(Almost Red/Black. 18 numbers red/black, almost=12 numbers:)
Here is a way to bet red or black and get the 2 to 1 winning advantage.
First, you will be studing either red or black numbers.
Write down the first six reds or blacks that are not repeaters.
The bet will then be the remaining 12 numbers.
If a number hits, repeat getting next six different reds/blacks
As for progressions, I like the 111112222233333xmultiplier fibonicci
Enjoy.
Hi, Proofreaders2000 ,
can you give us some smples please?
Best regards
kawa4711
Ok.. Here is an example
(Dublinbet numbers 8/30/08, 11:10 am CST)
1. 23 (Red)
2. 12 (Red) *that's 2 reds that are different*
3. 21 (Red)
4. 13 (Black) *that's 1 black, 5 to go*
5. 31 (Black) *2 blacks that are different*
6. 12 (Red) *can't use this one, it's a repeater*
7. 15 (Black) *that's 3, need three more*
8. 31 (Black) *can't use this one, either*
9. 14 (Red) *red number 4*
10. 24 (Black) *black number 4*
11. 34 (Red) *red number 5*
12. 19 (Red) *red number 6* (still keep an eye on black numbers)
----------------------------------
Ok...it time to bet on the 12 'sleeping' reds...1,3,5,7,9,16,18,25,27,30,32,36
Start with $0.50 on each number (suggested) with a bankroll of $300; $1 on each number (suggested) with a bank of $600.
Bet those numbers until a number hits at $0.50 or $1 for five spins, then increase bets by $0.50.
(Fibonicci part: If the number hits after a progression, stay at the same bet after you get 6 more reds/blacks that are non-repeaters and decrease after 2 or 3 spins hit. Again, increase by 0.50 or $1 if numbers do not hit after the next 5 spins.)
Dublinbet Numbers. 8-30-08 11:40am CST
(I recorded the red different numbers, no black numbers or repeating reds)
1 red
3 red
18 red
16 red
25 red
27 red
I then bet reds:5, 7, 9, 12, 14, 19, 21, 23, 30, 32, 34, 36 at $0.50/number=$6.00 total bet
Next spin 19 hits...18-6=$12 profit
Hi, Proofreaders2000
thank you for sharing your excellent idea and your examples.
This is an interesting idea and I will play it on my OC.
I have a further question:
How will you play, if there are p.e.
R
R
R
R
R
all different red numbers
B
B
B
B
B
all different numbers
different red number
Begin playing the red numbers, which havent been thrown
next number ( 6 th black number)
B
Will you play now:
Red (because 6 different Reds)
and Black (too)because there are 6 different blacks).
Best regards
kawa4711
I wait until six different reds show if I will bet betting on 12 sleeping reds. Likewise 6 different black numbers & play the twelve unhit black numbers. (You could play red & black at the same time, but one may cancel out the other-or lose both, which could get expensive.)
I like 12 numbers, which win the same value as dozens or rows
(If you can't decide whether to play red or black, look at the most frequent hitting numbers, red or black.)
Example: Dublinbet 8-31-08 4:48pm CST
1. 3 red *red number 1*
2. 11 black *black number 1*
3. 23 red *red number 2*
4. 27 red *red number 3*
5. 7 red *red number 4*
6. 32 red *red number 5*
7. 34 red *red number 6*
(at this point, I could play the unhit reds 1,5,9,12,14,16,18,19,21,25,30,36--there should be twelve numbers
8. 33 black *black number 2* ---rebet same 12 reds at same $ amts
9. 16 red hits, $18-12=$6 profit (at $0.50/number). (Start red count again at 0)
10. 5 red *red number 1*
11. 7 red *red number 2*
12. 6 black *black number 3*
13. 8 black *black number 4*
14. 14 red *red number 3*
15. 12 red *red number 4*
16. 6 black *repeater, doesn't count*
17. 3 red *red number 5, *just one more red to start betting*
18. 16 red *red number 6*
-----------------
Unhit reds at this point: 1, 9, 18, 19, 21, 23, 25, 27, 30, 32, 34, 36--(should be 12 reds)
19. 27 red hits***$18-6=$12 profit(at $0.50/number)....(start red count at zero again)
20. 23 red *red number 1*
21. 25 red *red number 2*
22. 2 black *black number 5*
23. 14 red *red number 3*
24. 15 black *black number 6*(shown black numbers: 15,2,4,6,33,11)----we can now bet on the 12 unhit black numbers. (Unhit black #'s:8,10,13,17,20,22,24,26,28,29,31,35)
25. 33 black hits, *rebet same 12 black numbers*
26. 24 black hits, $18-12=$6 profit (at $0.50/number) start black numbers count at 0.
How is this going?
So...it's not good :)
How is this going?--Zabbot
Honestly haven't seen this system in years. :)
I'll take a look at it :ok:
Tried it and modified it a little, runs hot and cold. You might make a little or a lot just depends. In the long run it is a loser
:nono: :wacko:
Hi,
This is nearly the same as my method
Its called 711 mk1
I have tested it for weeks without any loss.
Ill post it for you all.
Proofreaders2000 :good:
Finally got to look at this one. Making some tweaks and testing version 2
(Let's see if it holds up). :ok: