(Inspired by J/L.)
Here is an idea I had for a matrix type bet.
1 = 1A. (The 1 represents the first dozen and the A represents the first column.)
2 = 1B. (The 1 represents the first dozen and the B represents the second column.)
3 = 1C. (The 1 represents the first dozen and the C represents the third column.)
4 = 1A.
5 = 1B.
6 = 1C.
7 = 1A.
8 = 1B.
9 = 1C.
10 = 1A.
11 = 1B.
12 = 1C.
13 = 2A. (The 2 represents the second dozen and the A represents the first column.)
14 = 2B. (The 2 represents the second dozen and the B represents the second column.)
15 = 2C. (The 2 represents the second dozen and the C represents the third column.)
16 = 2A.
17 = 2B.
18 = 2C.
19 = 2A.
20 = 2B.
21 = 2C.
22 = 2A.
23 = 2B.
24 = 2C.
25 = 3A. (The 3 represents the third dozen and the A represents the first column.)
26 = 3B. (The 3 represents the third dozen and the B represents the second column.)
27 = 3C. (The 3 represents the third dozen and the C represents the third column.)
28 = 3A.
29 = 3B.
30 = 3C.
31 = 3A.
32 = 3B.
33 = 3C.
34 = 3A.
35 = 3B.
36 = 3C.
Here are some numbers I recorded from my session at the casino earlier this afternoon.
25
2
2
6
16
16
22
6
16
9
15
10
21
35
13
6
7
25
25
35
15
11
7
10
23
11
15
32
32
1
14
32
7
12
11
19
19
3
23
24
6
3
17
3
18
7
32
35
Now using my chart from above, let's feed them into a matrix and see what happens.
3A 1B 1B 1C
2A 2A 2A 1C
2A 1C 2C 1A
2C 3B 2A 1C
1A 3A 3A 3B
2C 1B 1A 1A
2B 1B 2C 3B
3B 1A 2B 3B
1A 1C 1B 2A
2A 1C 2B 2C
1C 1C 2B 1C
2C 1A 3B 3B
So there you have it.
If you go to the second line in the matrix and read across, you will see 2A 2A 2A.
Now here is where I would look for the 2A to break.
That would require me to play the following numbers.
2/3, 5/6, 8/9, 11/12, 26/27, 29/30, 32/33, 35/36. That comes to 8 splits.
You will find all the possible bets for this method will come to 8 splits.
That bet would have won because after 2A 2A 2A, you will see that 1C arrived.
The only other bet in the matrix was in the second column going down vertical.
You will see 1C 1C 1C.
I would look for the 1C to break.
That would require me to play the following numbers.
13/14, 16/17, 19/20, 22/23, 25/26, 28/29, 31/32, 34/35. Once again, only 8 splits.
That bet would also have won because after 1C 1C 1C, you will see that 1A arrived.
I like this bet because the outlay is not too much. You could use a progression if you wanted. Also you are looking for a combination of a dozen AND a column together to keep repeating to defeat you.
Play after the third repeat (as shown in the examples above) and if it does not win, wait for another three of the same kind to appear and try again. Because you are only betting 8 splits, you should be able to stretch some kind of betting progression to allow this to be almost unbreakable.
I thought of this idea well doing my shopping early this morning when deciding between bananas or oranges. It just shows you my mind is pretty screwed up, lol.
Good luck.
p.s. It is also probably a good idea to cover the zero with a chip just in case.
Interesting
What about the costruction of your matrix?
First decision,OK,than you go horizontal or vertical?How many decisons Horiz and Vert,(4,5,6??),before continuing with the second line?
How are the columns of your matrix long?
Thanks
Did you have good winning sessions with this strategy. Also what progression you're using?
I will spend a few days doing some testing with this and come back with the results.
It is pretty much an open book at the moment. I just thought I would share the concept with you guys.
Feel free to post any tweaks or tests that you may do in the meantime.
cheers.
In the limited testing that I did do, I never went past step 1. I am yet to see something come up 4 times consecutively.
Thanks John Gold, nice idea! What I have noticed in your casino results is the amount of vertical repeats of either 1,2,3 or A,B,C.... usually catches 1 of them in a line or somtimes both:
3A 1B 1B 1C
2A 2A 2A 1C
2A 1C 2C 1A
2C 3B 2A 1C
1A 3A 3A 3B
2C 1B 1A 1A
2B 1B 2C 3B
3B 1A 2B 3B
1A 1C 1B 2A
2A 1C 2B 2C
1C 1C 2B 1C
2C 1A 3B 3B
From that It may be possible to bet on the single dozen/column to catch the repeat, should be a lot of wins there :biggrin:
Just flat betting 1,1,1,1 every line for the repeat came out like this:
3A 1B 1B 1C
2A 2A 2A 1C +1
2A 1C 2C 1A +5
2C 3B 2A 1C +6
1A 3A 3A 3B +4
2C 1B 1A 1A -1
2B 1B 2C 3B 0
3B 1A 2B 3B +7
1A 1C 1B 2A +5
2A 1C 2B 2C +12
1C 1C 2B 1C +19
2C 1A 3B 3B +20
John,
How is this going ?
Many thanks for sharing .
Cheers Pedro
Matrix Form
1A 1B 1C
01 02 03 1A 1B 1C
04 05 06 1A 1B 1C
07 08 09 1A 1B 1C
10 11 12 1A 1B 1C
2A 2B 2C
13 14 15 2A 2B 2C
16 17 18 2A 2B 2C
19 20 21 2A 2B 2C
22 23 24 2A 2B 2C
3A 3B 3C
25 26 27 3A 3B 3C
28 29 30 3A 3B 3C
31 32 33 3A 3B 3C
34 35 36 3A 3B 3C
1C 03 06 09 12
1B 02 05 08 11
1A 01 04 07 10
2C 15 18 21 24
2B 14 17 20 23
2A 13 16 19 22
3C 27 30 33 36
3B 26 29 32 35
3A 25 28 31 34
1A 01 04 07 10
2A 13 16 19 22
3A 25 28 31 34
1B 02 05 08 11
2B 14 17 20 23
3B 26 29 32 35
1C 03 06 09 12
2C 15 18 21 24
3C 27 30 33 36
1A 1B 1C
2A 2B 2C
3A 3B 3C
Quote from: libertydog on April 10, 2011, 05:43:58 PM
Just flat betting 1,1,1,1 every line for the repeat came out like this:
3A 1B 1B 1C
2A 2A 2A 1C +1
2A 1C 2C 1A +5
2C 3B 2A 1C +6
1A 3A 3A 3B +4
2C 1B 1A 1A -1
2B 1B 2C 3B 0
3B 1A 2B 3B +7
1A 1C 1B 2A +5
2A 1C 2B 2C +12
1C 1C 2B 1C +19
2C 1A 3B 3B +20
libertydog,
Please explain the bet on each horizontal line. Did you use pairs? can you provide an example with numbers.
Thanks,
John
after 1c 1c 1c - 1a would have been a loser for you jg ? or have I miss understood something. Interesting angle all the same
Quote from: John Gold on April 10, 2011, 02:26:15 PM
In the limited testing that I did do, I never went past step 1. I am yet to see something come up 4 times consecutively.
In my mind that translates as a conservative, 1 1 2 progression. risking only 4 u against a 7 in a row event, profiting only on the first step, recovering on 2nd and 3rd.
And as with any progression, I'd insure the zero.
Very interesting, worth testing.
Jeromin
Quote from: John Gold on April 10, 2011, 12:57:39 PM
(Inspired by J/L.)
Here is an idea I had for a matrix type bet.
1 = 1A. (The 1 represents the first dozen and the A represents the first column.)
2 = 1B. (The 1 represents the first dozen and the B represents the second column.)
3 = 1C. (The 1 represents the first dozen and the C represents the third column.)
4 = 1A.
5 = 1B.
6 = 1C.
7 = 1A.
8 = 1B.
9 = 1C.
10 = 1A.
11 = 1B.
12 = 1C.
13 = 2A. (The 2 represents the second dozen and the A represents the first column.)
14 = 2B. (The 2 represents the second dozen and the B represents the second column.)
15 = 2C. (The 2 represents the second dozen and the C represents the third column.)
16 = 2A.
17 = 2B.
18 = 2C.
19 = 2A.
20 = 2B.
21 = 2C.
22 = 2A.
23 = 2B.
24 = 2C.
25 = 3A. (The 3 represents the third dozen and the A represents the first column.)
26 = 3B. (The 3 represents the third dozen and the B represents the second column.)
27 = 3C. (The 3 represents the third dozen and the C represents the third column.)
28 = 3A.
29 = 3B.
30 = 3C.
31 = 3A.
32 = 3B.
33 = 3C.
34 = 3A.
35 = 3B.
36 = 3C.
Here are some numbers I recorded from my session at the casino earlier this afternoon.
25
2
2
6
16
16
22
6
16
9
15
10
21
35
13
6
7
25
25
35
15
11
7
10
23
11
15
32
32
1
14
32
7
12
11
19
19
3
23
24
6
3
17
3
18
7
32
35
Now using my chart from above, let's feed them into a matrix and see what happens.
3A 1B 1B 1C
2A 2A 2A 1C
2A 1C 2C 1A
2C 3B 2A 1C
1A 3A 3A 3B
2C 1B 1A 1A
2B 1B 2C 3B
3B 1A 2B 3B
1A 1C 1B 2A
2A 1C 2B 2C
1C 1C 2B 1C
2C 1A 3B 3B
So there you have it.
If you go to the second line in the matrix and read across, you will see 2A 2A 2A.
Now here is where I would look for the 2A to break.
That would require me to play the following numbers.
2/3, 5/6, 8/9, 11/12, 26/27, 29/30, 32/33, 35/36. That comes to 8 splits.
You will find all the possible bets for this method will come to 8 splits.
That bet would have won because after 2A 2A 2A, you will see that 1C arrived.
The only other bet in the matrix was in the second column going down vertical.
You will see 1C 1C 1C.
I would look for the 1C to break.
That would require me to play the following numbers.
13/14, 16/17, 19/20, 22/23, 25/26, 28/29, 31/32, 34/35. Once again, only 8 splits.
That bet would also have won because after 1C 1C 1C, you will see that 1A arrived.
I like this bet because the outlay is not too much. You could use a progression if you wanted. Also you are looking for a combination of a dozen AND a column together to keep repeating to defeat you.
Play after the third repeat (as shown in the examples above) and if it does not win, wait for another three of the same kind to appear and try again. Because you are only betting 8 splits, you should be able to stretch some kind of betting progression to allow this to be almost unbreakable.
I thought of this idea well doing my shopping early this morning when deciding between bananas or oranges. It just shows you my mind is pretty screwed up, lol.
Good luck.
p.s. It is also probably a good idea to cover the zero with a chip just in case.
You probably forgot about "SLEEPERS". It is possible in your first example, your first case 2A 2A 2A, to get several hits on the second dozen while your bets are on split numbers associated with first and third dozens.
Please explain what step do you take, progression? I have seen repeat of a dozen more than 22 times. No kidding...
Thanks,
John
Thank you for your system.
Here is some observation that has been excluded from your system:
In the majority of the time it will take a substantial amount of spins while the matrix is being build. The key is to get 3 alike before establish a bet.
Using a single zero roulette, by the time I got an entry to bet it took about 56 spins where I already experienced 4 times the number zero. While building the matrix and get spins going, I place 1 unit bet on Red and 1 on Black. So basically with the experience of getting 4 times the zero number I started my bet with a negative 8 units taken from my overall bankroll.
John
Quote from: jrhelp007 on April 24, 2011, 01:19:30 PM
While building the matrix and get spins going, I place 1 unit bet on Red and 1 on Black. So basically with the experience of getting 4 times the zero number I started my bet with a negative 8 units taken from my overall bankroll.
John
What I do to stay logged on is bet the minimum ( say 50 cents in bet365 or DB) on the last number. One repeat in 36 spins and you break even, hardly unusual. Worst case scenario, about 15 to 20 euro loss till I can bet. Then, when the opportunity comes, I bet 25/50 euro per dozen/column. Basically, at best you've made a couple of units, at worst you've lost a couple of units. If the system is any good, I've recovered them in a couple of spins.
Jeromin