cross off the 1,for every 1 we cross off the 0
For every 2 we cross off the 9,for every 9 we cross off the 2
For every 3 we cross off the 8,for every 8 we cross off the3
For every 4 we cross off the 7,for every 7 we cross off the 4
For every 5 we cross off the 6,for every 6 we cross off the 5
The above table applies ONLY to the cadences of the spun numbers
Example 33 cadence is 3
26 cadence is 6
13 cadence is 3
7 cadence is 7
Sorry Red,
Now to your numbers,
33 we ignore all numbers over 30
21 Tables 21-22-23
24-25-26
27-28-29
30 cross out number 21 and its partner in the cadence table 1=0,so that's
the number 30 as the only number in this table with a 0
15 Tables 11-12-13
14-15-16
17-18-19
20 cross out the number 15 and its partner in the cadence table5=6,so
that's the number 16 as the only number in this table with a 6
24 Tables 21-22-23
24-25-26
27-28-29
30 cross out the number 24 and its partner in the cadence table 4=7,so
that's the number 27 as the only number in this table with a 7
Now we have our betting opportunity as the 21 to 30 table has 4 numbers crossed off
35 L bet 22-23-25-26-28-29
28W
Hello Red
I hope that you can understand this way ,it took me a while LOL
I had trouble posting all information in one ,don't know why never had a problem before!
I have done some small tests on this not great so far :(
Hope you have better luck mate
btw you will have to tweak this for 00 as this designed for euro wheel maybe put 00 with table 1 - 00 and you will have to play 8 numbers instead of 7 just my 2p worth
All the best
Rob
Hello Red,
I will go through some of the numbers you posted to try and give you an idea.
It is certainly not the easiest method to understand at first glance. Well it seems very creative, I can only think that it is still probably a work in progress. The plus side to all this is that it may get someone elses creativity flowing.
so. . . .
33, does not qualify. (remember, we don't count numbers 30-36. )
21, In the section 20,21,22,23,24,25,26,27,28,29. Take out numbers 20+21.
15, In the section 10,11,12,13,14,15,16,17,18,19. Take out numbers 15+16.
24, In the section 20,21,22,23,24,25,26,27,28,29. Take out numbers 24+27. In this section you have already taken out numbers 20+21, and now that you have taken out 24+27 as well, you are just left with 6 numbers in this group and remember that is one of the rules for qualifying a bet. Therefore you are left with the numbers 22,23,25,26,28,29 to bet for the next 6 spins. You only flat bet these numbers. So the most you can lose is 36 units. If you get just one hit, you at least get your money back.
so. . . . .
35, once again that number does not qualify for anything. -6 units.
28, a winner here. you have laid out 12 units and have just got back 36 units here, so you are in profit by +24. I am not sure at this point if you are supposed to carry on and look for more hits or restart. To stay on the side of caution, let's stop at a win within the 6 spin betting opportunity or take our 36 unit loss and start retracking.
so. . .
11, In the section 10,11,12,13,14,15,16,17,18,19. Take out numbers 10+11.
32, does not qualify.
4, In the section 0,1,2,3,4,5,6,7,8,9. Take out numbers 4+7.
3, In the section 0,1,2,3,4,5,6,7,8,9. Take out numbers 3+8. So here in this section, you have now taken out numbers 4+7 and 3+8. You have a betting opportunity here and are required to bet numbers 0,1,2,5,6,9 for the next 6 spins.
so. . . . .
21, loss -6. total running count +18.
35, loss -12. total running count +12.
17, loss -18. total running count +6.
8, loss -24. total running count 0.
2, winner, +6. total running count +30.
So far you have had 2 winning coups from your 2 attempts.
I hope you get the idea now red.
How about when you get time, try the remaining numbers and I will let you know if you get them correct or not.
regards
sherminator
thank you 2bob and sherminator, I was about to post what I think how I played correctly and now I will go over your posts too. in the mean time does this seem right. . . I looked at the first 7 spins which were
33
21
15
24
35
28
11
so I took out 11 and 20 from that group and I took out the 28 and 23 from that group and I took out the 24 and 27 from that group which leaves the 21-22-25-26-29-30 from that group, we bet on those numbers and get a hit in 3 spins with the 21 (next 3 numbers were 4-3-21. ) I cannot even begin to think about if there is any logic at all to this system but at least it's different from most and I find it curious. . . will go over your posts and see how the rest of the numbers do. . . thanks much for the "tech support"!
I just had a quick visual scan through the remaining numbers you posted up red and can see that there were roughly another 6 betting opportunities, at first glance it looks like 4 won, 1 broke even and there was 1 loser. It actually becomes very easy to track it once you understand it.
It could just be that the numbers you provided are a good set of numbers for this method, but going on those results, I suppose it warrants further inspection.
I have five more sets of 75 consecutive spins from real wheels. I will post the second set here for now in case you want to check it out. still going over the first set. . .
