John Gold's Blog.

[bonanza]

Hello John,
I am new here and this is my first post;
this is a very interesting thread  , I read over and over and think that I have understand most things.  There are a few questions.  First: you said

You will be betting between 6-8 streets when you have a bet.  Very seldom there will be 5 to bet.

I think I found some more situations: e. g.  the sleepers are the streets number 7, 8 and 10; the last street come in is the 12.  When I go through from 12-1 to 12-12 there are only four combinations:
12-5
12-7
12-8
12-10
so we bet only four streets.  Is that correct?

Thank you in advance
bonanza

[John Gold]

Hello Bonanza (great name by the way )

You are right about the 4 streets in the situation you described.
There will be a few times when you are betting from street 12 that you only need to bet 4. It is rare.
I counted up all the possible bet combinations and it comes to 2070.
I will spend a few hours over the weekend typing them all out. I have them in a small jotter with paper clips seperating the 12 different streets. It makes it easy for when I am in the casino to just flick through to the required page I need.
I did not want to write them out here at first because then it would just look over complicated. I suppose one issue from working out the bet from the marking template+betting chart is that you could easily miss one out if you are new to it or under pressure due to time restrictions.

I have experienced some really good luck with this over the last few days. In most of the sessions that I am playing, the three group tends to go missing for very long periods. I am getting a lot of good runs with the 1 and 4 group giving me loads of consecutive wins on the second result of each pair.

Please feel free to ask any questions if you are not sure about anything and thanks for showing interest in the method itself. 

[John Gold]

I seem to be really good at jinxing myself just lately what with one thing and another.  I decided to have a quick game online and came across the following scenario.

10
–
7
11
3 (4)

2 (10)  5/8/12.  LOSS.
3 (3)  5/8/12.  WON.  (GROUP 4)

11  5/6/12.  WON.
11  5/6/7.  LOSS.  (GROUP 2)

1 (2,9) 5/6/7.  LOSS.
12 (1,11,12) 5/6/7.  LOSS.  (GROUP 3)  MISSING GROUP 1.

8 (8} 5/6/7.

I don't like to bet when the missing group is 1. At least it only took 11 spins to reach this stage.

One other thing some of you may be wondering. If I have to cross out a street because one of the sleeping streets appeared, I always cross out the furthest one back. BUT.... what if you go back and the furthest back streets look like (2,3,5) which one will I cross out because you could argue here that all three are the furthest back. Well to make it simple, I cross out the 2. It always seems to work out fine doing it like that so I would stick with doing it that way.

[bonanza]

Hello John,
thanks a lot for your quick reply, now it is sure that I got it right.  Another question is about the marking template: why is it street #12 that is the fewest street in that template? Is there a certain reason?
And: if you should advise a progression for your method, which would it be? The 'Pluscoup', too? I think a Paroli xould also be a good option.

Kind Regards
bonanza

[John Gold]

@bonanza,

You will notice that the 12 street is one of the missing three sleepers a lot of the time.
For a 12 to appear needs one of two things to happen.  Either the actual street 12 itself appears or the previous street repeats itself. So this is a 1/6 chance. All the other streets are more likely to appear than the 12 street.

The idea for all this came when I was messing around looking at a few different concepts. I have tried several variations but this idea that I am presenting now throws up the best win streaks by far of anything that I have tested.. That would be one reason why your idea of the paroli is a good way of tackling it. I am not a keen fan of negative progressions. However the paroli if used at the right times could take advantage of the win streaks. Just a gentle paroli of 1,2,3 etc.... would be my preferred way of going about it.

Another variation is to look out for the 'hot streets' This would mean playing the method in reverse.
Let's say the 4,5,12 are the three missing streets and the current street is 10.
So playing the method I am describing here would see you betting streets 2,3,4,5,6,8,10,12. (A total of 8 streets).
If you wanted to bet for the sleepers to continue sleeping, you would bet the 1,7,9,11 streets. (A total of 4 streets).
The only time I would look out for this is when the GROUP 1 is the missing group. You are more likely to see a LOSS in the first result of the pair and a LOSS in the second result of the pair if the GROUP 1 continues to be the furthest back group.
The problem with this is it can take a long time for the GROUP 1 to be the furthest back group. Also the win streaks don't tend to be as extended as with the other way around. I prefer playing the sleepers route as a more conservative approach. It does not experience as much deviations which can be harmful to a limited bankroll.

It was playing the hot streets variation first which opened my eyes to the really long winning streaks that the sleeper streets enjoyed. I was playing in a casino one day and the GROUP 3 became the furthest back group. It continued to sleep for another 27 pairs and I could have cleared the national debt if I took advantage of that.
So after a lot of testing, it became aparrent that the sleeper streets was the better opportunity of the two.

cheers.

