" System of Eights"
These Take Out All The Guess Work!
8X2
2 chips on numbers 13,15,22, and 24
4 chips on 2nd dozen, column B, and numbers 8, and 29
8 chips on numbers 4,6,31,33, and 17/20
12 chips on numbers 0, and 00
88 total 1,000 buy-in
8X3
2 chips on column A, column C, and numbers 14,23,16/19, and 18/21
4 chips on 2nd dozen, and numbers 8, and 29
8 chips on numbers 4,6,31,33, and 17/20
11 chips on numbers 0, and 0
86 total 1,000 buy-in
8X4
4 chips on numbers 8,14,23,29,16/19, and 18/21
4 chips on 1st dozen, 3rd dozen, column A, and column C
8 chips on numbers 4,6,31,33, and 17/20
12 chips on numbers 0, and 00
104 chips total, 1,000 buy-in
When you bet on 0, and 00, treat the two numbers as separate bets*. When one of them hits, double the amount of chips bet on the number. Then, when it hits again, cut it in half. This should work on all of the listings above.
so how to use this if the first bets lose start the second or other progression ? ???
I know. I changed it since after you had posted a response for me, but now answer is listed above so I do apologize. The answer that you are looking for is that you just need to press the number of chips bet on the 0, or 00 when the ball lands on one of those numbers. Then when the number comes up again, cut the winning bet in half again. You do this every time there is a winning bet on 0, or 00. The thing is you must treat each one of the bets as separate bets, only changing the chip count on the winning bet. Not on both. Only the number that the ball lands on...