John Gold's Blog.

[WARRIOR]

Maybe you could scan them.

[John Gold]

Maybe you could scan them.

That would be very time consuming.

It took me several hours a day over a month figuring them all out and writing them down in a book.  It would be a lot easier if someone can programme it. It would then be easy enough for anyone to just copy over the data.

[John Gold]

Somebody asked me what the deal is with the three rows of numbers. I will try and explain a bit.

A lot of players like to have a few different systems to play at the same time. They figure that one of them may bring some positive results and they can just hop on board and try and ride it. So the three rows is in effect playing three different systems/games at the same time or at least tracking them to see which is throwing up favourable results and which you would not want to touch with a barge pole.

I am going to show a couple of pictures below. This shows the numbers coming out for the three rows and then counts down all the original numbers. The count stays the same each time there is a repeat. The total of stars represent how many original numbers have come out for the three rows combined.

Looking at the first picture shows how original numbers are slower coming out in row 1 compared to row 2.  At spin 58, there are still 9 numbers left of the original 37 to appear in row 1 where as there are only 3 numbers left to appear of the original 37 in row 2. So following along these lines is at least giving you different options to follow.

These pictures are interesting to me because it shows that by using the three rows, a sleeper is coming out nearly every other spin. It appears also that you are getting a lot of double hits (original numbers coming out) between around spin 40-50. This would produce a 72 unit payout. 3 original numbers would be even better with a 108 unit payout.  It could be something worth looking into.


[attachthumb=#]

[attachthumb=#]

[John Gold]

The suggestion I mentioned in the above post looks promising.

Playing for one of the cold numbers to appear from all three rows from spin 38 produced the following results from the first picture.

chips placed = 860.

chips returned = 1044.

I would personally suggest however to stop once there is a combined total of 12 cold numbers left from the three rows and start again. This would have saved 36 chips in the above example.

The second picture produced the following.

chips placed = 751.

chips returned = 884.

So not a bad return over the 2 games. About 350 chips. There were 29 wins (including double and triple wins) in the first game and 24 wins (including double and triple wins) in the second game. It certainly deserves some extensive research and testing. I have recently become a big fan of progressions in the right circumstances. This one may suit some style of progression because the numbers are always counting down as the sleepers are appearing.

[John Gold]

I have added another game below. This time I have included how many chips are bet per round and a +/- running total. This game was a bit more volatile than some of my other testing and the sleepers were not in a hurry to come out.

There were still a combined total of 22 sleepers from all three rows yet to appear at spin 58. The average in my testing at this point is about 18 sleepers still to show. It still came out with a 45 unit gain when I stopped.

[attachthumb=#]

[John Gold]

One final game.

A few things to note here.....

Second row. From spin 38-52, there was only one sleeper appeared. Then 6 came out in the next 7 spins.

Third row. From spin 49-60. There was only one sleeper appeared.

Still managed to hit a 50 chip profit.

[attachthumb=#]

[John Gold]

Another game here.

[attachthumb=#]

That's around 500 chips up after 5 games.  There is maybe the starting point for something useful here.

[WARRIOR]

 No clue on how your system works.

[medo]

No clue on how your system works.

--Very simple Warrior mate.....if I tell you that it wins
would you then start from the post 1.....if you wana be
a winner,well do just that.

[John Gold]

Hello guys and thank you Medo for your encouragement. 

I suppose this blog has become a bit of a hodgepodge of ideas and is getting a bit confusing.

I will continue to write in here but I am considering starting a new thread titled 'John Gold's Methods'. That could be more structured with an index page which will give a brief description of each method.

Some of the methods in this blog are unfinished because nobody really showed any interest. I will say here and now that the method with the 8 number combinations is far and away the best of the bunch and is in my opinion as close to a long term winning method that I am likely to find. I will continue with that once anybody interested has completed the necessary charts they need in order to understand what's going on.

One thing that puts me off starting another thread is the seeming lack of interest on this forum. It looks to have died a death. It is probably a sympton of complacency and too many that have being around the block a few times but you only get out of it what you put in. I have had guys contact me with all sorts of great ideas and they don't post on here for the very reasons that I am talking about. Anyway, I will get down from my soapbox now.

@WARRIOR, I will attempt to explain more about the method above regarding the sleepers when I get back from my visit to the dentist this afternoon.

[WARRIOR]

--Very simple Warrior mate.....if I tell you that it wins
would you then start from the post 1.....if you wana be
a winner,well do just that.

I have been following this trying to absorb what has been written ,i understand the charts ,and the d and colums but no idea how to extract all the rest no clear to me thanks.

[John Gold]

I will attempt to explain in a bit more detail about the method involving the 'sleepers'.
Example:
The first number to appear is 10 and then follows 22. Here would be the markings. (this is for a single zero wheel)
10
22(36)  10 (36)  12(36)  ***
So what does all this tell us?
I am using my three columns approach here.
The first column is the actual number that appears on the wheel.
The second column number is the pockets the ball travelled from the previous number to the new number. So for 10 to reach 22, it has to pass the following numbers. 5,24,16,33,1,20,14,31,9 and then lands on 22. Counting these numbers up in total = 10. (you don't count from the original number which was the 10, you start counting from the 5.)