30
10
25
23
22
6
31
24
18
0
25
22
26
0
9
23
34
3
34
25
22
20
4
24
35
3
0
7
14
34
6
4
29
27
3
26
12
35
3
10
30
7
35
00
15
34
32
33
4
4
1
14
24
8
21
24
24
15
36
30
00
18
12
30
9
9
32
33
28
34
18
00
27
24
going back to first set it seems I have another win (I have the numbers in a spreadsheet showing spin numbers 1 through 75 with the outcomes in another column, which would make it easier to explain, but they get jumbled when I try and paste them here.) anyway we had these numbers 33 21 15 24 35 28 11 32 4 3 21 and won on the 21, then I looked back starting at the 3, took out 3 and 8 and 4 and 7, leaves 0-1-2-5-6-9-10 in that set to bet on (and I suppose 00) and after the 3 we got 21 35 17 8 2 so won on the 2 ($5 anyway.) think I'm doing this right (?) again, no clue why this would work but it's interesting that someone would come up with this for whatever reasons... will contiinue to test... (at least it's flat bet which is always good)... later...........
hmmm... looks like another win eight spins after the hit on the 2... after the 2 we get 26 11 28, maybe I am not supposed to use this set again (?) but it is easy to see with the 26 and 28 if I am doing this right we can eliminate 26 25 and 28 23 which leaves 21 22 24 27 29 30 to bet and after 28 we get 15 14 4 14 27 so we win six units when 27 hits. hmmm, interesting, well I really gotta get back to the real world but will continue to look at this... my little holy grail.... yeah I wish....
Once you get used to it, you only really need to look through the numbers from a session to see if it would have shown a profit or not.
I have did an awful lot of testing over the last few years using 6 numbers as a base for looking at creating different kinds of methods, especially using RXextreme. What got me was seeing the sharp spikes going both up and down. Probability would dictate that you should be hitting 1 in 6. Unfortunately as we all know, this is roulette and weird and wonderful things happen, lol. THere are a periods of say 500-750 spins where the spike just continues to go downwards. Then you will often see a correction where if you were playing for them 500-750 spins, you would have made an absolute killing.
So the point I am making and especially regarding this system is that I don't necessarily think it is such a good idea to be just betting the 6 spins "blind" every time you get a qualifier. To be a long term winner at roulette, you really need to look into things a bit deeper than that and that is why I said in an earlier post that I thought this was a good starter concept, but it could definately be improved.
I think one way of doing that and this is just a suggestion, would be to introduce the law of the third into the equation.
For example, let's say you get wins on the 1st, 3rd, 5th and 6th spins of every betting opportunity, you may then just like to continue to play on these particular spins looking for continued wins and hoping the 2nd and 4th spin continue to be absent for a while.
You will amazed how long the winning runs can be using this kind of plan and it conserves your chips.
Once you hit a loss playing like this, you could then just track and look for a winner again before continuing in this fashion.
As an example, let's say you are just playing on spins 1,3,5,6 and you get the following win sequence.
1
1
3
5
5
3
6
2
The 2 has broken the winning run, but it would have paid for itself because of the chips you saved. Now the missing "win spins" would be 1 and 4.
However, because the last one was a loser, you may like to wait for a winner like I said before starting again because often the losses can come rapid fire just like ther winners can.
For example carying on, you might now see the following.
4 so this would also be a loser and now the 2 furthest back win spins are 1 and 5.
5 so now there has being three losers in a row. Now the 2 furthest back win spins are 1 and 3.
5 so this starts of the winning sequence again and now you can start betting again. you might see something like...
5
4
6
4
6
2
2
Everything in roulette seems to mostly happen in cycles and you need to be looking out for it and trying to take advantage, sure it is not an excact science and there are times when it can go choppy, however, there is nobody saying you have to bet every spin and it is my experience that a bit of good judgement and patience will nearly always pay off and boost your long term winnings.
thanks for the input. what do you do if a system does great (on a couple hundred spins) and there is no reason for it to? this system does killer on my spins. I know how to play it quickly now and feel like I would like to give it a go at the casino... where it will probably let me down. but you should check out how amazingly well it does on the spins above (in reply 19.) it is just hit after hit.
Hi,
So I worked the solution to the 75 spins and you'd come out ahead around 90 bucks, however, I would suggest doing 5 spins per set. Most casinos pay 35 to 1 and if you did 6 spins you'd be losing a dollar. With 5 you break even. Not sure if this skews the results? Also, the easiest way to get your 6 numbers is to take two of the numbers in a range (not in the 30's) and "reduce" those.
(e.g 13, 4, 34, 32, 26, 31, 2)
Take the 4 and 2 and reduce. The reason is that each group has 10 numbers 1-10, 11-20, 21-30 and all you are looking to do is cut 4
Hope this helps.