[bonanza]

@John Gold,

thanks again for your direct answer.

Either the actual street 12 itself appears or the previous street repeats itself.  So this is a 1/6 chance.  All the other streets are more likely to appear than the 12 street.

So, when you devised the template it would be possible to change things, e. g.  street #1 become the 1/6 chance?
Looking out for 'hot streets' is an interesting alternative, I will follow this.

One very important thing that was not mentioned yet: What about 'the green one', how do you handle the Zero?

Cheers
bonanza

P. S. 

Bonanza (great name by the way )

Thanks, do you know the meaning of 'Bonanza'?

a source, usually sudden and unexpected, of luck or wealth

So this definition matches for your method, too?! 

[John Gold]

Hello bonanza,
I spent a few hours today turning everything inside out and upside down to see what would happen.
Using my three column approach will always leave the 12 street as a 6/1 chance for appearing.
Some of the different variations seemed to have a negative effect on the win streaks for the sleepers.
This was evident just by studying the different marking templates I came up with.

I regard the zero as just another losing number if it hits. You can choose to either mark it down or ignore it.

Let's hope this method becomes a real 'bonanza' for anybody that tracks and plays it. 

[John Gold]

 Hello guys.
I posted a type of matrix bet up in the full systems section a few months ago. Unfortunately I made a few miscalculations.
Anyway, I decided to take another look and see what I could come up with.

I have found something that looks pretty solid in so far that the wins seems to come at a steady enough rate compared to the losses. I was winning by just flat betting in my tests but I think a conservative progression could help it along.

The bet can only ever be 10 splits. (no more, no less) It is based on dozens and columns.

The matrix will go 5 lines across, it does not really matter how deep it goes.

Number 1 = 1A because it is in the first dozen and the first column.
Number 2 = 1B because it is in the first dozen and the second column.
Number 3 = 1C because it is in the first dozen and the third column.

So here is the full listing.

1 = 1A.
2 = 1B.
3 = 1C.
4 = 1A.
5 = 1B.
6 = 1C.
7 = 1A.
8 = 1B.
9 = 1C.
10 = 1A.
11 = 1B.
12 = 1C.
13 = 2A.
14 = 2B.
15 = 2C.
16 = 2A.
17 = 2B.
18 = 2C.
19 = 2A.
20 = 2B.
21 = 2C.
22 = 2A.
23 = 2B.
24 = 2C.
25 = 3A.
26 = 3B.
27 = 3C.
28 = 3A.
29 = 3B.
30 = 3C.
31 = 3A.
32 = 3B.
33 = 3C.
34 = 3A.
35 = 3B.
36 = 3C.

I will show you a matrix that I played earlier today. Next to each marking is a (Y) or an (N).
The (Y) stands for a repeat and the (N) stands for a non-repeat.

2C          2C          1A          2C          2A
3B (N)   1A (N)    1A (Y)   1A (N)    2B (Y)
1A (N)   1A (Y)    3A (Y)   2A (Y)    3C (N)
2C (N)   1A (Y)    2A (Y)   2B (Y)    3C (Y)

Here is where the first bet would be. WHY? Because I am looking for 3 lots of (N) and then betting that the next result will be a (Y). So there is a bet in the first column from the example above.
For a (Y) to appear, there will need to be either a SECOND DOZEN or a THIRD COLUMN appear.
So a WIN would be represented by any of the following.
2A, 2B, 2C, 1C, 3C.
It would be a loss if any of the following appeared.
1A, 1B, 3A, 3B. (and of course the zero however you could cover the zero if you were in the latter stages of some type of progression)
I can cover the bet in the example above by playing the following 10 splits.
13/16, 14/17, 15/18, 19/22, 20/23, 21/24, 3/6, 9/12, 27/30, 33/36.

It is as simple as that. If the bet wins, great. I pocket 8 chips profit.
If it loses, I lose 10 chips.

After any loss, STOP and wait for another betting opportunity to present itself. Don't chase the same bet because the (N) could go on an extended run for that particular bet.

What I found was that it is very common for a (Y) to appear after 3 lots of (N).
You are covering 20 numbers (10 splits) which is covering more than half the board.

Please let me know your results if you decide to test it. I have did limited testing which has held up well. More testing will give a better  idea on what is the best way to tackle it.

[John Gold]

One point that I should have explained better is in the matrix itself regarding the (N) and the (Y)

Let's say you have 3B.

A (Y) can be anything from the THIRD DOZEN or COLUMN B.

This will be either 3A, 3B, 3C, 1B, 2B.