The third column number is a count which resembles a clock face. So for 10 to reach 22, it has to pass the following numbers. 11,12,13,14,15,16,17,18,19,20,21 and finally 22. Counting these numbers in total = 12. (once again, you don't count from the original number which was 10, you would start from the 11.)

The (36) represents the countdown from a full set of 37 numbers. This procedure is done for all three columns.
The *** is showing that there were original numbers appearing in all three columns with no repeats.

The next number out is 34. Here are the new markings.

10
22(36)  10(36)  12(36)  ***

34(35)  18(35)  12(36)  **

So the 34 in the first column is another original number and not a repeat. The counter now drops to (35)

How did I get the 18 in the second column. The ball travelled from the previous number which was 22 to the new number which is 34. So the ball travelled over the following pockets.
18,29,7,28,12,35,3,26,0,32,15,19,4,21,2,25,17 and finally 34. (Add them all up and you get 18)
The 18 is also an original number in the second column and so the counter in the second column is down to (35)

How did I get the 34 in the third column. Count from the 22 to the 34.
23,24,25,26,27,28,29,30,31,32,33 and finally 34. (This adds up to 12 numbers in total)
The 12 is a repeat number in the third column. Therefore the counter remains at (36)
You have only had 2 original numbers on this line and that's why you see the **.

If you look at the snapshots and read the above, you should be able to see clearly what is going on regarding the markings.

I will explain more about the betting side of things in my next post.

[John Gold]

Now to explain the betting side of things.  

The game in the snapshot below will help you understand the explanation.

The following occurred in this game after 37 spins.

There were 13 sleepers in the first column.
There were 17 sleepers in the second column.
There were 14 sleepers in the third column.

I am going to start betting on all these from the 38th spin onwards.
The sleepers from the first column are as follows.
1,2,4,8,10,13,15,18,19,25,26,32,33.
So this is a simple case of just placing a single chip on each of these 13 numbers.
Working out the bets for the sleepers in the second and third column is slightly more complicated as you will see.
The sleepers from the second column are as follows.
2,5,6,9,10,11,12,13,14,19,20,24,25,29,32,33,36.
Ok, now the previous number out on spin 37 was 36 .
So you have to match up the sleepers with the pocket distances from the 36.
So the first sleeper in the second column was 2.
Going two pockets from the 36 will take you to number 30. So you would place a chip on 30.
The second sleeper in the second column was 5.
Going 5 pockets from the 36 will take you to number 10. So you place a chip on 10.
I will list the sleepers from column 2 and the corresponding numbers you would be betting on.
2 = 30.
5 = 10.
6 = 5.
9 = 33.
10 = 1.
11 = 20.
12 = 14.
13 = 31.
14 = 9.
19 = 28.
20 = 12.
24 = 0.
25 = 32.
29 = 21.
32 = 17.
33 = 34.
36 = 13.

What is important to remember is that the numbers to bet on for the second and third columns will adjust every spin. (you can easily make tracking wheels for use in a B+M casino. A tracking programme would be ideal for online play)

The sleepers from the third column are as follows.
2,5,6,7,8,14,17,19,20,23,25,26,27,29.
Now the previous number was 36. So you bet using the clock face analogy.
The first sleeper in the third column was 2.
Counting two forward from 36 (including the zero) takes you to 1. So would you place a chip on the 1.
The second sleeper in the third column was 5.
Counting 5 forward from 36 takes you to 4. So you would you place a chip on the 4.
I will now list the sleepers from column three and the corresponding numbers you would be betting on.
2 = 1.
5 = 4.
6 = 5.
7 = 6.
8 = 7.
14 = 13.
17 = 16.
19 = 18.
20 = 19.
23 = 22.
25 = 24.
26 = 25.
27 = 26.
29 = 28.

So you are placing 44 chips down in total. Some numbers will have 2 chips on and some may have 3 on.
Looking at the snapshots, you get a double hit when you see **.  This  results in a 72 chips return.
*** is a triple hit returning 108 chips.

Studying what I have written over the last few posts and looking at the snapshots will make things clear.
This particular method is still in the early stages of testing. It looks promising. I am going to put some time in and see if it can be improved.

[attachthumb=#]

cheers

[John Gold]

One thing to note with all three columns is how they all comply with the so-called 'rule of the third'.

You can chart 37 numbers using the 3 column approach and the average missing numbers in all three columns is the same over thousands of spins. The average after the second cycle of 37 spins seems to be around 4 numbers left still to come out in each of the three columns. That's why I suggested stopping when there was a combined total of 12 sleepers still to come out across the three columns. A triple hit (***) at this stage would be nice but you could throw away a 50 chip profit in 4 spins.

Another point to note which is a bit difficult to explain is regarding the sleepers in the second and third column.
Take the example above. The first sleeper in the second row was 2 which had me betting number 30. Well yes, the 2 in terms of the pocket distances was a sleeper but the actual number 30 had already appeared twice in the first 37 spins. Maybe that is one of the reasons why this appears to work well. You are sometimes playing 'hot sleepers'  Talk about a contradiction in terms.

[John Gold]

This game was an interesting one because a lot of numbers had already come by the 37th spin.

There were only 10 sleepers left in the first column and 11 sleepers left in the second column.

It still got into a healthy profit by the 48th spin.

[attachthumb=#]

[John Gold]

-