A (N) will be either 1A, 1C, 2A, 2C.

[John Gold]

1A         2A         3A         1B         1A
3B (N)   3A (Y)    1C (N)   1A (Y)    3C (N)
2B (Y)   3C (Y)    3A (N)   3C (N)    3B (Y)
2A (Y)   2B (N)    2B (N)   2C (Y)    3A (Y)
2B (Y)   1B (Y)    2C (Y)   1A (N)    1A (Y)

In the example above, the numbers were as follows.

1, 19, 31, 11, 1,
29, 25, 6, 4, 36,
17, 33, 25, 27, 26,
22, 23, 20, 18,, 28,
14, 2, 15, 1,  4,

The only bet came in the third column. After the 20 appeared, I was looking for a (Y) to appear on the next result of that particular column.

The actual bet was 10 splits, 13/16, 14/17, 15/18, 19/22, 20/23, 21/24, 2/5, 8/11, 26/29, 32/35.
These splits cover all the SECOND DOZEN and the SECOND COLUMN which would give me a (Y) should either the second dozen or second column appear. If an (N) appeared, I would abandon that bet and wait for the next opportunity where three lots of (N) appeared and bet for the next result in the column to be a (Y).

[John Gold]

I decided to pay one of my local casinos a visit with this 'matrix' idea this afternoon. I could not resist after my testing results. 

I decided to play the live wheel and autowheel both at the same time. All the wheels are playable from a touch screen terminal. 2 wheels is the most you could physically handle with this method. I made a few mistakes but nothing that cost me any money.

There were 2 bets on the autowheel and 2 bets on the live wheel over the course of 55 spins which took about 75 minutes. All 4 bets won. I have suffered a few losses in this casino lately, so it was nice just to come out with a profit. 32 units is nothing to be sniffed at.

Taking my testing into account as well, it appears there is roughly a betting opportunity every 20 odd spins or so.
That is a bit slow and is why I was happy to play the 2 wheels.

[John Gold]

Here is an example I loaded on to a spreadsheet.

2 bets. One in the second column and one in the fourth column. Both won.

[attach=#]

[John Gold]

Here is another 100 spin example.

There were 5 bets. 3 wins + 2 losses. (W, W, L, L, W)

There is a good example here in the first column and the second column of why you only take one shot at a win and don't continue chasing looking for the (Y). The (N) went on a streak of 5. I have seen streaks of 10 and that would be damaging.

[attach=#]

[John Gold]

Back to the previous method and another example.
(I included the actual numbers in this one before converting them into streets)
15 = 5.
–
10 = 4.
30 = 10.
12 = 4.
9 = 3.
22 = 8.
33 = 11.

28 = 10.  7/9/12. LOSS.
24 = 8.  7/9/12. WON.  (GROUP 4)

36 = 12.  1/7/9. LOSS.
10 = 4.  1/7/9. LOSS.  (GROUP 3)

29 = 10.  1/7/9. LOSS.
4 = 2.  1/7/9.  LOSS.  (GROUP 3)

29 = 10.  1/7/9. WON.
10 = 4.  7/9/12. WON.  (GROUP 1)  MISSING GROUP = 2. (LOOK FOR L/W)

12 = 4.  5/7/9. WON.
34 = 12. (12) 5/9/11. WON.  (GROUP 1)

27 = 9.  2/5/11. LOSS.
18 = 6. (6) 2/5/11. WON.  (GROUP 4)

2 = 1. (2) 5/10/11. LOSS.
23 = 8.  5/10/11. WON.  (GROUP 4)

1 = 1. (1,3,5) 4/10/11.

So the missing group ended up as the 2 which means I was expecting the pairs to come out in a L/W formation.

The next three pairs returned WW,  LW,  LW.  So 5 out of 6 is not bad forecasting. (enough to pay for dinner, lol)

[John Gold]

Hello.   

I think it's time to draw a line now with the 'sleeping streets' method. I am starting to sound like a broken record.

I was going to look at constructing a method for just betting one number. On reflection, I think that idea can wait.

What I will do instead is show you how you can use the 'rule of the third' to your advantage.

A lot of guys on different forums say that the rule of the third is useless and can't work. I disagree with that myself. There are many different ways to exploit it.

Just for anybody that does not know what the 'rule of the third is'.

In any set of 37 spins, it is very likely that 2/3 of the numbers will be a combination of repeats and the other 1/3 of the numbers will be sleepers. This is pretty much a constant. Someone on a french forum did a billion spin computer simulation to proof the point. If I can find it again, I will post up the figures.

The stuff that I am going to share here is a bit of a headscratcher. I will try my best to explain it as clearly as I can.

cheers.

[John Gold]

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